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11Sources of magnetic field: what a current makes, and how to add it up
Hold two lengths of flex side by side, a hand apart, and run twelve amps down both of them in the same direction. They pull towards each other, with a force you can measure on a laboratory balance. Nothing is charged, nothing is magnetised, and neither one is moving. Ten sections of this course can tell you the current, the resistance, the power and the heat in those wires, and how hard a magnet would push on them, and not one of them can say where that pull comes from.
By the end of this section you can start from currents and geometry alone and produce the magnetic field at a point, by three routes that you choose between deliberately: the straight wire result with superposition, a closed loop and the , or an integral over the wire itself.
In 60 seconds
Currents make magnetic fields; the field of a long straight wire falls off as one over the distance and wraps around the wire in circles, everything else on this page is that result plus superposition, and the two shortcuts, Ampere's law and Biot and Savart, exist so that you do not have to add the contributions up by hand.
the wire is long compared with the distance r, which is the perpendicular distance from the wire's axis; the direction is the grip rule, thumb along the current and fingers curling the way the field goes
Force per unit length between two parallel currents
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$
two long parallel wires a distance d apart; the same direction attracts and opposite directions repel, which is the opposite of the habit two like charges leave you with
Ampere's law
$$\oint \vec B\cdot d\vec l = \mu_0 I_{\rm enc}$$
steady currents, any closed path; it is always true, but it only gives you B when the symmetry lets you pull a constant B out of the integral
the current is spread uniformly over the cross section; the field rises straight from zero at the axis and falls as one over r outside, the two meeting at the surface
Solenoid and toroid
$$B_{\rm sol} = \mu_0 n I,\qquad B_{\rm tor} = \frac{\mu_0 N I}{2\pi r}$$
well inside a long solenoid, where n is turns per metre and the field does not depend on where you stand; and inside a toroid, where r is your distance from the centre of the ring and the field does depend on it
$$d\vec B = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\hat r}{r^{2}}$$
no symmetry to exploit: arcs, finite segments, loops, anything bent; the segments pointing straight at the field point contribute nothing
Centre of a loop, of an arc, and a point on the axis
$$B_{\rm centre} = \frac{\mu_0 N I}{2R},\qquad B_{\rm arc} = \frac{\mu_0 I\phi}{4\pi R},\qquad B_{\rm axis} = \frac{\mu_0 N I R^{2}}{2(R^{2}+x^{2})^{3/2}}$$
the three results that come out of Biot and Savart often enough to be worth knowing; phi is the angle of the arc in radians, and the axis result reduces to the centre one when x is zero
a coil is wound on a core; B_0 is the field the same coil would make in vacuum, and for iron the multiplier is not a constant but a number that collapses as the metal saturates
Three most common mistakes
Using an inverse square for the wire. A point charge gives $1/r^{2}$; a long straight wire gives $1/r$, because the far parts of the wire keep contributing. Doubling the distance from a wire halves the field, it does not quarter it.
Reading $\oint\vec B\cdot d\vec l = 0$ as $\vec B = 0$. A loop that encloses no net current has zero , and the field on that loop can still be strong everywhere. The integral is a statement about a sum around a path, not about a value at a point.
Mixing $\mu_0 I/(2\pi R)$ and $\mu_0 I/(2R)$. The first is a straight wire at distance R, the second is the centre of a circular loop of radius R, and they differ by a factor of $\pi$. Look at the shape of the wire before you pick.
The published weights for this course are two midterms at twenty per cent each, quizzes at ten, homework at five, the final at twenty five and laboratory work at twenty. Nothing tells us how many questions on any paper come from this material, so treat it the way the weights do: it is one week of a course whose written assessment is worth eighty per cent in total, and the parts of it that transfer everywhere are the straight wire result, superposition and the grip rule.
How much time do you have?
10 minutes
You leave able to do the thing almost every question here starts with: the field a straight wire makes at a point, in size and in direction, and two of them added as vectors.
The 60 second card · Formula card · What a current makes: the field of a long straight wire · Two currents, one force, and the ampere · Mistake ledger
45 minutes
You add the two machines that turn geometry into a field without an integral, and the two devices they are always asked about: Ampere's law with a well chosen loop, and the solenoid and toroid.
The 60 second card · What a current makes: the field of a long straight wire · Two currents, one force, and the ampere · Ampere's law: what it says and what it does not · Choosing the loop: inside a thick wire and inside a cable · The solenoid and the toroid · Method box: which of the three routes to use · Method box: getting the direction right every time · Exam level example · Practice set C · Mistake ledger
full read
You can also handle arcs, finite segments and off centre points with Biot and Savart, say what an iron core really does to a coil and where it stops doing it, and defend every direction and every sign you wrote.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pair · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the magnitude and the direction of the magnetic field a long straight current makes at a given point, and add the fields of several straight wires as vectors.
Determine the force per unit length between two parallel currents, state which way it acts, and use it either to find a current from a measured force or to balance a wire against gravity.
State Ampere's law, identify the current enclosed by a given path with the correct sign, and explain why zero circulation does not mean zero field.
Apply Ampere's law with a deliberately chosen path to get the field inside and outside a thick current carrying wire and in each region of a .
Calculate the field inside a solenoid from its turns per metre and inside a toroid from its number of turns and the radius, and say where each formula stops being trustworthy.
Integrate the Biot and Savart law over simple geometries to get the field at the centre of a loop or an arc and at a point on the axis of a loop, and recognise the segments that contribute nothing.
Explain what a core of matter does to the field of a coil, distinguish ferromagnetic, paramagnetic and diamagnetic behaviour by the size and the sign of the effect, and read a curve.
Syllabus coverage
Sources of Magnetic Field
The magnetic field produced by a long straight current and the constant that fixes its size; the force between two parallel currents and the definition of the ampere; Ampere's law, the sign of the enclosed current and what the law can and cannot deliver; the field inside and outside a solid conductor and in the regions of a coaxial cable; the solenoid and the toroid; the Biot and Savart law with the centre of a loop, the centre of an arc and a point on the axis; and the magnetic properties of matter, covering ferromagnetism and domains, the of a core, hysteresis, and the much weaker paramagnetic and diamagnetic effects.
The week line is the phrase Sources of Magnetic Field and carries no chapter number, so no chapter or section number of the textbook is quoted anywhere on this page.
covered
Fields and currents that change with time
What happens to Ampere's law when the current is not steady, and the voltage a changing field drives round a loop.
Deferred to the later week lines on induction and on the complete set of field equations.
deferred
The energy stored in a coil
How much energy sits in the field of a solenoid and what that costs to build up.
Deferred to the later week line on coils and alternating currents. Here a solenoid is a way of making a known field and nothing more, and every question about it asks for a field, a current or a number of turns.
deferred
The magnetic field of the Earth
Where the planet's own field comes from and how big it is.
Not named on the week line and not examinable, but one number for it, a few times ten to the minus five tesla, is used throughout as a ruler: almost every field computed here is compared with it, so a plausible answer can be told from an impossible one at a glance.
off_syllabus
Recall first
The force a field exerts on a current carrying wire
A straight wire of length $L$ carrying current $I$ in a uniform field feels $\vec F = I\vec L\times\vec B$, of size $F = BIL\sin\theta$, where $\theta$ is the angle between the wire and the field. The vector $\vec L$ points along the current.
The whole of the force between two wires is this result with the field of the other wire written in.
The magnetic force on a moving charge, and the right hand rule
A charge $q$ moving with velocity $\vec v$ through a field feels $\vec F = q\,\vec v\times\vec B$. The direction is found by pointing $\vec v\times\vec B$ with the right hand and then reversing the answer if $q$ is negative.
Every direction on this page is a cross product taken with the same hand, so the habit transfers directly.
The of a coil, and the torque on it
A flat coil of $N$ turns, area $A$, carrying current $I$, has $\vec\mu = NI\vec A$ with $\vec A$ perpendicular to the plane of the coil, and a uniform field puts a torque $\vec\tau = \vec\mu\times\vec B$ on it.
It is needed twice here: once to recognise that the far field of a small loop is written in terms of the same $\vec\mu$, and once to see that a loop is both a thing that makes a field and a thing that feels one.
The closed path integral of an electrostatic field
For any electrostatic field, $\oint\vec E\cdot d\vec l = 0$ around every closed path, which is the same statement as saying that a potential difference exists and that the work done going round a loop and back to the start is zero.
It is the contrast that makes this section's central law worth stating.
Choosing a surface by symmetry, from Gauss's law
In electrostatics, $\oint\vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$ holds for any closed surface, but it delivers $E$ only when a surface can be chosen on which $E$ is constant in magnitude and either along or across the surface everywhere.
The magnetic version on this page works the same way and fails the same way.
Current and current density
The current through a cross section is $I$, and if it is spread uniformly over an area $A$ the current density is $J = I/A$. The current passing through part of that cross section, of area $A'$, is then $I' = JA' = I\,A'/A$.
It is exactly what is needed to say how much current a small inside a thick wire encloses.
Equilibrium of a wire under gravity
A wire in equilibrium has zero net force. Per unit length that reads $F/L = \lambda g$ upwards, where $\lambda$ is the mass per unit length in kilograms per metre and $g = 9.80\ \mathrm{m/s^{2}}$.
The levitation problem, which is a standard exam question on the force between two currents, is solved by setting the magnetic force per unit length equal to this and nothing else.
The field produced by several sources at a point is the vector sum of the fields each source would produce there on its own, each computed as though the others were absent.
Almost every multi wire problem here is superposition plus one formula.
Try it yourself first (3 questions)
1§11.0 — a stationary charge beside a live wire●●○○○
This one is worth doing before you read on, because the wrong answer is the natural one and it will get in your way for the whole section. A long wire carries a steady current of several amperes. A small charged bead is held at rest a centimetre away from it.
Given
the wire carries a steady current
the bead is charged and is held at rest
there is no electric field from the wire, which is uncharged overall
Find
(a) Does the bead feel a magnetic force, and why?
Hint 1/4
Do not ask what the wire does; ask what the force law on a charge needs as inputs, and check that list against what the situation supplies.
Hint 2/4
The magnetic force on a charge is $\vec F = q\,\vec v\times\vec B$, so it needs a charge, a field, and a velocity, all three.
Hint 3/4
Here the charge is present, the field is present because a current is flowing, and the bead is held at rest so $\vec v = \vec 0$.
Hint 4/4
No force: the field is there, but with zero velocity the cross product vanishes.
Show solutionPut the given values into the force law
$$\vec F = q\,\vec v\times\vec B = q\,(\vec 0)\times\vec B = \vec 0$$
the cross product of the zero vector with anything is the zero vector, so no property of the field can rescue it
Separate the two claims that are being confused
$$\vec B \neq \vec 0 \ \text{at the bead}$$
true, and it is what the rest of this section computes
$$\vec F = \vec 0 \ \text{on the bead}$$
also true, and independent of the first: a field existing somewhere and a force acting on something are different statements
Answer $$\boxed{\,\vec F = \vec 0\ \text{while}\ \vec B \neq \vec 0\,}$$
Check
Independent check: release the bead and let it fall, and a sideways force appears at once, which shows the field was there all along and that motion was the only missing ingredient.
2§11.0 — force on a wire in a field that is handed to you●●○○○
Straight recall of the result the previous work on magnetism ended with, because the force between two wires on this page is built directly on top of it. A straight wire of length $0.250\ \mathrm{m}$ carries $4.00\ \mathrm{A}$ through a uniform field of $0.600\ \mathrm{T}$, with the wire making an angle of $30.0^{\circ}$ with the field.
Given
$L = 0.250\ \mathrm{m}$
$I = 4.00\ \mathrm{A}$
$B = 0.600\ \mathrm{T}$
$\theta = 30.0^{\circ}$ between the wire and the field
Find
(a) Find the magnitude of the force on the wire.
(b) State the angle at which that force would be largest, and its value there.
Hint 1/4
Nothing needs deriving: pick the one formula that turns a current, a length and a field into a force, and read off which angle it wants.
Hint 2/4
$F = BIL\sin\theta$, with $\theta$ measured between the wire and the field.
Hint 3/4
With $B = 0.600\ \mathrm{T}$, $I = 4.00\ \mathrm{A}$, $L = 0.250\ \mathrm{m}$ and $\theta = 30.0^{\circ}$, the product $BIL$ comes first and the sine last.
Hint 4/4
The force is $0.300\ \mathrm{N}$, and it would be $0.600\ \mathrm{N}$ with the wire at right angles to the field.
Show solutionCompute the maximum first, then bring in the angle
grouping the three quantities that carry the units keeps the arithmetic and the geometry in separate lines, so an angle error cannot hide inside a unit error
Unit check: tesla times ampere times metre is $(\mathrm{N/(A\,m)})\,\mathrm{A}\,\mathrm{m} = \mathrm{N}$, so the answer is a force. Plausibility: a third of a newton is the weight of a thirty gram object, a firm but ordinary push for a quarter metre of wire in a strong laboratory field.
3§11.0 — walking a closed loop in an electrostatic field●●○○○
The last of the three, and the one that this section is about to overturn for magnetic fields. Take any arrangement of fixed charges, pick any closed path through the region, and walk all the way round it back to where you started, adding up $\vec E\cdot d\vec l$ as you go.
Given
the charges are fixed and the field is electrostatic
the path is closed, ending where it began
the path may be any shape and may pass as close to the charges as you like
Find
(a) True or false: the total is always zero. Give the reason in one sentence.
Hint 1/4
Translate the integral into something you already have a word for, rather than trying to picture the walk.
Hint 2/4
The work per unit charge along a path is the drop in potential, $\int\vec E\cdot d\vec l = V_{\rm start} - V_{\rm end}$.
Hint 3/4
For a closed path the start and the end are the same point, so the two potentials are the same number.
Hint 4/4
True: the total is zero for every closed path in an electrostatic field.
Show solutionRewrite the integral as a potential difference
$$\int_a^b \vec E\cdot d\vec l = V_a - V_b$$
this is what it means for the electrostatic field to have a potential: the integral depends on the endpoints only and not on the route
$$a = b \;\Rightarrow\; \oint \vec E\cdot d\vec l = V_a - V_a = 0$$
closing the path forces the two endpoints to coincide, so whatever the potential is there, it cancels against itself
Answer $$\boxed{\,\oint\vec E\cdot d\vec l = 0\ \text{for every closed path}\,}$$
Check
Independent check from the picture rather than the algebra: electrostatic field lines start on positive charges and end on negative ones, so no field line ever closes on itself, and a path cannot be chosen that runs downhill the whole way round.
Notation
symbol
reads as
means
watch out
$\mu_0$
mu nought
the , $4\pi\times10^{-7}\ \mathrm{T\,m/A}$, the constant that fixes how much field a given current makes
it sits in the numerator here, where the electric constant $\varepsilon_0$ sat in the denominator of the electric formulas; a field that grows with $\mu_0$ and one that shrinks with $\varepsilon_0$ are not a contradiction, they are two constants defined in opposite places
$\mu_r$
mu r
the relative permeability of a material, the factor by which it multiplies the field a coil would otherwise make
a pure number, close to one for almost everything and in the hundreds or thousands for iron; for iron it is not a constant of the metal but a value at a stated field
$\vec\mu$
mu vector
the magnetic dipole moment of a loop or coil, $NI\vec A$, carried over from the previous work on magnetism
the same Greek letter as the two above and a completely different quantity, with units of ampere metre squared; here it appears only in the far field of a loop, and it always carries an arrow or a subscript that tells you which one it is
$r$
r
the perpendicular distance from a wire's axis to the field point, or the distance from a to the field point in the Biot and Savart law
in the toroid it means your distance from the centre of the whole ring, not from the wire, which is why the field there is not uniform
$R$
R
a radius belonging to the object itself: the radius of a thick wire, of a circular loop, or of an arc
keep it separate from $r$; a great many wrong answers here are the right formula with the two swapped
$I_{\rm enc}$
I enclosed
the net current passing through any surface bounded by the chosen closed path, counted with sign
it is the net current through the loop, not the total current in the problem and not the current in the nearest wire; a wire outside the loop contributes nothing to it however close it is
$n$
little n
of a solenoid, $N/L$, in turns per metre
a solenoid formula written with $N$ instead of $n$ is out by a factor of the length in metres, which is the single most common slip in this material
$d\vec l$
d l vector
a short piece of the wire, pointing along the direction of the current in it
it points along the wire, never towards the field point; the vector towards the field point is $\hat r$ and the two are crossed together
$\phi$
phi
the angle subtended by a circular arc at its centre, in radians
radians, so a semicircle is $\pi$ and not $180$; putting degrees into the arc formula gives an answer about fifty seven times too large
Conventions used here
How the axes and the page are set up in this section
Throughout this section $x$ points to the right across the page, $y$ points up the page, and $z$ points out of the page towards you. A field into the page is therefore the $-z$ direction and is drawn as a grid of small crosses; a field out of the page is $+z$ and would be drawn as dots. Every cross product is taken in this right handed order.
Which way round the field of a current goes
The grip rule fixes it: point the thumb of your right hand along the direction of the conventional current, and your fingers curl the way the field circles. For a current out of the page the field circles anticlockwise as you look at it, so at a point directly to the right of the wire the field points up the page. Every direction on this page is obtained that way and never by memory of a picture.
The sign of the current enclosed by an Amperian loop
Before evaluating $\oint\vec B\cdot d\vec l$ you must choose a direction to walk round the loop. Curl the fingers of your right hand that way; your thumb then points in the direction that counts as positive current. A current threading the loop the other way is negative and subtracts. Reversing the walking direction reverses the sign of both sides of the equation and changes nothing physical.
Which constants this page uses, and the two grouped forms of mu nought
Answers are quoted to three significant figures. The permeability of free space is $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$, and the combination that appears in almost every straight wire calculation is $\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$ exactly, which is worth carrying separately because it removes the $\pi$ from the arithmetic. Also used are the elementary charge $e = 1.60\times10^{-19}\ \mathrm{C}$ and $g = 9.80\ \mathrm{m/s^{2}}$.
What the distance in the straight wire formula means
In $B = \mu_0 I/(2\pi r)$ the quantity $r$ is the perpendicular distance from the axis of the wire to the field point, measured in metres, and not the distance from either end and not the distance along the wire. Where a wire has a thickness, $r$ is still measured from the axis, so the surface of a wire of radius $R$ sits at $r = R$.
What long enough means for a wire, a solenoid and a toroid
Every straight wire result on this page assumes the wire is long compared with the distance to the field point, and every solenoid result assumes the length is large compared with the diameter and that you are well inside, away from the ends. Where a problem gives you a finite length, that length is either much larger than the other distances in the problem, in which case the idealisation is used and said to be used, or the problem is solved with Biot and Savart instead. Near an end of a real solenoid the field falls to roughly half of the value inside.
How a field made by matter is reported
When a coil is wound on a core, $B_0$ always means the field the same coil with the same current would make with nothing inside it, and $B$ means the total field in the material. Their ratio is the relative permeability $\mu_r$, a pure number with no units. For a ferromagnet that number is written as a value at the stated working point and never treated as a property of the metal that holds at every current.
11.1What a current makes: the field of a long straight wire
A current wraps field around itself in closed circles, and the field halves when you double your distance from it.
Every field so far arrived as a given, a number in the wording of the problem. From here you calculate it, and the cheapest source to start from is one long straight wire.
Solvable with what we have
Force on a charge moving through a field you were given.
Force on a current carrying wire lying in a field you were given.
Torque on a coil, and the radius of a bent particle path.
Not solvable yet
What field a current carrying wire makes at a point beside it.
Why two live wires pull on each other when neither is magnetised.
What field sits at the centre of a coil or inside a solenoid.
Copy the electric case. A charged rod pushes a test charge straight away from itself, so guess the same for a wire: a field pointing radially outwards and falling as $1/r^{2}$.
Why it fails
Both halves are wrong and a compass shows it in a second. Ring a vertical current with compasses and none points at the wire; they line up head to tail in a closed circle.
RuleRule 11.1: the magnetic field of a long straight current
Conditions
the wire is straight and long compared with $r$, so that the ends are far away from the field point
$r$ is the perpendicular distance from the axis of the wire, and the point is outside the wire
the current is steady, that is, not changing with time
there is nothing magnetic nearby, so the field is the wire's alone
The field beside a long straight wire is proportional to the current and inversely proportional to your distance, so doubling the distance halves it. The direction is neither towards the wire nor away: grip the wire with your right hand, thumb along the current, and your fingers curl the way the field runs.
The field of a single straight wire, drawn for a current of $12\ \mathrm{A}$ coming out of the page. Twice the distance gives half the field, not a quarter: an inverse square would have made the outer arrow a quarter as long. Note too that the arrows never point outwards — these lines close on themselves.
Looks like this, but is not
The lines run outwards and thin out as they spread, which is why the field weakens. It sounds like a charged rod and it even gets the weakening right.
They run around, not outwards, and the weakening has a different cause. A line that spreads from a source must start somewhere, and a magnetic line never does: it closes on itself. These are circles of growing circumference, and the field is weaker on the big ones because the same circulation is spread along a longer path. Different mechanism, hence $1/r$ and not $1/r^{2}$.
current in the wire
field at $5.0\ \mathrm{cm}$
compared with what
$1.0\ \mathrm{A}$, a phone charger
$4.0\ \mathrm{\mu T}$
about a twelfth of the Earth's field
$12\ \mathrm{A}$, a kettle
$48\ \mathrm{\mu T}$
about the Earth's own field
$100\ \mathrm{A}$, a welder
$400\ \mathrm{\mu T}$
about eight times the Earth's field
$1000\ \mathrm{A}$, a busbar
$4.0\ \mathrm{mT}$
about a tenth of a fridge magnet
Read down the first two columns and the proportionality is visible without algebra: every factor of ten in the current is a factor of ten in the field. Read the third column and something more useful appears.
The field 5.0 cm from a kettle lead carrying 12 A
A long straight wire carries a steady current of $12.0\ \mathrm{A}$ vertically upwards. Find the magnitude of the magnetic field it produces at a point $5.00\ \mathrm{cm}$ due east of the wire, give the direction of that field, and compare its size with the Earth's field of about $5\times10^{-5}\ \mathrm{T}$.
using the grouped constant $\mu_0/2\pi$ rather than $\mu_0$ itself removes the $\pi$ from the arithmetic, and there is nothing else in the problem that needs it
a ratio rather than a difference, because the Earth's field is only known here to one significant figure
Answer $$\boxed{\,B = 4.80\times10^{-5}\ \mathrm{T},\ \text{horizontal and perpendicular to the wire, pointing north at a point east of it}\,}$$
Check
Independent check with a compass, which tests the direction rather than repeating the arithmetic. Due east of the wire the field found above is horizontal and points north, and the Earth's horizontal field points north as well, so the two are parallel: a compass placed there does not turn at all, it simply sits in a horizontal field about twice its usual strength. Move the same $5.00\ \mathrm{cm}$ round to a point due north of the wire and the wire's field there runs west, at right angles to the Earth's and of the same size, so the needle swings by about forty five degrees. The deflection depending on where the point is placed, and not only on the size of $B$, is the sign that the direction has actually been used.
One multiplication and one division, provided the constant is carried as $2\times10^{-7}$ and the centimetres are turned into metres before anything else happens.
Keep the habit rather than the number: household currents at household distances make fields of the order of the Earth's. Any answer here that gives a whole tesla from a bare wire is wrong by tens of thousands.
Two parallel wires, 10 A and 15 A: the field at the midpoint and the point where it vanishes
Two long parallel wires are $20.0\ \mathrm{cm}$ apart and lie perpendicular to the page. The left one carries $10.0\ \mathrm{A}$ out of the page and the right one carries $15.0\ \mathrm{A}$, also out of the page. Find (a) the magnetic field at the midpoint of the line joining them, and (b) the point on that line, if there is one, where the total field is zero.
Given
$I_1 = 10.0\ \mathrm{A}$ out of the page, at $x = 0$
$I_2 = 15.0\ \mathrm{A}$ out of the page, at $x = 0.200\ \mathrm{m}$
both wires are long and straight
$x$ runs to the right, $y$ up the page, $z$ out of the page
Find
the field at the midpoint, and the location of the null point
same distance, one and a half times the current, one and a half times the field
Turn each into a direction before adding anything
$$\hat B_1 = \hat z\times\hat x = +\hat y$$
the midpoint lies to the right of wire 1, and the current is out of the page, so this contribution points up the page
$$\hat B_2 = \hat z\times(-\hat x) = -\hat y$$
the midpoint lies to the left of wire 2, so the same rule with the direction reversed sends this one down the page: between two currents in the same direction the two fields always oppose
$$\vec B = (2.00 - 3.00)\times10^{-5}\,\hat y = -1.00\times10^{-5}\,\hat y\ \mathrm{T}$$
opposing vectors subtract, and the stronger wire wins, so the net field points down the page
cross multiplying and collecting; the constants cancelled at the previous line, which is why the answer contains neither $\mu_0$ nor $\pi$
Answer $$\boxed{\,(a)\ \vec B = 1.00\times10^{-5}\ \mathrm{T}\ \text{down the page};\qquad (b)\ x = 8.00\ \mathrm{cm}\ \text{from the}\ 10.0\ \mathrm{A}\ \text{wire}\,}$$
Check
Independent check on (b) by evaluating both fields there rather than trusting the algebra: $B_1 = (2\times10^{-7})(10.0)/(0.0800) = 2.50\times10^{-5}\ \mathrm{T}$ and $B_2 = (2\times10^{-7})(15.0)/(0.120) = 2.50\times10^{-5}\ \mathrm{T}$. Equal sizes, opposite directions, so the total really is zero.
Two applications of one formula and one linear equation. The only place a calculator was needed was the last division.
The transferable part is the order of operations: size, then direction, then add. Adding $2.00$ and $3.00$ into $5.00$ first and worrying about signs afterwards gives the right answer to the wrong configuration.
Checkpoint
§11.1 — what happens when you step back from the wire●●○○○
Thirty seconds, no calculator. You are holding a magnetometer $3.0\ \mathrm{cm}$ from a long straight wire and reading a steady value. You walk out to $9.0\ \mathrm{cm}$ from the same wire, with the current unchanged.
Given
a long straight wire with a steady current
first reading taken at $3.0\ \mathrm{cm}$
second reading taken at $9.0\ \mathrm{cm}$
Find
(a) What does the reading do?
Hint 1/4
You are being asked for a ratio, not a value, so nothing in the problem except the two distances can matter.
Hint 2/4
$B = \mu_0 I/(2\pi r)$, so at fixed current $B$ is inversely proportional to $r$ to the first power.
Hint 3/4
Here $r$ goes from $3.0\ \mathrm{cm}$ to $9.0\ \mathrm{cm}$, a factor of three, and $B_2/B_1 = r_1/r_2$.
Hint 4/4
Three times the distance means one third of the field.
Show solutionCancel everything that did not change
and because it is a ratio of two lengths, the centimetres need not even be converted
Answer $$\boxed{\,B_2 = \tfrac{1}{3}B_1\,}$$
Check
Independent check by putting in numbers that were never given. Take $I = 6.0\ \mathrm{A}$: then $B_1 = (2\times10^{-7})(6.0)/(0.030) = 4.0\times10^{-5}\ \mathrm{T}$ and $B_2 = (2\times10^{-7})(6.0)/(0.090) = 1.3\times10^{-5}\ \mathrm{T}$, whose ratio is one third.
⚠ Using an inverse square law for a wire
every field met before this one came from point charges and fell off as $1/r^{2}$, and the exponent gets copied across without being checked against the geometry
wrong$$B = \frac{\mu_0 I}{2\pi r^{2}}$$
right$$B = \frac{\mu_0 I}{2\pi r}$$
⚠ Reaching for the wrong grouped constant
two constants live in this material, $\mu_0/2\pi = 2\times10^{-7}$ and $\mu_0/4\pi = 1\times10^{-7}$, they differ by a factor of two, and both look equally plausible written down
the arrows of a charged rod point outwards and the picture is strong enough to survive being told otherwise, especially when only the magnitude is asked for
Each wire sits in the other one's field, so currents in the same direction pull together and opposite ones push apart.
The hook is one line away: we have the field a wire makes, and from the earlier work the force a field puts on a wire, and putting the first inside the second is the whole derivation.
TheoremResult 11.2: the force per unit length between two long parallel currents
Conditions
both wires are long, straight and parallel, and $d$ is the perpendicular distance between them
$d$ is small compared with the lengths, so that neither wire sees the other's ends
both currents are steady
the force is quoted per unit length because two infinite wires would otherwise give an infinite total
Multiply the two currents together, divide by the separation, and multiply by two times ten to the minus seven; that is how many newtons act on every metre of either wire. The two forces are equal in size and opposite in direction no matter how different the currents are, and the sign is the reverse of the electrostatic habit: currents running the same way attract.
Proof
Nothing new is needed. Wire 1 makes a field at the place where wire 2 sits, and wire 2 is a current carrying wire in a field.
The field of wire 1 at the position of wire 2, a distance $d$ away, is $B_1 = \mu_0 I_1/(2\pi d)$. By the grip rule it is perpendicular to both wires, so it meets wire 2 broadside.
The force on a length $L$ of wire 2 is $F = B_1 I_2 L\sin 90^{\circ} = B_1 I_2 L$, because the field is at right angles to that wire.
Substituting, $F = \dfrac{\mu_0 I_1 I_2 L}{2\pi d}$, and dividing by $L$ gives the result. The length cancels, which is why the answer is always quoted per metre.
For the direction, do the cross product once and trust it. With both currents out of the page, the field of wire 1 at wire 2 points up the page, and $I_2\vec L\times\vec B_1$ with $\vec L$ out of the page and $\vec B_1$ up the page points back towards wire 1. Reversing $I_2$ reverses that force and nothing else.
Doing the same calculation the other way round, the force on wire 1 from the field of wire 2, gives the same size with the opposite direction.
Why the sign comes out the way it does. In each panel the $\textcolor{#1f6feb}{\text{field of the left wire}}$ is drawn where the right wire sits, and the $\textcolor{#8250df}{\text{force}}$ on that wire is the current in it crossed into that field. The field of one wire, crossed into the current of the other, fixes the sign both times.
Looks like this, but is not
The wire with the bigger current gets pushed harder. Ten amps in one wire against one amp in the other, and it feels obvious which one moves.
The formula contains the product of the two currents and nothing else, so it is symmetric: both wires feel the same number of newtons per metre. What differs is what that force does to them, since the thin one accelerates more, but the force itself is shared. If you find yourself computing two different forces for the two wires, you have used one current where the product belongs.
current in each wire
force per metre
what that force is comparable with
$1.0\ \mathrm{A}$
$2.0\ \mathrm{\mu N/m}$
the weight of a fifth of a milligram, undetectable by hand
$10\ \mathrm{A}$
$0.20\ \mathrm{mN/m}$
the weight of a small grain of rice per metre
$100\ \mathrm{A}$
$20\ \mathrm{mN/m}$
the weight of two grams per metre, clearly visible on thin wire
$1000\ \mathrm{A}$
$2.0\ \mathrm{N/m}$
the weight of two hundred grams per metre, enough to wreck a busbar
The product of the currents is what appears, so every factor of ten in both wires together is a factor of a hundred in the force.
One ampere, one metre apart: the force that used to define the ampere
Two long straight parallel wires are $1.00\ \mathrm{m}$ apart in vacuum and each carries $1.00\ \mathrm{A}$ in the same direction. Find the force per unit length on either of them, state whether it is attractive or repulsive, and say what total force acts on a $1.00\ \mathrm{m}$ length.
Given
$I_1 = I_2 = 1.00\ \mathrm{A}$, in the same direction
$d = 1.00\ \mathrm{m}$
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
the force per metre, its sense, and the force on one metre
Plausibility: $2\times10^{-7}\ \mathrm{N}$ is the weight of about twenty micrograms, a fraction of a grain of fine sand.
One substitution. The arithmetic is trivial by construction, because the numbers were chosen to make it so.
This arrangement was for many years the official definition of the ampere, which is why $\mu_0$ is exactly $4\pi\times10^{-7}$. The ampere is now defined by counting charges per second instead, but not to a precision that touches any figure used here.
Floating one wire above another: the current that cancels gravity
A straight horizontal wire of mass per unit length $5.00\times10^{-3}\ \mathrm{kg/m}$ is to be held floating $5.00\ \mathrm{mm}$ directly above a second, fixed, parallel wire. The same current $I$ passes through both. Find the current needed, and say which way it must run in the upper wire relative to the lower one.
Given
$\lambda = 5.00\times10^{-3}\ \mathrm{kg/m}$
$d = 5.00\times10^{-3}\ \mathrm{m}$
the same current $I$ in both wires
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the current, and the relative direction of the two currents
SolutionDecide the direction before touching any numbers
$$\vec F_{\rm mag}\ \text{must point up}$$
the only other force per unit length is the weight, which points down, so the magnetic force has to be a repulsion pushing the upper wire away from the lower one
$$\text{repulsion} \Rightarrow \text{currents in opposite directions}$$
and this settles the answer to the second half of the question before any arithmetic.
Balance the two forces per unit length
$$\frac{\mu_0 I^{2}}{2\pi d} = \lambda g$$
both sides are per unit length, so the length never enters; equating a total force to a force per metre is the standard way this problem goes wrong
taking the positive root, since the direction has already been fixed by the physics and a negative current here would mean nothing
Answer $$\boxed{\,I = 35.0\ \mathrm{A}\ \text{in each wire, running in opposite directions}\,}$$
Check
Independent check by going forwards instead of backwards: with $I = 35.0\ \mathrm{A}$ and $d = 5.00\ \mathrm{mm}$, $F/L = (2\times10^{-7})(35.0)^{2}/(5.00\times10^{-3})$, that is $(2\times10^{-7})(1225)/(5.00\times10^{-3}) = 4.90\times10^{-2}\ \mathrm{N/m}$, and the weight per metre is $\lambda g = (5.00\times10^{-3})(9.80) = 4.90\times10^{-2}\ \mathrm{N/m}$. Plausibility: thirty five amperes through a wire of five grams per metre, which is roughly a one millimetre copper wire, is a lot but not absurd for a short demonstration.
One physical decision and one square root, with the direction settled first because it is worth a mark on its own.
Say a sentence about stability in an exam answer. Vertically this equilibrium is stable, since the repulsion grows as the gap closes; sideways it is not, which is why the demonstration keeps the upper wire in a groove.
Checkpoint
§11.2 — the sign that everyone gets backwards●●○○○
Thirty seconds, and the wrong answer here is the one your electrostatics training supplies. Two long straight parallel wires each carry a steady current, and the two currents run in the same direction.
Given
two long parallel wires
steady currents in both
the currents run in the same direction along the wires
Find
(a) What do the wires do to each other?
Hint 1/4
Do not try to recall the rule; build it in two steps, asking what field wire one makes where wire two sits and what a field does to a current.
Hint 2/4
$B_1 = \mu_0 I_1/(2\pi d)$ at wire two, perpendicular to both wires, and then $\vec F = I_2\vec L\times\vec B_1$.
Hint 3/4
Put both currents out of the page. Then $\vec B_1$ at wire two points up the page, $\vec L$ points out of the page, and $\hat z\times\hat y = -\hat x$.
Hint 4/4
The force on wire two points back towards wire one, so parallel currents attract.
Show solutionField of wire one where wire two sits
$$\vec B_1 = \frac{\mu_0 I_1}{2\pi d}\,\hat y$$
wire two lies to the right of wire one, and for a current out of the page the grip rule sends the field up the page there
Force that field puts on wire two
$$\vec F = I_2 L\,\hat z\times B_1\hat y = I_2 L B_1(\hat z\times\hat y) = -I_2 L B_1\,\hat x$$
using the fixed unit vector identity rather than a hand, so that the result can be checked on paper afterwards
$$-\hat x \ \text{points from wire two back towards wire one}$$
which is an attraction
Answer $$\boxed{\,\text{attraction, of size}\ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}\,}$$
Check
Independent check by reversing one current and seeing that the answer reverses: with $I_2$ into the page, $\vec L = -L\hat z$ and the force becomes $+I_2LB_1\hat x$, pointing away.
⚠ Importing the electrostatic sign rule
like charges repel is one of the most heavily drilled facts in the course, and like currents looks like the same sentence
⚠ Setting a force per unit length equal to a whole force
the formula produces newtons per metre and the balancing quantity is often written as a plain weight, so the two get equated without the length being reconciled
wrong$$\frac{\mu_0 I^{2}}{2\pi d} = mg$$
right$$\frac{\mu_0 I^{2}}{2\pi d} = \lambda g = \frac{m}{L}g$$
⚠ Giving the two wires different forces
when the currents are unequal it feels as though the larger one ought to push harder, and the formula is read as belonging to one wire rather than to the pair
11.3Ampere's law: what it says, and what it does not
Walk any closed path, add up the field along it, and the total depends only on the current threading the path.
Adding wires one at a time works while there are two or three of them, and stops working for a solenoid with six hundred turns; what follows is the shortcut, and it also supplies the derivation that was promised and postponed in the first block.
TheoremLaw 11.3: Ampere's law
Conditions
the currents are steady, that is, constant in time; a current that is switching on or off is outside what this form of the law can handle
the path is closed and may be any shape and any size
$I_{\rm enc}$ is the net current through any surface bounded by that path, counted with the sign fixed by the right hand rule
the law is always true under those conditions, but it only yields $B$ when the symmetry is good enough to take $B$ outside the integral
$$\boxed{\,\oint \vec B\cdot d\vec l = \mu_0 I_{\rm enc}\,}$$
Choose any closed path you like, walk all the way round it, and at every step multiply the length of your step by the component of the field along it; the running total when you get back to the start equals the permeability of free space times the net current passing through the loop. Currents outside the loop change the field at every point of it and change the total by nothing at all.
Proof
The one experimental input is the measured field of a long straight wire from the first block: it circles the wire, and its magnitude is $\mu_0I/(2\pi r)$. Everything below is a consequence.
Take a circle of radius $r$ centred on the wire and walk it the way the field points. Then $\vec B$ is parallel to $d\vec l$ at every step and constant in size, so $\oint\vec B\cdot d\vec l = B\oint dl = \dfrac{\mu_0 I}{2\pi r}(2\pi r) = \mu_0 I$.
Notice what happened: the $r$ cancelled. A bigger circle has a weaker field and a longer path, in exactly compensating amounts, so every circle gives the same answer. That cancellation is the whole reason a law of this shape can exist.
Now take a path of any shape that still goes round the wire. Split each step $d\vec l$ into a piece pointing away from the wire and a piece going around it. The field is purely circular, so the radial piece contributes nothing at all. The circular piece has length $r\,d\theta$, so $\vec B\cdot d\vec l = \dfrac{\mu_0 I}{2\pi r}\,r\,d\theta = \dfrac{\mu_0 I}{2\pi}d\theta$, with no $r$ left in it.
Going once round the wire means $\theta$ increases by $2\pi$, so the total is $\mu_0 I$ again, for a path of any shape whatever. The result depends on going round the wire and on nothing else about the route.
If the path does not enclose the wire, $\theta$ increases on the way out and decreases by the same amount on the way back, ending where it began. The net change in $\theta$ is zero, so the circulation is zero, even though $\vec B$ was not zero at a single point of the path.
Finally, fields from several wires add, so the circulations add too, and the total is $\mu_0$ times the sum of the currents that are enclosed, each with the sign of the direction it threads the loop. That is the law as stated.
The two cases that define the law. Around the $\textcolor{#2da44e}{\text{path that encircles the current}}$ the running total comes to $\mu_0 I$ however the path is shaped; around the one that does not, it comes to zero, and the $\textcolor{#1f6feb}{\text{field}}$ on that second path is nowhere near zero. The second path is the whole point: zero circulation and zero field are different statements.
Looks like this, but is not
The circulation around that loop came out zero, so the field must be zero on it. The integral of something that vanishes usually means the something vanished.
Not for a signed sum around a closed path. On the loop that misses the wire, the field is strong on the near side and weak on the far side, but the near side is short and the far side is long, and the two contributions cancel exactly. A magnetometer carried around that loop reads a nonzero value at every point. Zero circulation is a statement about a total, and a total of zero is not the same as a set of zeros.
Getting the straight wire field back out of the law
Take Ampere's law as given, together with the two facts that follow from the symmetry of a long straight wire: the field can depend only on the distance $r$ from the wire, and it can only point around the wire. Derive $B = \mu_0 I/(2\pi r)$.
Given
a long straight wire carrying steady current $I$
$\oint\vec B\cdot d\vec l = \mu_0 I_{\rm enc}$
by symmetry $B$ depends only on $r$ and points along the circle around the wire
Find
the field magnitude at distance $r$
SolutionChoose the path so the integral becomes a multiplication
$$\text{path: a circle of radius}\ r\ \text{centred on the wire, in the plane perpendicular to it}$$
chosen because it is the only path on which the symmetry guarantees both that $B$ is constant and that it is parallel to $d\vec l$.
the first equality uses the field being parallel to the path, the second uses it being constant in size, and the third is just the circumference.
Put in the enclosed current and solve
$$I_{\rm enc} = I$$
the whole wire threads the circle, and nothing else is present
$$B(2\pi r) = \mu_0 I \Rightarrow B = \frac{\mu_0 I}{2\pi r}$$
which is the rule quoted without proof in the first block, now with its origin visible
Answer $$\boxed{\,B = \frac{\mu_0 I}{2\pi r}\,}$$
Check
Independent check on the shape of the answer rather than its size: the $2\pi r$ that appears in the denominator is the circumference of the path, so the formula is really saying that the circulation $\mu_0 I$ is shared evenly along the path.
Three lines, and the entire difficulty is in the first one. Choosing the path is the physics; the rest is arithmetic that the choice has made trivial.
This is the pattern for every use of the law: pick a path on which the field is constant and parallel to the walk, so the integral collapses to $B$ times a length. Without such a path the law is true and useless.
Four wires, one loop: which currents count and with what sign
A closed path is drawn on the page, and four long wires run perpendicular to it. Inside the path there are wires carrying $5.00\ \mathrm{A}$ out of the page and $3.00\ \mathrm{A}$ into the page. Outside the path, close beside it, there are wires carrying $8.00\ \mathrm{A}$ out of the page and $2.00\ \mathrm{A}$ into the page. The path is walked anticlockwise as seen by the reader. Find $\oint\vec B\cdot d\vec l$, and state what the answer says about the field at any single point on the path.
Given
inside the path: $5.00\ \mathrm{A}$ out of the page and $3.00\ \mathrm{A}$ into the page
outside the path: $8.00\ \mathrm{A}$ out of the page and $2.00\ \mathrm{A}$ into the page
the path is walked anticlockwise as seen by the reader
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
the circulation, and what it tells you about the field
SolutionFix the sign convention before counting anything
$$\text{anticlockwise walk} \Rightarrow \text{positive current is out of the page}$$
curl the right hand the way you are walking and the thumb comes out of the page; this is a choice.
Count only what threads the loop
$$I_{\rm enc} = +5.00 - 3.00 = 2.00\ \mathrm{A}$$
the two outside wires are discarded, not because they are weak but because they pass through no surface bounded by this path.
the full $\mu_0$ is needed here rather than $\mu_0/2\pi$, because no circumference is being divided out
Say what has and has not been determined
$$B(\text{at a point}) : \text{not determined}$$
the outside wires make a large contribution to the field everywhere on the path while contributing nothing to the total.
Answer $$\boxed{\,\oint\vec B\cdot d\vec l = 2.51\times10^{-6}\ \mathrm{T\,m},\ \text{and}\ \vec B\ \text{at a point on the path is not determined}\,}$$
Check
Independent check by reversing the walking direction: going clockwise makes into the page positive, so $I_{\rm enc} = +3.00 - 5.00 = -2.00\ \mathrm{A}$ and the circulation comes out as $-2.51\times10^{-6}\ \mathrm{T\,m}$.
One subtraction and one multiplication. The marks in a question like this sit in the two wires that were correctly ignored and in the last line.
Carry away the last line rather than the number. Ampere's law constrains a sum, not a field, and becomes a formula only when symmetry lets that sum be written as $B$ times a length. A messy arrangement can still be asked for its circulation.
Checkpoint
§11.3 — zero circulation, and what it does not mean●●●○○
Thirty seconds, and it is the single most common misreading of this law. A closed path is drawn near a long straight current carrying wire, entirely to one side of it.
Given
a long straight wire with a steady current
a closed path lying entirely to one side of the wire
no current passes through the path
Find
(a) What can you conclude?
Hint 1/4
Two different quantities are on the table: a number attached to the whole path, and a vector attached to each point of it. Decide which one the law constrains.
Hint 2/4
$\oint\vec B\cdot d\vec l = \mu_0 I_{\rm enc}$ fixes the total around the path and says nothing about any individual point.
Hint 3/4
Here $I_{\rm enc} = 0$ because the wire is outside the path, so the total is zero; but the wire is close, so the field on the path is easily measurable.
Hint 4/4
Circulation zero, field not zero: the strong near side and the long weak far side cancel.
no current passes through any surface bounded by the path, and how close the wire is does not enter
Ask separately about the field
$$B = \frac{\mu_0 I}{2\pi r} \neq 0 \ \text{at every point of the path}$$
the field of the wire is nonzero everywhere except at infinity, and the path is at a finite distance
Answer $$\boxed{\,\oint\vec B\cdot d\vec l = 0\ \text{while}\ \vec B\neq\vec 0\ \text{everywhere on the path}\,}$$
Check
Independent check by an explicit cancellation. Take a path made of two semicircular arcs at radii $r$ and $2r$ centred on the wire, joined by two radial segments; the wire itself lies outside the region this path bounds. On the near arc $B = \mu_0 I/(2\pi r)$ and the length walked along the field is $\pi r$, so that leg contributes $\mu_0 I/2$. On the far arc $B = \mu_0 I/(4\pi r)$, half as strong, but its length is $2\pi r$, twice as long, so it contributes $\mu_0 I/2$ as well — and it is walked in the opposite sense, so it enters with a minus sign. On the two radial segments $\vec B\perp d\vec l$ and the dot product is zero. Total: $\mu_0 I/2 - \mu_0 I/2 + 0 + 0 = 0$, exactly what the law demands, while $B$ was nonzero at every single point of the walk. The far arc is longer by precisely the factor that makes it weaker, which is why the cancellation is exact and not accidental.
⚠ Reading zero circulation as zero field
the integral sign suggests a sum of positive things, and the habit from ordinary integrals of positive functions is that a zero total means a zero integrand
wrong$$\oint\vec B\cdot d\vec l = 0 \Rightarrow \vec B = \vec 0$$
right$$\oint\vec B\cdot d\vec l = 0 \ \text{only constrains the signed total around the path}$$
⚠ Counting a current that does not thread the loop
a nearby wire visibly dominates the field on the path, and it feels wrong to discard the largest number in the problem
right$$I_{\rm enc} = \sum_{\text{wires through the loop}} \pm I_i$$
⚠ Adding the enclosed currents without their signs
the sign depends on a walking direction that was chosen silently and then never written down, so there is nothing on the page to check the sign against
11.4Choosing the loop: inside a thick wire and inside a cable
Pick a circular path, ask how much current is inside it, and the law hands you the field in one line.
The law is true for every path and useful for very few, so the skill is choosing; here are the two geometries where the choice is easiest and where the exam questions come from.
MethodResult 11.4: the field inside and outside a uniform solid conductor
Conditions
the conductor is a long straight cylinder of radius $R$ carrying total current $I$
the current is spread uniformly over the cross section, so the current density is $J = I/(\pi R^{2})$ everywhere in the metal
$r$ is measured from the axis
the two expressions agree at $r = R$, which is the check that they were derived consistently
Inside the metal the field grows in direct proportion to how far out you are, because a bigger circle catches a bigger share of the current; outside it falls off as one over the distance, because now the whole current is caught and only the path length keeps growing. The largest field anywhere is at the surface, and from outside the wire behaves exactly as though all its current ran along the axis.
Proof
Outside is the calculation already done: a circle of radius $r>R$ encloses the whole current, so $B(2\pi r) = \mu_0 I$ and $B = \mu_0 I/(2\pi r)$. A thick wire looks like a thin one from outside.
Inside, take a circle of radius $r<R$ inside the metal. The symmetry is the same as before, so the left hand side is still $B(2\pi r)$.
The current is spread uniformly, so the share inside radius $r$ is the share of the area: $I_{\rm enc} = I\dfrac{\pi r^{2}}{\pi R^{2}} = I\dfrac{r^{2}}{R^{2}}$.
Then $B(2\pi r) = \mu_0 I r^{2}/R^{2}$, and dividing by $2\pi r$ leaves $B = \mu_0 I r/(2\pi R^{2})$, one power of $r$ having cancelled.
Check the join. At $r=R$ the inside formula gives $\mu_0 I R/(2\pi R^{2}) = \mu_0 I/(2\pi R)$, and the outside formula gives the same thing. No jump at the surface.
Check the ends. At $r=0$ the inside formula gives zero, which is right: a circle of no size encloses no current. As $r$ grows without limit the outside formula goes to zero as well. The maximum is at the surface and nowhere else.
The whole result in one line. Inside the $\textcolor{#d1690a}{\text{metal}}$ the $\textcolor{#1f6feb}{\text{field}}$ rises straight from zero at the axis; outside it falls as one over the distance, and the two pieces meet at the surface with no jump, which is the check worth doing every time.
Looks like this, but is not
Inside a hollow copper pipe carrying a current, the metal is all around you, so the field in the hollow must be strong. Being surrounded by current sounds like the worst possible place to stand.
It is exactly zero. Take a circle inside the hollow: it encloses no metal at all, so $I_{\rm enc} = 0$, and the symmetry still forces $B$ to be constant on that circle, so $B(2\pi r) = 0$ gives $B = 0$. The contributions of all that surrounding current cancel one another everywhere inside. It is the magnetic twin of the electrostatic result that a uniformly charged shell has no field inside it, and the reason both work is symmetry rather than distance.
where you are
current enclosed
field there
$r = 0.50\ \mathrm{mm}$, in the core
$0.75\ \mathrm{A}$
$3.0\times10^{-4}\ \mathrm{T}$
$r = 2.0\ \mathrm{mm}$, in the gap
$3.0\ \mathrm{A}$
$3.0\times10^{-4}\ \mathrm{T}$
$r = 3.5\ \mathrm{mm}$, in the sleeve
$1.61\ \mathrm{A}$
$9.2\times10^{-5}\ \mathrm{T}$
$r = 5.0\ \mathrm{mm}$, outside
$0\ \mathrm{A}$
$0\ \mathrm{T}$
The last row is the point of the whole design: outside the cable the two currents cancel exactly, so it leaks no field into the room and no field from the room reaches the signal inside.
A 10 A current in a 2.0 mm wire: the field at three radii
A long straight copper wire of radius $2.00\ \mathrm{mm}$ carries a steady $10.0\ \mathrm{A}$ spread uniformly over its cross section. Find the magnitude of the magnetic field at $r = 1.00\ \mathrm{mm}$, at $r = 2.00\ \mathrm{mm}$ and at $r = 4.00\ \mathrm{mm}$ from the axis.
Given
$R = 2.00\times10^{-3}\ \mathrm{m}$
$I = 10.0\ \mathrm{A}$, uniform over the cross section
field wanted at $r = 1.00$, $2.00$ and $4.00\ \mathrm{mm}$
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
the field at the three radii
SolutionSort the three points into inside and outside first
$$r = 1.00\ \mathrm{mm} < R,\quad r = 2.00\ \mathrm{mm} = R,\quad r = 4.00\ \mathrm{mm} > R$$
the two formulas apply in different places and the classification decides which one is used.
The point inside the metal
$$B = \frac{\mu_0 I r}{2\pi R^{2}} = (2\times10^{-7})\frac{(10.0)(1.00\times10^{-3})}{(2.00\times10^{-3})^{2}}$$
the enclosed current is only a quarter of the total here, since the area inside $r$ is a quarter of the full cross section
Independent check from the shape of the graph rather than the arithmetic: half way in and twice as far out both give half the surface value, one because the field is proportional to $r$ and the other because it is proportional to $1/r$. Plausibility: a millitesla is about a fortieth of a fridge magnet and twenty times the Earth's field, which is what ten amperes at two millimetres should give.
One classification and three substitutions, with the same grouped constant used each time.
The line worth carrying away is the classification step. Almost every wrong answer here is the right formula used in the wrong region, and writing down which side of $R$ each point falls on removes that whole class of error.
A coaxial cable: the field in the gap, in the sleeve and outside
A coaxial cable has a solid inner conductor of radius $1.00\ \mathrm{mm}$ carrying $3.00\ \mathrm{A}$ out of the page, and an outer sleeve of inner radius $3.00\ \mathrm{mm}$ and outer radius $4.00\ \mathrm{mm}$ carrying the same $3.00\ \mathrm{A}$ back into the page. Both currents are spread uniformly over their own cross sections. Find the field at $r = 2.00\ \mathrm{mm}$, at $r = 3.50\ \mathrm{mm}$ and at $r = 5.00\ \mathrm{mm}$.
Given
inner conductor: radius $a = 1.00\ \mathrm{mm}$, current $3.00\ \mathrm{A}$ out of the page
outer sleeve: from $b = 3.00\ \mathrm{mm}$ to $c = 4.00\ \mathrm{mm}$, current $3.00\ \mathrm{A}$ into the page
both currents uniform over their cross sections
field wanted at $r = 2.00$, $3.50$ and $5.00\ \mathrm{mm}$
Find
the field at the three radii
SolutionIn the gap between the conductors
$$I_{\rm enc} = 3.00\ \mathrm{A}$$
a circle at $2.00\ \mathrm{mm}$ encloses the whole inner conductor and none of the sleeve, so the return current has not started to count yet
the two currents are equal and opposite, so beyond the sleeve there is nothing left to enclose and the symmetry then forces the field itself to vanish, not merely its circulation
Independent check by bracketing the middle answer: at the inner face of the sleeve none of the return current is enclosed yet, giving $2.00\times10^{-4}\ \mathrm{T}$, and at the outer face the field must be zero.
Three regions, three enclosed currents, one formula. The only real work was the annular area fraction in the middle region.
Note the difference between the two zeros in this section. Zero circulation plus enough symmetry gives a zero field, as it does outside this cable; zero circulation on its own, as around a loop that merely misses a wire, does not.
Checkpoint
§11.4 — where the field of a thick wire is largest●●○○○
Thirty seconds with the graph in your head. A long straight wire of radius $R$ carries a steady current spread evenly over its cross section, and you may put your magnetometer anywhere from the axis outwards.
Given
a solid cylindrical wire of radius $R$
steady current, uniform over the cross section
you may measure at any distance from the axis
Find
(a) Where is the field strongest?
Hint 1/4
You are looking for a maximum of a function that is built in two pieces, so check each piece separately and then look at where they join.
Hint 2/4
Inside, $B = \mu_0 I r/(2\pi R^{2})$, which increases with $r$; outside, $B = \mu_0 I/(2\pi r)$, which decreases with $r$.
Hint 3/4
The first piece rises all the way from $r = 0$ to $r = R$, and the second falls from $r = R$ onwards, both reaching $\mu_0 I/(2\pi R)$ at $r = R$.
Hint 4/4
A rise up to the surface and a fall after it puts the maximum exactly at the surface.
strictly decreasing outside, so no maximum out there either; the only remaining candidate is the join
Evaluate at the join
$$B(R) = \frac{\mu_0 I}{2\pi R}$$
both formulas give this, which also confirms there is no jump at the surface
Answer $$\boxed{\,B_{\max} = \frac{\mu_0 I}{2\pi R}\ \text{at}\ r = R\,}$$
Check
Independent numerical check with the wire from the worked example, $I = 10.0\ \mathrm{A}$ and $R = 2.00\ \mathrm{mm}$: at $r = 1.00\ \mathrm{mm}$ the field is $5.00\times10^{-4}\ \mathrm{T}$, at $r = 2.00\ \mathrm{mm}$ it is $1.00\times10^{-3}\ \mathrm{T}$, and at $r = 4.00\ \mathrm{mm}$ it is back to $5.00\times10^{-4}\ \mathrm{T}$.
⚠ Using the whole current at a point inside the metal
the current in the wire is the number printed in the problem, and the idea that only part of it counts requires noticing that the Amperian circle is inside the conductor
Wind the same wire many times over and the field inside becomes uniform and set by turns per metre alone.
A single wire at ten amps gives about the Earth's field, which is useless for anything; the way to a strong field is not more current but more turns, and this block is what happens when you take that seriously.
TheoremResult 11.5: the field inside a long solenoid and inside a toroid
Conditions
solenoid: the length is much greater than the diameter, the turns are close together and evenly spaced, and the field point is well inside, far from either end
solenoid: $n = N/L$ is the number of turns per metre, so the field does not depend on the total number of turns by itself
toroid: the winding is closely and evenly spaced all the way round, and $r$ is the distance from the centre of the ring to the field point
in both cases the interior is empty, so these are the fields with nothing but air inside the winding
$$\boxed{\,B_{\rm solenoid} = \mu_0 n I = \mu_0\frac{N}{L}I\,,\qquad B_{\rm toroid} = \frac{\mu_0 N I}{2\pi r}\,}$$
Inside a long solenoid the field is the same everywhere, points along the axis, and is fixed by how tightly the wire is wound rather than by how much of it there is; a metre of solenoid with two thousand turns gives the same field as a centimetre of solenoid with twenty. Inside a toroid the field is not uniform: it is strongest on the inner rim and weakest on the outer one, falling off as one over your distance from the centre of the ring, exactly as though the whole winding had been replaced by a straight wire down the middle of the hole.
Proof
For the solenoid, take a rectangular path with one long side of length $\ell$ running along the axis inside, and the opposite side outside the winding. The other two sides run across from one to the other.
The two crossing sides contribute nothing, because inside the solenoid the field is along the axis and these sides run across it, so $\vec B\cdot d\vec l = 0$ on them.
The outside side contributes nothing either, because the field outside a long solenoid is very much weaker than inside.
So the whole circulation is the inside leg alone: $\oint\vec B\cdot d\vec l = B\ell$.
That path encloses one turn for every $1/n$ metres of its length, so $I_{\rm enc} = nI\ell$. Setting the two sides equal gives $B\ell = \mu_0 nI\ell$, and $\ell$ cancels.
The cancellation of $\ell$ is the interesting part: the answer cannot depend on which piece of the solenoid you chose.
For the toroid the path is a circle of radius $r$ going round inside the winding, on which symmetry makes $B$ constant and along the path. Each of the $N$ turns crosses that circle once, so $I_{\rm enc} = NI$ and $B(2\pi r) = \mu_0 NI$.
Here nothing cancels: $r$ survives in the answer, so the toroid field is not uniform. A circle drawn in the hole of the doughnut encloses no turns at all and gives zero, and so does one drawn outside the whole ring, where the windings cross it twice in opposite senses.
The choice of path that makes the solenoid easy. Of the four legs of the $\textcolor{#2da44e}{\text{rectangle}}$ only the one lying along the $\textcolor{#1f6feb}{\text{field}}$ inside contributes anything: two run across the field and the fourth sits outside, where the field is negligible.
Looks like this, but is not
More turns means more field, so a solenoid with a thousand turns beats one with two hundred. Every turn adds its own contribution, so the total ought to grow with the count.
Only if the extra turns are packed into the same length. The field depends on $n = N/L$, not on $N$, so a thousand turns spread over a metre and two hundred turns spread over twenty centimetres give exactly the same field inside. Winding the same coil onto a former twice as long halves the field while the number of turns has not changed at all. The number that matters is how tightly it is wound.
A 600 turn solenoid at 2.50 A, and what stretching it does
A solenoid $30.0\ \mathrm{cm}$ long is wound with $600$ turns and carries $2.50\ \mathrm{A}$. Find the field well inside it. Then the same coil is stretched, without unwinding or changing the current, until it is $60.0\ \mathrm{cm}$ long; find the new field.
Given
$L = 0.300\ \mathrm{m}$, $N = 600$ turns
$I = 2.50\ \mathrm{A}$
then stretched to $L' = 0.600\ \mathrm{m}$ with $N$ and $I$ unchanged
the number of turns is unchanged and the length has doubled, so the density halves
$$B' = \mu_0 n' I = \tfrac12 B = 3.14\times10^{-3}\ \mathrm{T}$$
no need to redo the arithmetic: the field is proportional to $n$ and everything else was held fixed
Answer $$\boxed{\,B = 6.28\times10^{-3}\ \mathrm{T}\ \text{and, after stretching,}\ B' = 3.14\times10^{-3}\ \mathrm{T}\,}$$
Check
Independent check against a bare straight wire carrying the same current. At $1.00\ \mathrm{cm}$ from a single straight wire, $B = (2\times10^{-7})(2.50)/(1.00\times10^{-2}) = 5.00\times10^{-5}\ \mathrm{T}$, while the same wire and the same $2.50\ \mathrm{A}$ wound into this coil give $6.28\times10^{-3}\ \mathrm{T}$: a factor of $126$. That factor is what winding buys, and it is the whole reason coils exist. Plausibility: six millitesla is about a fifth of a fridge magnet, entirely ordinary for a small laboratory solenoid.
Two divisions and one multiplication, and the second half of the question needed no arithmetic at all once the proportionality was noticed.
The stretching half separates understanding from recall. When a change leaves $N$ and $I$ alone and alters only the length, do not recompute: name the quantity in the formula that moved, and scale.
A toroid with 800 turns: how much the field varies across the winding
A toroid is wound with $800$ turns and carries $5.00\ \mathrm{A}$. Its inner radius is $12.0\ \mathrm{cm}$ and its outer radius is $15.0\ \mathrm{cm}$. Find the field at the inner rim, at the mean radius and at the outer rim, and state the field in the hole in the middle and outside the whole ring.
the outer rim, twenty per cent below the inner one
The two empty regions
$$\text{hole in the middle}: I_{\rm enc} = 0 \Rightarrow B = 0$$
a circle drawn inside the hole passes through no turns at all, and the symmetry then turns zero circulation into zero field
$$\text{outside the ring}: I_{\rm enc} = NI - NI = 0 \Rightarrow B = 0$$
a circle drawn beyond the outer rim catches every turn twice, once going in and once coming back, so the two cancel
Answer $$\boxed{\,B_{\rm inner} = 6.67\times10^{-3}\ \mathrm{T},\ B_{\rm mean} = 5.93\times10^{-3}\ \mathrm{T},\ B_{\rm outer} = 5.33\times10^{-3}\ \mathrm{T},\ B = 0\ \text{in the hole and outside}\,}$$
Check
Independent check by ratio rather than by arithmetic: the field should be inversely proportional to $r$, so $B_{\rm inner}/B_{\rm outer}$ ought to equal $0.150/0.120 = 1.25$, and $6.67/5.33 = 1.25$. Plausibility: a toroid this size with these numbers gives about six millitesla, the same order as the solenoid in the previous example, which it should be since the ampere turns per metre of path are comparable.
One grouped constant and five short lines. The two zeros at the end are worth as much as the three numbers and take no calculation.
The twenty per cent spread across the winding is the practical point. A toroid is chosen to keep the field inside the device, never to make it uniform; for uniformity you use a long solenoid, where the geometry cancels the length.
Checkpoint
§11.5 — stretching a solenoid without unwinding it●●○○○
Thirty seconds, and it is the standard way this formula is tested. A solenoid is connected to a fixed current source, and without unwinding a single turn or changing the current, you pull it out until it is twice as long as before.
Given
the number of turns $N$ is unchanged
the current $I$ is unchanged
the length is doubled, so the turns are half as tightly packed
Find
(a) What happens to the field well inside it?
Hint 1/4
Decide which single quantity in the formula the stretching changed, and leave everything else alone.
Hint 2/4
$B = \mu_0 nI$ with $n = N/L$, so $B = \mu_0 NI/L$ and the field is inversely proportional to the length.
Hint 3/4
Here $N$ and $I$ are fixed and $L$ has doubled, so $B' / B = L/L' = 1/2$.
Hint 4/4
The field halves, because the turns are now half as tightly packed.
Show solutionPut the formula in terms of what changed
$$B = \mu_0 nI = \frac{\mu_0 NI}{L}$$
rewriting in terms of $N$ and $L$ separately is what makes the effect of stretching visible.
the constants and the two fixed quantities cancel in the ratio, so no numbers are needed
Answer $$\boxed{\,B' = \tfrac12 B\,}$$
Check
Independent check with the numbers from the worked example: $600$ turns on $0.300\ \mathrm{m}$ at $2.50\ \mathrm{A}$ gives $6.28\ \mathrm{mT}$, and the same coil on $0.600\ \mathrm{m}$ gives $n = 1000\ \mathrm{turns/m}$ and $B = (4\pi\times10^{-7})(1000)(2.50) = 3.14\ \mathrm{mT}$, which is half.
⚠ Putting the total number of turns into the solenoid formula
$N$ is the number printed in the question and $n$ has to be computed, so the eye reaches for the one that is already there
wrong$$B = \mu_0 N I$$
right$$B = \mu_0 n I = \mu_0\frac{N}{L}I$$
⚠ Using the toroid formula as though the field were uniform
a toroid looks like a solenoid bent into a ring, and the solenoid field is uniform, so the property is carried over with the picture
wrong$$B_{\rm toroid} = \mu_0 n I \ \text{everywhere inside}$$
right$$B_{\rm toroid} = \frac{\mu_0 NI}{2\pi r},\qquad \text{different at every radius}$$
⚠ Using the solenoid formula near an end
the formula carries no position in it, so it reads as though it applies everywhere inside the coil
wrong$$B(\text{at the mouth of the coil}) = \mu_0 nI$$
right$$B(\text{at the mouth of the coil}) \approx \tfrac12\mu_0 nI,\qquad B = \mu_0 nI\ \text{only well inside}$$
11.6Biot and Savart: adding up a bent wire piece by piece
Chop the wire into pieces, give each a contribution with a cross product in it, and add the pieces up.
Ampere's law needs a symmetry and a loop, arc or finite segment has none, so this last tool is the general one: slower, always available, and the only way to reach the centre of a coil.
TheoremLaw 11.6: the Biot and Savart law, and the three results it is usually used for
Conditions
$d\vec l$ is a short piece of the wire pointing along the current in it, and $\hat r$ points from that piece to the field point, a distance $r$ away
the contributions of all the pieces are added as vectors, which is why symmetry arguments about which components cancel do most of the work
the centre and arc results are for a circular wire of radius $R$, with the angle $\phi$ of the arc measured in radians
the axis result is for a point a distance $x$ from the centre of a circular coil of $N$ turns, measured along the axis.
Every short piece of a current carrying wire makes its own small field at a point, proportional to the current and to the length of the piece, falling off as the inverse square of the distance to it, and pointing at right angles to both the piece and the line joining the two. Add up those small fields as vectors along the whole wire and you have the answer. A piece that points straight at the field point contributes nothing at all, because the cross product of two parallel directions is zero.
Proof
Take the centre of a full circular loop of radius $R$. Every piece of the wire is the same distance $R$ from the centre.
At every piece, $d\vec l$ runs along the circle and $\hat r$ points inwards to the centre, so the two are perpendicular and $\vert d\vec l\times\hat r\vert = dl$. The sine factor is one everywhere and never has to be tracked.
The direction is the same for every piece as well: by the right hand rule each contribution points along the axis, out of the plane on the same side. Nothing cancels, so the vector sum is a plain sum of magnitudes.
Hence $B = \dfrac{\mu_0 I}{4\pi R^{2}}\displaystyle\oint dl = \dfrac{\mu_0 I}{4\pi R^{2}}(2\pi R) = \dfrac{\mu_0 I}{2R}$, and $N$ turns wound together multiply this by $N$.
The arc follows immediately. If the wire covers only an angle $\phi$ instead of the full $2\pi$, the integral $\oint dl$ becomes $R\phi$ instead of $2\pi R$, and $B = \dfrac{\mu_0 I\phi}{4\pi R}$. A semicircle, $\phi = \pi$, gives exactly half the full loop.
For a point on the axis the distances are still all equal, to $\sqrt{R^{2}+x^{2}}$, but the directions are not. Each contribution tilts, and the components perpendicular to the axis cancel in pairs from opposite sides of the loop while the axial components add.
Keeping only the axial part introduces a factor $R/\sqrt{R^{2}+x^{2}}$, and putting that together with $\oint dl = 2\pi R$ and the inverse square gives the axis result. Setting $x = 0$ returns the centre result.
The geometry that makes the standard hairpin problem a one line calculation.
Looks like this, but is not
A wire is a wire, so a straight piece of it a distance $r$ away gives $\mu_0 I/(2\pi r)$. The formula is the first one in the section and it has a straight wire in it.
That formula was derived for a wire long enough that its ends are far away, and a short segment is the opposite case. Adding up the Biot and Savart contributions of a finite segment always gives less than the infinite answer, because the missing parts of the wire were contributing something. In the extreme case of a very short segment, a distance $r$ away, the field is smaller by roughly the ratio of the segment length to $r$. Use the infinite formula only when the wire really is long compared with $r$.
The centre of a 20 turn coil of radius 5.0 cm carrying 3.0 A
A flat circular coil of $20$ turns and radius $5.00\ \mathrm{cm}$ carries $3.00\ \mathrm{A}$. Find the field at its centre, and compare it with the field a single straight wire carrying the same current would make at the same distance of $5.00\ \mathrm{cm}$.
Independent check on the ratio, obtained algebraically instead of numerically: $\dfrac{\mu_0 NI/(2R)}{\mu_0 I/(2\pi R)} = N\pi$, and with $N = 20$ that is $62.8$, matching the arithmetic exactly. Plausibility: three quarters of a millitesla is fifteen times the Earth's field and a fiftieth of a fridge magnet, which is the right scale for a small coil on a bench.
Two substitutions. Notice that the second one exists only to give the first a meaning, and it is the reason the answer is memorable.
The factor $N\pi$ is the reason nobody makes fields with straight wires. Bending the same metre of copper into twenty turns costs no extra current and delivers sixty three times the field, and every electromagnet takes that trade as far as it will go.
A hairpin: two long leads and a semicircle of radius 4.0 cm
A wire carrying $6.00\ \mathrm{A}$ comes in along a straight line, bends into a semicircle of radius $4.00\ \mathrm{cm}$ centred on a point $\mathrm{C}$ that lies on that same line, and leaves along the line again on the far side. Find the magnetic field at $\mathrm{C}$.
Given
$I = 6.00\ \mathrm{A}$
semicircle of radius $R = 4.00\times10^{-2}\ \mathrm{m}$
both straight leads lie along the line through $\mathrm{C}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
the field at the centre of the semicircle
SolutionKill the straight pieces first
$$d\vec l \parallel \hat r \Rightarrow d\vec l\times\hat r = \vec 0$$
on each straight lead the current runs directly towards or directly away from $\mathrm{C}$, so the angle in the cross product is zero or a hundred and eighty degrees and the sine vanishes at every point of them
$$B_{\rm leads} = 0$$
however long they are; length cannot rescue a contribution that is zero at every point.
substituting only after the geometry has been reduced to a single fraction
The direction
$$\hat B = \widehat{d\vec l\times\hat r},\ \text{the same for every piece of the arc}$$
so the answer is perpendicular to the plane of the hairpin, and which side it comes out of is fixed by the sense in which the current runs round the arc
Answer $$\boxed{\,B_{\mathrm{C}} = \frac{\mu_0 I}{4R} = 4.71\times10^{-5}\ \mathrm{T},\ \text{perpendicular to the plane of the hairpin}\,}$$
Check
Independent check by a limiting case: a full loop of the same radius and current would give $\mu_0I/(2R) = 9.42\times10^{-5}\ \mathrm{T}$, and this answer is exactly half of that, as a semicircle must be.
The whole problem was two decisions and one substitution, and the first decision, that the leads give nothing, is where the marks are.
Reduce every arc problem to a fraction of a loop before touching a calculator: a quarter circle gives a quarter of $\mu_0I/(2R)$. Done that way, the general arc formula and its radians never have to be used at all.
Checkpoint
§11.6 — the piece of wire that contributes nothing●●○○○
Thirty seconds, and it is the step that turns a hard looking problem into a one liner. A straight length of wire carries a current, and you want the field at a point $\mathrm{P}$ that lies on the same straight line as the wire, beyond its end.
Given
a straight segment of wire carrying a steady current
the field point $\mathrm{P}$ lies on the line of the wire, past its end
no other current is present
Find
(a) What field does that segment produce at $\mathrm{P}$?
Hint 1/4
Do not estimate a size; ask instead what angle the current element makes with the line running from it to the field point.
Hint 2/4
$\vert d\vec B\vert \propto \vert d\vec l\times\hat r\vert = dl\sin\theta$, where $\theta$ is the angle between the element and the direction to the point.
Hint 3/4
Here $\mathrm{P}$ lies on the line of the wire, so every element points directly at it or directly away and $\theta$ is $0$ or $180^{\circ}$.
Hint 4/4
The sine is zero at every point of the segment, so the whole segment contributes nothing.
Independent check from the picture rather than the algebra: the field of a straight current circles it, and $\mathrm{P}$ sits on the axis of those circles, which is the one place where a circle has no direction to offer.
⚠ Using the straight wire formula at the centre of a loop
both formulas are $\mu_0 I$ divided by a length with a small number in front, and $R$ appears in both, so they are easy to interchange while writing quickly
wrong$$B_{\rm centre} = \frac{\mu_0 I}{2\pi R}$$
right$$B_{\rm centre} = \frac{\mu_0 I}{2R}$$
⚠ Putting degrees into the arc formula
the angle of an arc is naturally spoken of in degrees, and the formula does not look like one that cares
11.7What a core does: ferromagnetism, and the two weak effects
Matter multiplies the field a coil makes, by a hair for most substances and by thousands for iron.
Every field so far has been made in air, and every real electromagnet has something inside it; this block is what that something does and where the simple description breaks down.
RuleRule 11.7: a core multiplies the field of a coil
Conditions
$B_0$ is the field the same coil with the same current would produce with nothing inside it, for a solenoid $\mu_0 nI$
$\mu_r$ is the relative permeability of the material, a pure number with no units
for paramagnetic and diamagnetic substances $\mu_r$ really is a constant of the material to the accuracy anyone needs.
for ferromagnets $\mu_r$ is not a constant: it depends on the applied field, on the temperature, and on what the sample has been through before
Put something inside a coil and the field changes by a factor that belongs to the material. For almost everything that factor is one to five decimal places, so the core might as well be air; a paramagnetic substance nudges the field up by a few parts in a hundred thousand and a diamagnetic one nudges it down by about as much. For iron, nickel, cobalt and their alloys the factor is in the hundreds or thousands, and it is not a fixed number: it falls away as the metal approaches .
Why $\mu_r$ for iron is not a number you can look up once. The field in the metal depends on which way the applied field is moving: $\textcolor{#6b6f76}{\text{raising it}}$ and $\textcolor{#1f6feb}{\text{lowering it}}$ trace different curves. Both branches flatten off near one and a half tesla, and beyond that more current buys almost nothing.
Looks like this, but is not
The relative permeability of soft iron is fifteen hundred, the way its density is seven thousand nine hundred kilograms per cubic metre. Both are numbers in a table with the material's name beside them.
The density really is a property of the metal; the permeability is a property of the metal at a stated working point. Quoting fifteen hundred for iron means it multiplies the field by that much when the applied field is small, and if you use the same number at a hundred times the current you predict a field of tens of tesla.
class and example
what it does to the field
size of the effect
diamagnetic, for example copper or bismuth
weakens it slightly, and is pushed out of a strong field
$\chi_m$ about $-1\times10^{-5}$ for copper and about $-1.7\times10^{-4}$ for bismuth
paramagnetic, for example aluminium
strengthens it slightly, and is pulled into a strong field
$\chi_m$ about $+2\times10^{-5}$, and it falls as the temperature rises
ferromagnetic, for example iron
strengthens it enormously, and is pulled in hard
$\mu_r$ from about $10^{2}$ to about $10^{4}$, depending on the field and the history
ferromagnetic above its critical temperature
behaves as an ordinary paramagnet
$\mu_r$ drops to just above $1$, and the material stops being a magnet
The gap between the first two rows and the third is five to eight powers of ten, and that is the whole practical story of magnetic materials.
An iron cored solenoid, and the answer that cannot be right
A solenoid with $2000$ turns per metre carries $0.200\ \mathrm{A}$ and is wound on a soft iron core whose relative permeability is $1500$ at this working point. Find the field inside. Then a student, wanting more field, raises the current to $2.50\ \mathrm{A}$ and keeps using $\mu_r = 1500$. Find what that predicts, and say what is wrong with it.
Given
$n = 2000\ \mathrm{turns/m}$
$I = 0.200\ \mathrm{A}$ first, then $2.50\ \mathrm{A}$
$\mu_r = 1500$ quoted at the first working point
iron saturates at roughly $1.5$ to $2\ \mathrm{T}$
Find
the field at the first current, the prediction at the second, and the fault in it
SolutionThe field the coil would make on its own
$$B_0 = \mu_0 nI = (4\pi\times10^{-7})(2000)(0.200) = 5.03\times10^{-4}\ \mathrm{T}$$
computing the empty coil field first keeps the two multiplications separate, so that $\mu_0$ can never be applied twice
iron cannot produce a field of nine tesla at any current, so the prediction is not merely imprecise, it is impossible, and the fault must be in the constant rather than in the arithmetic
Answer $$\boxed{\,B = 0.754\ \mathrm{T}\ \text{at}\ 0.200\ \mathrm{A};\ \text{the}\ 9.42\ \mathrm{T}\ \text{prediction is impossible, since}\ \mu_r\ \text{collapses as the iron saturates}\,}$$
Check
Plausibility check on the first answer, which is the one being trusted: $0.754\ \mathrm{T}$ is about the field of a strong laboratory electromagnet and about fifteen thousand times the Earth's field, produced by a fifth of an ampere.
Three multiplications, and the only difficult part is the willingness to reject the last one.
Every ferromagnetic calculation should end with a look at the number. Above about two tesla the answer is telling you the model has run out, not that you have a strong magnet; past that ceiling the iron has to go.
Aluminium and bismuth cores: an effect you cannot see at three figures
A coil produces a field of exactly $0.500\ \mathrm{T}$ with nothing inside it. Find the field inside if the space is filled with aluminium, for which the is about $+2.2\times10^{-5}$, and if it is filled with bismuth, for which it is about $-1.7\times10^{-4}$. State how many significant figures you would need to quote to see the difference.
Given
$B_0 = 0.500\ \mathrm{T}$ with an empty core
aluminium: $\chi_m \approx +2.2\times10^{-5}$
bismuth: $\chi_m \approx -1.7\times10^{-4}$
$\mu_r = 1+\chi_m$
Find
the two fields, and the number of digits needed to distinguish them
SolutionTurn each susceptibility into a multiplier
a decrease of eighty five microtesla, about eight times bigger than the aluminium effect and still invisible at three figures
Answer the last part honestly
$$0.500\ \mathrm{T}\ \text{to three figures in every case}$$
five figures are needed to separate the aluminium from the empty coil, and four to separate the bismuth, so at the precision used everywhere else on this page the three answers are the same number
Independent check of the ratio between the two effects: bismuth's susceptibility is about $1.7\times10^{-4}/2.2\times10^{-5} \approx 7.7$ times aluminium's in size, and the two changes computed above, $85\ \mathrm{\mu T}$ and $11\ \mathrm{\mu T}$, stand in the ratio $7.7$ as well. Plausibility: both changes are of the order of the Earth's field, which is what a part in ten thousand of half a tesla should be.
Two multiplications, and the interesting work is entirely in the last line.
This is why the rest of this section could ignore matter completely. Air, copper, aluminium, plastic and water are all magnetically within a hundredth of a per cent of vacuum, so only iron, nickel and cobalt change a calculation.
Checkpoint
§11.7 — doubling the current in a core that is already saturated●●●○○
Thirty seconds, and it is the question that separates using a table from understanding it. An electromagnet with an iron core is producing $1.6\ \mathrm{T}$, close to the saturation field of iron. You double the current through its winding.
Given
an iron cored electromagnet producing about $1.6\ \mathrm{T}$
iron saturates at roughly $1.5$ to $2\ \mathrm{T}$
the current through the winding is doubled
Find
(a) What happens to the field in the core?
Hint 1/4
The formula has two factors in it, so ask which of them you are entitled to hold fixed while the current changes.
Hint 2/4
$B = \mu_r B_0$ with $B_0 = \mu_0 nI$; doubling $I$ doubles $B_0$ exactly, but $\mu_r$ for a ferromagnet is a function of the applied field.
Hint 3/4
Here the core is already near the flat part of the curve at about $1.6\ \mathrm{T}$, which is where $\mu_r$ has already begun to collapse.
Hint 4/4
The field creeps up a little; it does not come close to doubling.
Show solutionSeparate what scales from what does not
$$B_0 = \mu_0 nI \Rightarrow B_0' = 2B_0$$
the empty coil field is exactly proportional to the current, and nothing about the core changes that
$$\mu_r = \mu_r(B_0)\ \text{and it is falling here}$$
on the flat part of the curve the field in the metal barely responds, so the ratio of the two fields is dropping as fast as the applied field is rising
Combine them
$$B' = \mu_r(2B_0)\,(2B_0) \ll 2B$$
a product of a factor of two and a factor well below one half
Answer $$\boxed{\,B' \ \text{rises only slightly above}\ 1.6\ \mathrm{T}\,}$$
Check
Independent check by pushing the wrong answer to an absurdity: if the field really doubled each time the current did, then four doublings from $1.6\ \mathrm{T}$ would give $25.6\ \mathrm{T}$ from an iron core, which is several times the strongest field iron can support and about the strongest steady field ever produced by any means.
⚠ Treating the relative permeability of a ferromagnet as a constant
it is printed in tables next to genuine material constants, and nothing in the symbol suggests that it depends on the field it is being used in
wrong$$B = \mu_r\mu_0 nI \ \text{with the same}\ \mu_r\ \text{at every current}$$
right$$B = \mu_r(B_0)\,\mu_0 nI,\qquad B \lesssim 2\ \mathrm{T}\ \text{for iron}$$
⚠ Applying the permeability twice
the working is often written as $B = \mu_r B_0$ after $B_0$ has already been computed with a permeability in it, and the two steps look like one
Which of the three routes to use, decided before you start
Any question that asks for a magnetic field produced by a current. Spending thirty seconds here saves the ten minutes that an integral you did not need would have cost.
Is it one of the four shapes you already know?
Long straight wire, solid or hollow cylinder, long solenoid, toroid, circular loop at its centre or on its axis, circular arc at its centre. If yes, quote the result and stop.
If not, is there a symmetry good enough for a loop?
Ask whether there is a closed path on which you can argue, before any calculation, that $B$ has the same magnitude everywhere and is either along the path or across it.
If not, use Biot and Savart
Anything finite, bent, or off the axis of symmetry needs the integral.
Several sources: do each one separately, then add as vectors
Superposition applies whichever route was used for each piece. Compute each magnitude, fix each direction, and only then add.
Finish with a size check
Compare the answer with the Earth's field, about $5\times10^{-5}\ \mathrm{T}$, and with a fridge magnet, a few hundredths of a tesla. A bare wire at a few centimetres should land near the first.
Where it goes wrong
Reaching for Ampere's law because it is the newest tool, on a geometry with no symmetry, and then being unable to take $B$ out of the integral.
Integrating a case that is on the list of known results, which costs time and introduces errors without earning anything.
Adding magnitudes from two sources without checking that they point the same way.
Skipping the size check, which is the only step that catches a power of ten.
Adding the fields of several straight wires at one point
Two, three or four long straight wires perpendicular to the page, and a field wanted at some point among them. This is the most common computational question in the whole topic.
Draw it and mark every perpendicular distance
Put the field point on the diagram and draw the line from each wire to it. That length is the $r$ for that wire and nothing else is.
Compute each magnitude separately
$B_i = (2\times10^{-7})I_i/r_i$ for each wire, written on its own line and labelled with which wire it belongs to. Do not add anything yet.
Fix each direction with the grip rule, and write it as an angle or as components
Each field is perpendicular to the line from its wire to the point, in the sense given by the grip rule.
Add the components, then take the magnitude
$B_x = \sum B_{ix}$ and $B_y = \sum B_{iy}$, then $B = \sqrt{B_x^{2}+B_y^{2}}$ and the direction from the two components. Look for cancellations before you reach for the calculator; symmetric arrangements usually kill one whole component.
Check the answer against a limiting case
Push one current to zero in your head and see whether the answer reduces to the single wire result.
Where it goes wrong
Adding magnitudes as though the two fields were parallel, which is right only in the special case where they happen to be.
Using the distance between the wires instead of the distance from a wire to the field point.
Getting one grip rule right and the other backwards, which turns a difference into a sum and roughly doubles the answer.
Forgetting that the field of each wire is perpendicular to its own radius, not to the line joining the two wires.
Getting the direction right every time, in two separate moves
Every direction question in this section, including the ones where you already feel sure. It costs one extra line and it is the difference between a sign error and a full mark.
Name the axes on the page before anything else
Write down that $x$ is to the right, $y$ is up the page and $z$ is out of it.
For the field of a wire, grip it
Thumb along the current, fingers curl the way the field goes.
For the force on a current, cross it
$\vec F = I\vec L\times\vec B$ with $\vec L$ along the current. Do not try to combine this with the previous step in one movement of the hand; two clean steps beat one clever one.
For an Amperian loop, curl the walk
Fingers curl the way you are walking round the loop, thumb gives the direction that counts as positive current. Write the choice down, because the sign of the answer means nothing without it.
Check by reversing something
Reverse one current in your head. Every direction in the problem that should flip must flip, and every magnitude must stay the same.
Where it goes wrong
Using the left hand, or using the right hand with the fingers along the current instead of the thumb.
Applying the sign of a reversed current both by turning the arrow round in the picture and by putting a minus into the algebra.
Choosing a walking direction for an Amperian loop silently, so that the sign of the enclosed current has nothing to be checked against.
Treating a field into the page as a negative magnitude rather than as a direction.
Ampere's law does the whole job: a long solenoid
A long solenoid has $1200$ turns per metre and carries $2.00\ \mathrm{A}$. Find the field well inside it.
Given
$n = 1200\ \mathrm{turns/m}$
$I = 2.00\ \mathrm{A}$
long solenoid, field point well inside
Find
the field inside
SolutionChoose the path and collapse the integral
$$\oint\vec B\cdot d\vec l = B\ell$$
the rectangle with one leg inside along the axis: two legs cross the field and the fourth is outside where the field is negligible, so only one term survives
$$\mu_0 I_{\rm enc} = \mu_0 n I \ell$$
the inside leg of length $\ell$ catches $n\ell$ turns, each carrying $I$
$$B\ell = \mu_0 nI\ell \Rightarrow B = \mu_0 nI = (4\pi\times10^{-7})(1200)(2.00) = 3.02\times10^{-3}\ \mathrm{T}$$
the arbitrary length cancels, which is the same statement as the field being uniform
Plausibility: three millitesla, about sixty times the Earth's field, from a coil wound at roughly one turn per millimetre carrying two amperes.
Ampere's law is true and useless: one circular loop
A single circular loop of radius $4.00\ \mathrm{cm}$ carries $2.00\ \mathrm{A}$. Find the field at its centre.
Given
$R = 4.00\times10^{-2}\ \mathrm{m}$
$I = 2.00\ \mathrm{A}$
field wanted at the centre of the loop
Find
the field at the centre
SolutionTry Ampere's law first, and see it fail
$$\oint\vec B\cdot d\vec l = \mu_0 I \ \text{for a path threading the loop}$$
the law is perfectly true here, and it stays true whatever path is chosen
$$B \ \text{cannot be taken outside the integral}$$
there is no path through the centre on which the magnitude of the field is constant and its direction fixed relative to the path, so the left hand side never collapses to $B$ times a length
Independent check by comparison with a straight wire: the same $2.00\ \mathrm{A}$ in a straight wire $4.00\ \mathrm{cm}$ away would give $1.00\times10^{-5}\ \mathrm{T}$, and the loop result should exceed it by a factor of $\pi$, since $\mu_0I/(2R)$ divided by $\mu_0I/(2\pi R)$ is $\pi$.
Two problems with the same law available, the same kind of source and comparable currents, and the difference is entirely one of symmetry: for the solenoid a path exists on which the field is constant and the answer falls out in three lines, and for the single loop no such path exists anywhere, so the same law gives a true statement that contains no usable information about the field at the centre.
How to tell them apart
Before writing anything, try to say out loud what the field looks like on some closed path, using only the symmetry of the source and not any calculation. If you can say that it has one magnitude everywhere on that path and is either along it or across it, Ampere's law will work.
Scaffolding comes off
The common skeleton
Draw the arrangement, put the field point on it, and mark the perpendicular distance from each wire to that point.
Write the magnitude of each wire's contribution on its own line: $B_i = (2\times10^{-7})I_i/r_i$, with no directions yet.
Fix each direction with the grip rule, using the vector form $\hat B = \hat I\times\hat r$ when the picture is ambiguous.
Resolve each contribution into components along the page axes, even when one component is obviously zero.
Add the components, then recombine into a magnitude and a direction.
Look for the cancellation the geometry was built to produce, and say which component died and why.
Check: compare the size with the Earth's field, and check that pushing one current to zero returns the single wire answer.
1 · fully worked
Two antiparallel currents, and the point that sees both from 5.0 cm
Two long straight wires run perpendicular to the page. Wire 1 is at the origin and carries $8.00\ \mathrm{A}$ out of the page; wire 2 is at $x = 6.00\ \mathrm{cm}$ and carries $8.00\ \mathrm{A}$ into the page. Find the magnetic field at the point $\mathrm{P} = (3.00\ \mathrm{cm},\ 4.00\ \mathrm{cm})$.
Given
wire 1 at the origin, $8.00\ \mathrm{A}$ out of the page
wire 2 at $x = 6.00\ \mathrm{cm}$, $8.00\ \mathrm{A}$ into the page
current into the page reverses the sense, and the geometry is mirrored, so both reversals happen at once and the result is the mirror image of the first
Add and read off the cancellation
$$B_x = (3.20\times10^{-5})(-0.800+0.800) = 0$$
the horizontal parts are equal and opposite, which is exactly what the symmetric placement of $\mathrm{P}$ was arranged to produce
the vertical parts reinforce, because reversing the second current turned what would have been a cancellation into an addition
Answer $$\boxed{\,\vec B = 3.84\times10^{-5}\ \mathrm{T}\ \text{in the}\ +y\ \text{direction, straight up the page}\,}$$
Check
Independent check by a limiting case: switch wire 2 off and the answer must drop to $3.20\times10^{-5}\ \mathrm{T}$ at $36.9^{\circ}$ above the $-x$ direction, which is what $B_1$ alone is; the computed total is larger than either contribution but smaller than their sum $6.40\times10^{-5}\ \mathrm{T}$, as it must be for two vectors at an angle. Plausibility: the answer is close to the Earth's field, which is the right scale for eight amperes at five centimetres.
One right triangle, one magnitude, two unit vectors and one addition. The trigonometry was avoided entirely by choosing a point that makes a three four five triangle with both wires.
Notice what the antiparallel arrangement did: with both currents out of the page the vertical parts would have cancelled instead. Predicting which component dies, before computing anything, is the most valuable habit in this kind of problem.
2 · you write the reasoning
Easier than the last one: no components, no triangle, three lines. Two long parallel wires are $10.0\ \mathrm{cm}$ apart and each carries $5.00\ \mathrm{A}$ in the same direction. The field is wanted at the midpoint of the line joining them. All three lines below are correct. Before opening the model reasons, say in your own words why each one is allowed, and pay particular attention to what the third line is really claiming.
reasoning
The first line uses half the separation, not the whole of it, because the field point is the midpoint and $r$ is always measured from the wire to the field point. The distance between the wires never appears in the formula at all.
reasoning
The second line is identical to the first, and that is a statement about the geometry rather than a repetition. Equal currents at equal distances give equal magnitudes, so anything that distinguishes the two contributions has to be direction, and the whole answer now depends on the third line.
reasoning
The third line is the only one doing physics. At a point between two currents flowing the same way, the two fields circle their own wires in the same rotational sense, which means that at a point between them they point in opposite directions. Equal magnitudes plus opposite directions gives exactly zero. Note what this does not say: the field is zero at that one point only, and a centimetre to either side it is not.
3 · find the buried error
Harder than the last one: two different shapes of wire in the same problem, so two different formulas, and a direction question at the end. A long straight wire lies in the plane of the page carrying $12.0\ \mathrm{A}$ to the right. A circular loop of radius $3.00\ \mathrm{cm}$ carrying $4.00\ \mathrm{A}$ clockwise, as seen by the reader, lies in the same plane with its centre $10.0\ \mathrm{cm}$ above the wire. A student produces the four lines below and reports a field of $5.07\times10^{-5}\ \mathrm{T}$ at the centre of the loop. Exactly two of the four lines are faulty. Find them.
the two buried errors (2)
⚠ step 2
The centre of a circular loop is not a straight wire problem. The correct result is $B = \mu_0 I/(2R)$, not $\mu_0 I/(2\pi R)$, so the field is larger by a factor of $\pi$: $B_{\rm loop} = (4\pi\times10^{-7})(4.00)/(2\times0.0300) = 8.38\times10^{-5}\ \mathrm{T}$, not $2.67\times10^{-5}\ \mathrm{T}$.
Both formulas are mu nought times a current over a length with a small number in front, both use the symbol for a radius, and the wire formula is the one that has been used four times already by the time this line is written.
right
Ask what shape the wire is before choosing the formula, and keep the factor of $\pi$ as the tell: $\mu_0I/(2\pi r)$ has a $\pi$ in it and belongs to a straight wire, $\mu_0I/(2R)$ has none and belongs to the centre of a loop.
⚠ step 4
The two contributions point in opposite directions, one out of the page and one into it, as lines 1 and 2 both correctly state. Combining them means subtracting, not adding. With the corrected loop field the answer is $8.38\times10^{-5} - 2.40\times10^{-5} = 5.98\times10^{-5}\ \mathrm{T}$, into the page.
Line 3 has just established that both fields lie along the same axis, and the natural next move is to combine them, at which point the two directions written down two lines earlier stop being consulted. The word total invites a sum.
right
Turn the directions into signs the moment they are found, rather than carrying them as words: write $B_{\rm wire} = +2.40\times10^{-5}\ \mathrm{T}$ and $B_{\rm loop} = -8.38\times10^{-5}\ \mathrm{T}$ with out of the page as positive, and the addition then does its own bookkeeping and gives $-5.98\times10^{-5}\ \mathrm{T}$, that is, into the page.
4 · the bare problem
§11.1 — three wires at the corners of a square●●●●○
No scaffolding this time. Three long straight wires run perpendicular to the page through three corners of a square of side $8.00\ \mathrm{cm}$, and the field is wanted at the fourth corner. Work to three significant figures and give a direction as well as a magnitude.
Given
wire A at the top left corner, $6.00\ \mathrm{A}$ out of the page
wire B at the top right corner, $6.00\ \mathrm{A}$ into the page
wire C at the bottom right corner, $6.00\ \mathrm{A}$ out of the page
the field point is the bottom left corner, and the square has side $8.00\ \mathrm{cm}$
$x$ to the right, $y$ up the page, $z$ out of the page
Find
(a) Find the field at the empty corner from wire A alone, and from wire C alone, with directions.
(b) Find the field at that corner from wire B alone, with its direction.
(c) Combine the three and give the magnitude and direction of the total field.
Hint 1/4
This is the same recipe as before with one extra wire: three distances, three magnitudes, three directions, one vector sum. The only new thing is that one of the wires sits on the diagonal.
Hint 2/4
$B = (2\times10^{-7})I/r$ for each wire, with $\hat B = \hat I\times\hat r$, and $\hat r$ running from that wire to the field point.
Hint 3/4
The two adjacent wires are $8.00\ \mathrm{cm}$ away and the diagonal one is $8.00\sqrt2 = 11.3\ \mathrm{cm}$ away. All three currents are $6.00\ \mathrm{A}$: A out of the page at the top left, B into the page at the top right, C out of the page at the bottom right.
Hint 4/4
The adjacent wires give $1.50\times10^{-5}\ \mathrm{T}$ each and the diagonal one $1.06\times10^{-5}\ \mathrm{T}$; combining the components, not the three magnitudes, gives $1.06\times10^{-5}\ \mathrm{T}$ directed at $45.0^{\circ}$ below the $+x$ axis.
Independent check by symmetry: A and C sit symmetrically about the diagonal through the field point and carry equal currents, so their sum must lie along it, and $(1.50,\,-1.50)$ does; B lies on that same diagonal and contributes along it too.
Full exam-style question
Exam level: a 500 turn solenoid with a wire down its axisexam format
A solenoid $25.0\ \mathrm{cm}$ long is wound with $500$ turns and carries $3.00\ \mathrm{A}$. A long straight wire runs down its axis carrying $15.0\ \mathrm{A}$. Find (a) the field the solenoid alone produces well inside itself, (b) the field the axial wire alone produces at a point $2.00\ \mathrm{cm}$ from the axis, (c) the magnitude and direction of the total field at that point, (d) the percentage by which the wire changes the magnitude found in (a), and (e) with the axial wire switched off and an iron core of relative permeability $800$ inserted, the solenoid current needed to reach $1.20\ \mathrm{T}$ inside.
axial wire: $I_w = 15.0\ \mathrm{A}$, field point $2.00\ \mathrm{cm}$ from the axis
iron core for part (e): $\mu_r = 800$ at the working point, target $B = 1.20\ \mathrm{T}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$ and $\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
the two separate fields, their combination in size and direction, the percentage change, and the current needed with a core
Solution
Parts (a) and (b) use different formulas on purpose and part (c) is the only place they meet. Treating the two fields as parallel and adding them would put part (d) near two per cent instead of two hundredths of one.
the ordinary straight wire result; the solenoid winding around it does not alter the field the wire itself makes
$$\vec B_w \perp \vec B_{\rm sol}$$
the wire's field circles the axis while the solenoid's runs along it, so the two are perpendicular at every point and neither can add to or subtract from the other directly
five times less current than in part (a), for a field a hundred and sixty times larger.
Answer $$\boxed{\begin{aligned} &\text{(a)}\ 7.54\times10^{-3}\ \mathrm{T}\ \text{along the axis} &&\text{(b)}\ 1.50\times10^{-4}\ \mathrm{T}\ \text{around the axis} \\ &\text{(c)}\ 7.54\times10^{-3}\ \mathrm{T}\ \text{at}\ 1.14^{\circ}\ \text{to the axis} &&\text{(d)}\ 0.0198\% \\ &\text{(e)}\ I = 0.597\ \mathrm{A} \end{aligned}}$$
Check
Independent check tying (a) and (e) together, since they were computed in opposite directions: feed the current found in (e) back through the formula of (a) with the core factor in place, $B = \mu_0 n I \mu_r = (4\pi\times10^{-7})(2000)(0.597)(800) = 1.20\ \mathrm{T}$, which is the target, so the division in (e) was done the right way up and not inverted. Plausibility on (c): a tilt of about one degree from a field fifty times weaker than the main one is right, since $\arctan(1/50)$ is about $1.1^{\circ}$.
Five parts, and only three of them needed a calculator. Part (d) is arithmetic that exists to make a point about geometry, and part (e) is one division done backwards.
Part (d) separates marks. Adding a small perpendicular field to a large one is first order in the direction and second order in the magnitude, so questions about tilt are sensitive and questions about size are not.
Practice
A · concept 4 questions
1§11.1 — the midpoint between two equal parallel currents●●○○○
One mark, and the claim sounds like the safest kind of arithmetic. Two long parallel wires a fixed distance apart carry equal currents in the same direction, and someone tells you that the field at the midpoint between them is twice what one wire alone would give there.
Given
two long parallel wires, equal currents
the currents run in the same direction
the field point is the midpoint of the line joining the wires
Find
(a) True or false: the field there is twice one wire's value. Give your reason in one sentence.
Hint 1/4
Two equal magnitudes can add to twice one of them, to zero, or to anything between. Decide the directions before deciding the total.
Hint 2/4
Each field circles its own wire by the grip rule, and superposition adds them as vectors, not as numbers.
Hint 3/4
At the midpoint, the point lies to the right of one wire and to the left of the other, so the same grip rule gives opposite directions there.
Hint 4/4
False: the two contributions are equal and opposite, so the field at the midpoint is exactly zero.
with both currents out of the page, the midpoint lies in the $+x$ direction from one wire and the $-x$ direction from the other, and the same rule then gives opposite answers
$$\vec B = \vec B_1+\vec B_2 = \vec 0$$
equal sizes and opposite directions, which is total cancellation rather than partial
Independent check: move a centimetre towards wire one and its contribution grows while the other shrinks, so the total is no longer zero.
2§11.2 — unequal currents, and who pushes harder●●○○○
One mark, and the intuition it tests is a strong one. Two long parallel wires lie side by side. The left one carries $2.00\ \mathrm{A}$ and the right one carries $8.00\ \mathrm{A}$, both in the same direction.
Given
left wire: $2.00\ \mathrm{A}$
right wire: $8.00\ \mathrm{A}$
same direction, long parallel wires, separation $d$
Find
(a) How do the forces per unit length on the two wires compare?
Hint 1/4
Ask whether the formula you are about to use describes one wire or describes the pair of them together.
Hint 2/4
$F/L = \mu_0 I_1 I_2/(2\pi d)$, which contains the product of the two currents and is symmetric in them.
Hint 3/4
Here $I_1 I_2 = (2.00)(8.00) = 16.0\ \mathrm{A^{2}}$, and swapping which wire is called one and which is called two changes nothing.
Hint 4/4
The forces are equal in size and opposite in direction, attractive because the currents are parallel.
Show solutionThe force on wire two from the field of wire one
Independent check by a thought experiment rather than algebra: if the forces were unequal, the pair of wires taken together would have a net force on itself and would accelerate off the bench with nothing pushing it.
3§11.3 — what a wire outside the loop changes●●●○○
One mark, and it is the distinction the whole law rests on. A closed path is drawn in a region, and a long straight current carrying wire is brought up close to it but kept entirely outside.
Given
a closed path in a region of space
a long straight wire brought close to the path but staying outside it
the wire carries a large steady current
Find
(a) What does bringing that wire up change?
Hint 1/4
Two different quantities are in play, one attached to each point of the path and one attached to the whole path. Ask about each separately.
Hint 2/4
$\oint\vec B\cdot d\vec l = \mu_0 I_{\rm enc}$, so the circulation depends only on the current threading the path, while the field at a point depends on every current anywhere.
Hint 3/4
Here the new wire adds $\mu_0 I/(2\pi r)$ at every point of the path, and it adds nothing to $I_{\rm enc}$, which stays at whatever it was.
Hint 4/4
The field on the path changes and the circulation does not.
the added field contributes zero circulation around a path it does not thread.
Answer $$\boxed{\,\Delta\vec B \neq \vec 0\ \text{at every point},\qquad \Delta\oint\vec B\cdot d\vec l = 0\,}$$
Check
Independent check by pushing the wire to a limit: slide it right up against the path from outside. The two quantities are as independent as the law says.
4§11.7 — the sample that is pushed out of the field●●●○○
One mark on the three classes of magnetic behaviour. Four small samples of equal size are hung one at a time on a thread near the gap of a very strong electromagnet, where the field is strong and getting stronger towards the gap.
Given
samples of iron, aluminium, copper and bismuth, all the same size
a strong and strongly varying magnetic field near the gap of an electromagnet
each sample is hung on a thread and released in turn
Find
(a) Which sample is pushed away from the gap most strongly?
Hint 1/4
Sort the four into classes first, then ask which class is repelled, and only then ask which member of that class is repelled hardest.
Hint 2/4
A positive susceptibility means the sample is drawn into a strong field, and a negative one means it is pushed out; the size of the effect goes with the size of the susceptibility.
Hint 3/4
Iron is ferromagnetic and drawn in hard, aluminium is paramagnetic with $\chi_m$ about $+2\times10^{-5}$, copper is diamagnetic at about $-1\times10^{-5}$, and bismuth is diamagnetic at about $-1.7\times10^{-4}$.
Hint 4/4
Bismuth: the only strongly diamagnetic one of the four, and pushed out about seventeen times harder than copper.
more than a factor of ten, so the comparison is not close and no precise values are needed
Answer $$\boxed{\,\text{bismuth, the strongly diamagnetic sample}\,}$$
Check
Independent check against a well known demonstration: bismuth and pyrolytic graphite are the two everyday materials that can be seen to be repelled by a strong magnet, and a small bismuth block will visibly push away from a rare earth magnet on a balance.
B · computation 7 questions
1§11.1 — reading a current off a field measurement●●○○○
The straight wire result used backwards, which is how a clamp meter works. A magnetometer placed $8.00\ \mathrm{cm}$ from a long straight wire reads $3.00\times10^{-5}\ \mathrm{T}$ for the wire's own contribution, with the Earth's field already subtracted out.
Given
$r = 8.00\times10^{-2}\ \mathrm{m}$
$B = 3.00\times10^{-5}\ \mathrm{T}$ from the wire alone
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the current in the wire.
(b) Find the distance at which that same wire would produce a field equal to the Earth's, taken as $5.00\times10^{-5}\ \mathrm{T}$.
Hint 1/4
Both parts are the same single relation rearranged for a different unknown, so solve it symbolically once before substituting anything.
Hint 2/4
$B = \mu_0 I/(2\pi r)$, so $I = 2\pi r B/\mu_0$ and $r = \mu_0 I/(2\pi B)$.
Hint 3/4
For (a) use $B = 3.00\times10^{-5}\ \mathrm{T}$ at $r = 8.00\times10^{-2}\ \mathrm{m}$; for (b) use the current from (a) with $B = 5.00\times10^{-5}\ \mathrm{T}$.
Hint 4/4
The current is $12.0\ \mathrm{A}$, and it matches the Earth's field at $4.80\ \mathrm{cm}$.
using the current just found, so an error in (a) would propagate here.
Answer $$\boxed{\,I = 12.0\ \mathrm{A},\qquad r = 4.80\ \mathrm{cm}\,}$$
Check
Independent check by proportionality instead of substitution: the field must fall from $5.00\times10^{-5}$ to $3.00\times10^{-5}\ \mathrm{T}$ as $r$ grows from the answer in (b) to $8.00\ \mathrm{cm}$, so the ratio of distances should be $5.00/3.00 = 1.67$, and $8.00/4.80 = 1.67$.
2§11.2 — the pull on a fifty metre span of power line●●●○○
The force between wires at a scale where it matters to an engineer. Two long parallel conductors of an overhead line are $0.300\ \mathrm{m}$ apart and each carries $200\ \mathrm{A}$, in opposite directions since one is the outward conductor and the other the return.
Given
$I_1 = I_2 = 200\ \mathrm{A}$, in opposite directions
$d = 0.300\ \mathrm{m}$
span length $50.0\ \mathrm{m}$
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the force per unit length between them and say whether it is attractive or repulsive.
(b) Find the total force on one $50.0\ \mathrm{m}$ span.
(c) State what the answer to (b) becomes during a fault in which the current briefly reaches $20.0\ \mathrm{kA}$.
Hint 1/4
Part (c) is not a fresh calculation, so notice which power of the current the force carries and scale part (b).
Hint 2/4
$F/L = \mu_0 I_1I_2/(2\pi d)$, and with equal currents this is proportional to $I^{2}$.
Hint 3/4
Here $I = 200\ \mathrm{A}$, $d = 0.300\ \mathrm{m}$ and $L = 50.0\ \mathrm{m}$; the fault current of $20.0\ \mathrm{kA}$ is a hundred times larger.
Hint 4/4
The line gives $2.67\times10^{-2}\ \mathrm{N/m}$ and $1.33\ \mathrm{N}$ on a span, rising to about $13.3\ \mathrm{kN}$ in the fault.
Show solutionForce per unit length, with the sense
Independent check of the ordinary service figure against the definition of the ampere: one ampere in each of two wires one metre apart gives $2\times10^{-7}\ \mathrm{N/m}$. Plausibility: thirteen kilonewtons is the weight of well over a tonne, which is why fault forces, not working forces, size the hardware.
3§11.3 — counting the enclosed current with its sign●●○○○
Pure bookkeeping, and the marks are all in the signs. A closed path is drawn on the page and walked anticlockwise as seen by the reader. Passing through it are wires carrying $7.00\ \mathrm{A}$ out of the page and $4.00\ \mathrm{A}$ into the page. A third wire, carrying $9.00\ \mathrm{A}$ out of the page, runs just outside the path.
Given
through the path: $7.00\ \mathrm{A}$ out of the page and $4.00\ \mathrm{A}$ into the page
outside the path: $9.00\ \mathrm{A}$ out of the page
the path is walked anticlockwise as seen by the reader
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the circulation of the magnetic field around the path.
(b) State what it becomes if the path is walked the other way instead.
(c) State what happens to it if the outside wire's current is doubled.
Hint 1/4
Decide first which currents belong in the sum at all, then give each of the survivors a sign, and only then multiply.
Hint 2/4
$\oint\vec B\cdot d\vec l = \mu_0 I_{\rm enc}$, with the sign of each enclosed current set by curling the right hand along the direction of the walk.
Hint 3/4
Walking anticlockwise makes out of the page positive, so here $I_{\rm enc} = +7.00 - 4.00$, and the $9.00\ \mathrm{A}$ wire is outside.
Hint 4/4
The circulation is $3.77\times10^{-6}\ \mathrm{T\,m}$; reversing the walk changes its sign, and the outside wire never enters.
Show solutionFix the convention, then count
$$\text{anticlockwise} \Rightarrow \text{out of the page is}\ +$$
written down explicitly, because the sign of the final answer is meaningless without it
the zero is the outside wire, written in deliberately so that it is discarded on purpose rather than overlooked
Multiply
$$\oint\vec B\cdot d\vec l = (4\pi\times10^{-7})(3.00) = 3.77\times10^{-6}\ \mathrm{T\,m}$$
the full $\mu_0$, since no circumference is being divided out here
Answer $$\boxed{\,\oint\vec B\cdot d\vec l = 3.77\times10^{-6}\ \mathrm{T\,m}\,}$$
Check
Independent check: replace the two enclosed wires by a single $3.00\ \mathrm{A}$ wire and the law gives $\mu_0(3.00) = 3.77\times10^{-6}\ \mathrm{T\,m}$, the same answer, since only the net enclosed current can matter.
4§11.4 — inside and outside a 5.0 mm conductor●●●○○
The two region formula on a conductor thick enough that the inside really matters. A long straight busbar of circular cross section has radius $5.00\ \mathrm{mm}$ and carries $25.0\ \mathrm{A}$ spread uniformly over that cross section.
Given
$R = 5.00\times10^{-3}\ \mathrm{m}$
$I = 25.0\ \mathrm{A}$, uniform over the cross section
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the field at $r = 2.00\ \mathrm{mm}$ from the axis.
(b) Find the field at $r = 10.0\ \mathrm{mm}$ from the axis.
(c) Find the two radii, one inside the metal and one outside, at which the field is half its maximum value.
Hint 1/4
Classify each radius as inside or outside before choosing a formula, and find the maximum before attempting part (c).
Hint 2/4
$B_{\rm in} = \mu_0 Ir/(2\pi R^{2})$ and $B_{\rm out} = \mu_0 I/(2\pi r)$, with the maximum $\mu_0 I/(2\pi R)$ at the surface.
Hint 3/4
Here $R = 5.00\ \mathrm{mm}$ and $I = 25.0\ \mathrm{A}$, so the surface value is $(2\times10^{-7})(25.0)/(5.00\times10^{-3})$, and the two points asked for are at $2.00\ \mathrm{mm}$ inside and $10.0\ \mathrm{mm}$ outside.
Hint 4/4
The answers are $4.00\times10^{-4}\ \mathrm{T}$, $5.00\times10^{-4}\ \mathrm{T}$, and half the maximum occurs at $2.50\ \mathrm{mm}$ and at $10.0\ \mathrm{mm}$.
Show solutionGet the surface value first, since everything is a fraction of it
Independent check that the two formulas agree at the boundary: the inside expression at $r = R$ gives $(2\times10^{-7})(25.0)(5.00\times10^{-3})/(2.50\times10^{-5}) = 1.00\times10^{-3}\ \mathrm{T}$ and the outside expression at the same place gives $(2\times10^{-7})(25.0)/(5.00\times10^{-3}) = 1.00\times10^{-3}\ \mathrm{T}$.
5§11.5 — designing a solenoid to hit a stated field●●●○○
A design question rather than an evaluation, which is how solenoids usually arrive in an exam. A solenoid is to be wound on a former $32.0\ \mathrm{cm}$ long, using $800$ turns.
Given
$L = 0.320\ \mathrm{m}$, $N = 800$ turns
target field $B = 1.00\times10^{-2}\ \mathrm{T}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the turns per metre and the current required.
(b) The same wire is unwound and rewound onto a former $64.0\ \mathrm{cm}$ long.
(c) State which of the two coils dissipates more heat, given that they use the same wire.
Hint 1/4
You are solving for the current rather than the field, so rearrange first and substitute once.
Hint 2/4
$B = \mu_0 nI$ with $n = N/L$, so $I = B/(\mu_0 n) = BL/(\mu_0 N)$.
Hint 3/4
Here $N = 800$, $L = 0.320\ \mathrm{m}$ then $0.640\ \mathrm{m}$, and $B = 1.00\times10^{-2}\ \mathrm{T}$ in both cases.
Hint 4/4
The short coil needs $3.18\ \mathrm{A}$ and the long one $6.37\ \mathrm{A}$, and the longer coil runs far hotter.
the same wire has the same resistance, so the whole difference is in the current, and it enters squared
Answer $$\boxed{\,n = 2500\ \mathrm{turns/m},\ I = 3.18\ \mathrm{A};\quad I' = 6.37\ \mathrm{A};\quad P' = 4P\,}$$
Check
Independent check on (a) by running it forwards: $\mu_0 nI = (4\pi\times10^{-7})(2500)(3.18) = 1.00\times10^{-2}\ \mathrm{T}$, which is the target. Plausibility: three amperes through a coil of eight hundred turns is an ordinary bench experiment, and ten millitesla is about two hundred times the Earth's field, which is the right order for a coil of this size.
6§11.6 — a quarter circle with two radial leads●●●○○
The arc result on a fraction other than a half. A wire carrying $9.00\ \mathrm{A}$ runs in from a long way off straight towards a point $\mathrm{C}$, stops a distance $6.00\ \mathrm{cm}$ short of it, bends into a quarter circle of that radius centred on $\mathrm{C}$, and then runs straight away from $\mathrm{C}$ to a long way off again.
Given
$I = 9.00\ \mathrm{A}$
quarter circle of radius $R = 6.00\times10^{-2}\ \mathrm{m}$ centred on $\mathrm{C}$
both straight leads lie along lines through $\mathrm{C}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the contribution of the two straight leads to the field at $\mathrm{C}$.
(b) Find the total field at $\mathrm{C}$.
(c) State what the answer would be if the quarter circle were replaced by a complete circle of the same radius.
Hint 1/4
Deal with the straight pieces first and in one line, then treat the arc as a fraction of a full loop rather than as an integral.
Hint 2/4
A current element pointing at the field point contributes nothing, and an arc of angle $\phi$ gives $\mu_0 I\phi/(4\pi R)$, which for a quarter circle is a quarter of $\mu_0 I/(2R)$.
Hint 3/4
Here $I = 9.00\ \mathrm{A}$ and $R = 6.00\times10^{-2}\ \mathrm{m}$, and the arc covers a quarter of a full turn.
Hint 4/4
The leads give nothing, the arc gives $2.36\times10^{-5}\ \mathrm{T}$, and a full circle would give four times that.
Show solutionThe leads
$$d\vec l \parallel \hat r \Rightarrow d\vec B = \vec 0$$
the sine in the cross product is zero at every point of a radial lead.
Independent check through the general arc formula, taken the long way: $B = \mu_0 I\phi/(4\pi R)$ with $\phi = \pi/2$ gives $(4\pi\times10^{-7})(9.00)(\pi/2)/[4\pi(6.00\times10^{-2})] = 2.36\times10^{-5}\ \mathrm{T}$, agreeing with the fraction method. Plausibility: about half the Earth's field, reasonable for nine amperes bent round six centimetres.
7§11.7 — an iron core, and the field it cannot reach●●●○○
The core calculation, followed by the check that stops it being nonsense. A solenoid wound at $1500\ \mathrm{turns/m}$ carries $0.400\ \mathrm{A}$ on a soft iron core whose relative permeability is $600$ at this working point.
Given
$n = 1500\ \mathrm{turns/m}$
$I = 0.400\ \mathrm{A}$
$\mu_r = 600$ at this working point
iron saturates at roughly $1.5$ to $2\ \mathrm{T}$
Find
(a) Find the field the coil alone would produce with no core.
(b) Find the field in the iron.
(c) Find the relative permeability that would be needed to reach $2.50\ \mathrm{T}$ at the same current, and say whether iron can deliver it.
Hint 1/4
Keep the coil and the core in separate steps: one produces a field, the other multiplies it.
Hint 2/4
$B_0 = \mu_0 nI$ and $B = \mu_r B_0$, so $\mu_r = B/B_0$ when you want to work backwards.
Hint 3/4
Here $n = 1500\ \mathrm{turns/m}$, $I = 0.400\ \mathrm{A}$, $\mu_r = 600$ for parts (a) and (b), and the target for (c) is $2.50\ \mathrm{T}$.
Hint 4/4
The empty coil gives $7.54\times10^{-4}\ \mathrm{T}$, the iron gives $0.452\ \mathrm{T}$, and reaching $2.50\ \mathrm{T}$ would need about $3320$, which iron cannot supply.
Show solutionThe empty coil
$$B_0 = \mu_0 nI = (4\pi\times10^{-7})(1500)(0.400) = 7.54\times10^{-4}\ \mathrm{T}$$
computed on its own line so that $\mu_0$ appears exactly once in the whole calculation
the permeability of a ferromagnet collapses as it saturates, so a value measured at a small applied field cannot be carried up to a field the metal cannot produce at all
Independent plausibility check on (b): $0.452\ \mathrm{T}$ is about nine thousand times the Earth's field and roughly what a modest iron cored electromagnet produces, from less than half an ampere.
C · exam level 3 questions
1§11.4 — a coaxial cable, region by region●●●●○
A full exam question on one geometry, worth the marks of four short ones. A coaxial cable has a solid inner conductor of radius $2.00\ \mathrm{mm}$ carrying $8.00\ \mathrm{A}$ out of the page, and an outer sleeve running from $5.00\ \mathrm{mm}$ to $7.00\ \mathrm{mm}$ carrying the same $8.00\ \mathrm{A}$ back into the page. Both currents are spread uniformly over their own cross sections.
Given
inner conductor: radius $2.00\ \mathrm{mm}$, current $8.00\ \mathrm{A}$ out of the page, uniform
outer sleeve: from $5.00\ \mathrm{mm}$ to $7.00\ \mathrm{mm}$, current $8.00\ \mathrm{A}$ into the page, uniform
all radii are measured from the common axis
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the field at $r = 1.00\ \mathrm{mm}$.
(b) Find the field at $r = 3.50\ \mathrm{mm}$.
(c) Find the field at $r = 6.00\ \mathrm{mm}$.
(d) Find the field at $r = 10.0\ \mathrm{mm}$.
(e) State where in the cable the field is largest, and its value there.
Hint 1/4
Four radii, four regions: for each, the only question is how much current a circle of that radius encloses, and the rest is one formula used four times.
Hint 2/4
$B = \mu_0 I_{\rm enc}/(2\pi r)$, with $I_{\rm enc}$ scaled by the area fraction inside a conductor and with the return current subtracting.
Hint 3/4
Inner conductor radius $2.00\ \mathrm{mm}$ carrying $+8.00\ \mathrm{A}$; sleeve from $5.00$ to $7.00\ \mathrm{mm}$ carrying $-8.00\ \mathrm{A}$; the four radii asked for are $1.00$, $3.50$, $6.00$ and $10.0\ \mathrm{mm}$.
Hint 4/4
The four answers are $4.00\times10^{-4}$, $4.57\times10^{-4}$, $1.44\times10^{-4}$ and $0$ tesla, with the maximum $8.00\times10^{-4}\ \mathrm{T}$ at the surface of the inner conductor.
Independent check on continuity: at $r = 2.00\ \mathrm{mm}$ the core and gap formulas both give $8.00\times10^{-4}\ \mathrm{T}$, at $r = 5.00\ \mathrm{mm}$ the gap and sleeve formulas both give $3.20\times10^{-4}\ \mathrm{T}$, and at $r = 7.00\ \mathrm{mm}$ the sleeve gives zero like the outside.
2§11.5 — a toroid, and the solenoid that would replace it●●●●○
An exam question that ends in a judgement rather than a number. A toroid of inner radius $8.00\ \mathrm{cm}$ and outer radius $12.0\ \mathrm{cm}$ is wound with $1200$ turns and carries $4.00\ \mathrm{A}$.
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$ and $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the field at the inner rim, at the mean radius and at the outer rim.
(b) Express the difference between the inner and outer values as a percentage of the value at the mean radius.
(c) Find the turns per metre a long solenoid would need to produce the mean field at the same current.
(d) State which of the two you would choose if the field had to be uniform to better than one per cent, and why.
Hint 1/4
Compute the constant part of the toroid formula once and divide three times, then treat parts (c) and (d) as questions about which formula contains a radius.
Hint 2/4
$B_{\rm toroid} = \mu_0 NI/(2\pi r)$, which depends on $r$, and $B_{\rm solenoid} = \mu_0 nI$, which does not.
Hint 3/4
Here $\mu_0 NI/(2\pi) = (2\times10^{-7})(1200)(4.00) = 9.60\times10^{-4}\ \mathrm{T\,m}$, and the three radii are $0.0800$, $0.100$ and $0.120\ \mathrm{m}$.
Hint 4/4
The three fields are $12.0$, $9.60$ and $8.00\ \mathrm{mT}$, a spread of $41.7\%$; the solenoid would need about $1910\ \mathrm{turns/m}$, and it is the one to choose for uniformity.
the whole recommendation is contained in which of the two formulas has a position in it
Answer $$\boxed{\,1.20\times10^{-2},\ 9.60\times10^{-3},\ 8.00\times10^{-3}\ \mathrm{T};\ 41.7\%;\ n = 1.91\times10^{3}\ \mathrm{turns/m};\ \text{choose the solenoid}\,}$$
Check
Independent check on part (b) without using any of the three field values: the field goes as $1/r$, so calling the outer one $X$ makes the inner one $(0.120/0.0800)X = 1.50X$ and the one at the mean radius $1.20X$, and the spread as a fraction of the mean is then $0.500/1.20 = 0.417$, matching.
3§11.6 — the axis of a loop, at both ends of the scale●●●●○
A derivation question of the kind that carries several marks with no arithmetic in it. Start from the field on the axis of a flat circular coil of $N$ turns and radius $R$ carrying current $I$, at a distance $x$ from its centre along the axis.
Given
$B(x) = \mu_0 NIR^{2}/[2(R^{2}+x^{2})^{3/2}]$ on the axis
the magnetic dipole moment of the coil is $\mu = NIA = NI\pi R^{2}$
$N$, $R$ and $I$ are all fixed
Find
(a) Show that the expression gives the centre of the loop result when $x = 0$.
(b) Show that for $x \gg R$ it becomes $B = \mu_0\mu/(2\pi x^{3})$ with $\mu$ the dipole moment of the coil.
(c) Find the distance along the axis at which the field has fallen to one eighth of its value at the centre.
Hint 1/4
All three parts are the same expression looked at in three ways: at one particular value, in a limit, and solved for a ratio. No new physics is needed.
Hint 2/4
For the limit use $(R^{2}+x^{2})^{3/2} \to x^{3}$ when $x \gg R$, and for part (c) take the ratio $B(x)/B(0) = [R^{2}/(R^{2}+x^{2})]^{3/2}$.
Hint 3/4
For (a) put $x = 0$; for (b) drop $R^{2}$ beside $x^{2}$ and use $\mu = NI\pi R^{2}$; for (c) set the ratio equal to $1/8$.
Hint 4/4
The centre gives $\mu_0 NI/(2R)$, the far field gives $\mu_0\mu/(2\pi x^{3})$, and the field is one eighth of its central value at $x = \sqrt{3}R$.
Independent numerical check of part (c) with the coil from the computation set, $R = 8.00\ \mathrm{cm}$: the prediction is $x = 13.9\ \mathrm{cm}$, and putting that into the full formula gives $[R^{2}/(R^{2}+x^{2})]^{3/2} = [0.00640/0.02560]^{3/2} = (0.250)^{3/2} = 0.125$, exactly one eighth.
D · interleaved 3 questions
1§11.1 — from a battery to a field, in one problem●●●●○
A $12.0\ \mathrm{V}$ battery of internal resistance $0.500\ \Omega$ supplies a $3.50\ \Omega$ heating element through a pair of long parallel leads held $1.50\ \mathrm{cm}$ apart. The leads are much longer than their separation.
Given
battery: $12.0\ \mathrm{V}$ with internal resistance $0.500\ \Omega$
load: $3.50\ \Omega$
the two leads are parallel, $1.50\times10^{-2}\ \mathrm{m}$ apart, and much longer than that
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the current in the leads.
(b) Find the field $4.00\ \mathrm{cm}$ from one lead, at a place where the other lead is far enough away to ignore.
(c) Find the force per unit length between the two leads and say whether they push apart or pull together.
Hint 1/4
Nothing about the field can be started until the current is known, and the current is a question about the circuit rather than about magnetism.
Hint 2/4
$I = \mathcal{E}/(R+r)$ for a single loop with internal resistance, then $B = \mu_0 I/(2\pi d)$ and $F/L = \mu_0 I_1I_2/(2\pi d)$.
Hint 3/4
Here $\mathcal{E} = 12.0\ \mathrm{V}$, $R = 3.50\ \Omega$, $r = 0.500\ \Omega$, the field point is $4.00\ \mathrm{cm}$ from one lead, and the leads are $1.50\ \mathrm{cm}$ apart.
Hint 4/4
The current is $3.00\ \mathrm{A}$, the field is $1.50\times10^{-5}\ \mathrm{T}$, and the leads repel with $1.20\times10^{-4}\ \mathrm{N/m}$.
the separation of the leads is used here, not the distance to the field point from part (b), and mixing the two is the standard way this question is lost
$$\text{out and back} \Rightarrow \text{antiparallel} \Rightarrow \text{repulsion}$$
a supply and return pair always carries opposite currents, so a two lead cable always tries to spread itself apart
Independent check on (a) through the energy rather than the loop rule: the battery delivers $\mathcal{E}I = 36.0\ \mathrm{W}$, of which $I^{2}r = 4.50\ \mathrm{W}$ is lost inside it and $I^{2}R = 31.5\ \mathrm{W}$ reaches the heater, and the two add back to $36.0\ \mathrm{W}$. Plausibility on (b): a third of the Earth's field from three amperes at four centimetres is the right order, since twelve amperes at five centimetres gave about the Earth's field earlier in this section.
2§11.1 — an electron travelling beside a live wire●●●○○
A long straight wire carries $15.0\ \mathrm{A}$. An electron is $3.00\ \mathrm{cm}$ away from it, moving parallel to the wire at $4.00\times10^{6}\ \mathrm{m/s}$, in the same direction as the conventional current.
Given
$I = 15.0\ \mathrm{A}$ in a long straight wire
the electron is $3.00\times10^{-2}\ \mathrm{m}$ from the wire
$v = 4.00\times10^{6}\ \mathrm{m/s}$, parallel to the wire and in the same sense as the current
electron charge magnitude $1.60\times10^{-19}\ \mathrm{C}$
Find
(a) Find the magnetic field at the electron's position.
(b) Find the magnitude of the force on the electron and say whether it is towards or away from the wire.
(c) State what happens to the electron's speed while this force acts.
Hint 1/4
Two separate results are being used in sequence: one from this material to get the field, one from the earlier work to get the force it produces.
Hint 2/4
$B = \mu_0 I/(2\pi r)$ for the field, then $\vec F = q\vec v\times\vec B$ with the sign of the charge applied after the cross product has been pointed.
Hint 3/4
Here $I = 15.0\ \mathrm{A}$ at $r = 3.00\times10^{-2}\ \mathrm{m}$, and the electron moves at $4.00\times10^{6}\ \mathrm{m/s}$ parallel to the current with charge $-1.60\times10^{-19}\ \mathrm{C}$.
Hint 4/4
The field is $1.00\times10^{-4}\ \mathrm{T}$, the force is $6.40\times10^{-17}\ \mathrm{N}$ directed away from the wire, and the speed does not change.
twice the Earth's field, which is the right order for fifteen amperes at three centimetres
The force, by components
$$\hat I = \hat y,\ \hat r = \hat x \Rightarrow \hat B = \hat y\times\hat x = -\hat z$$
putting the wire along $y$ and the electron at $+x$ makes the field at the electron point into the page
$$\vec v\times\vec B = (v\hat y)\times(-B\hat z) = -vB\,\hat x$$
using $\hat y\times\hat z = \hat x$, so the cross product points towards the wire before the charge is considered
$$\vec F = (-e)(-vB\hat x) = +evB\,\hat x = 6.40\times10^{-17}\ \mathrm{N}\ \text{away from the wire}$$
the negative charge reverses the cross product, which is the step that decides the whole answer
The speed
$$\vec F\perp\vec v \Rightarrow P = \vec F\cdot\vec v = 0 \Rightarrow \frac{dK}{dt} = 0$$
a magnetic force never changes a speed, whatever the source of the field
Answer $$\boxed{\,B = 1.00\times10^{-4}\ \mathrm{T},\quad F = 6.40\times10^{-17}\ \mathrm{N}\ \text{away from the wire},\quad v\ \text{unchanged}\,}$$
Check
Independent check of the direction by a completely different argument: an electron moving in the same direction as the conventional current constitutes, by itself, a current in the opposite direction, and antiparallel currents repel. Plausibility on the size: the force is about $10^{13}$ times the electron's weight, which is normal, and it gives an acceleration of $7\times10^{13}\ \mathrm{m/s^{2}}$, ordinary for a particle this light.
3§11.5 — a proton accelerated, then fired across a solenoid●●●●○
A proton starts from rest, is accelerated through a potential difference of $500\ \mathrm{V}$, and then enters the interior of a long solenoid through a small hole, moving perpendicular to the solenoid's axis. The solenoid has $4000\ \mathrm{turns/m}$, carries $2.00\ \mathrm{A}$, and its inside radius is $2.50\ \mathrm{cm}$.
Given
accelerating potential difference $500\ \mathrm{V}$, proton starting from rest
the proton enters perpendicular to the solenoid axis
Find
(a) Find the field inside the solenoid.
(b) Find the speed of the proton as it enters.
(c) Find the radius of the circular path the field would bend it into.
(d) State whether the proton completes a circle inside the solenoid.
Hint 1/4
Three different pieces of the course meet here in a fixed order: the coil gives a field, the accelerating stage gives a speed, and the two together give a radius.
Hint 2/4
$B = \mu_0 nI$, then $qV = \tfrac12 mv^{2}$ so $v = \sqrt{2qV/m}$, then $r = mv/(qB)$.
Hint 3/4
Here $n = 4000\ \mathrm{turns/m}$ and $I = 2.00\ \mathrm{A}$; $V = 500\ \mathrm{V}$ with $q = 1.60\times10^{-19}\ \mathrm{C}$ and $m = 1.67\times10^{-27}\ \mathrm{kg}$; the solenoid's inside radius is $2.50\ \mathrm{cm}$.
Hint 4/4
The field is $1.01\times10^{-2}\ \mathrm{T}$, the speed $3.10\times10^{5}\ \mathrm{m/s}$, the radius $0.321\ \mathrm{m}$, which is far larger than the solenoid, so it does not complete a circle.
Show solutionThe solenoid field
$$B = \mu_0 nI = (4\pi\times10^{-7})(4000)(2.00) = 1.005\times10^{-2}\ \mathrm{T}$$
an extra digit is kept here because it feeds a division two steps later
The speed, from the accelerating stage alone
$$qV = \tfrac12 mv^{2} \Rightarrow v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{1.60\times10^{-16}}{1.67\times10^{-27}}}$$
the magnetic field plays no part in this step, because a magnetic force does no work and cannot change a speed
the comparison is the answer to part (d), and it is worth making explicitly rather than leaving the reader to notice it
Answer $$\boxed{\,B = 1.01\times10^{-2}\ \mathrm{T},\ v = 3.10\times10^{5}\ \mathrm{m/s},\ r = 0.321\ \mathrm{m},\ \text{the path does not fit}\,}$$
Check
Independent check on (c) through the kinetic energy instead of the speed: $r = \sqrt{2mK}/(qB)$ with $K = qV = 8.00\times10^{-17}\ \mathrm{J}$ gives $\sqrt{2(1.67\times10^{-27})(8.00\times10^{-17})} = 5.17\times10^{-22}\ \mathrm{kg\,m/s}$ for the momentum, and dividing by $(1.60\times10^{-19})(1.005\times10^{-2}) = 1.61\times10^{-21}$ gives $0.321\ \mathrm{m}$, the same answer by a route that never computes $v$. Plausibility on (a): ten millitesla from four thousand turns per metre at two amperes is two hundred times the Earth's field, ordinary for a tightly wound solenoid.
Mistake ledger (23 entries)
⚠ Using an inverse square law for a wire
every field met before this one came from point charges and fell off as $1/r^{2}$, and the exponent gets copied across without being checked against the geometry
wrong$$B = \frac{\mu_0 I}{2\pi r^{2}}$$
right$$B = \frac{\mu_0 I}{2\pi r}$$
⚠ Reaching for the wrong grouped constant
two constants live in this material, $\mu_0/2\pi = 2\times10^{-7}$ and $\mu_0/4\pi = 1\times10^{-7}$, they differ by a factor of two, and both look equally plausible written down
the arrows of a charged rod point outwards and the picture is strong enough to survive being told otherwise, especially when only the magnitude is asked for
⚠ Setting a force per unit length equal to a whole force
the formula produces newtons per metre and the balancing quantity is often written as a plain weight, so the two get equated without the length being reconciled
wrong$$\frac{\mu_0 I^{2}}{2\pi d} = mg$$
right$$\frac{\mu_0 I^{2}}{2\pi d} = \lambda g = \frac{m}{L}g$$
⚠ Giving the two wires different forces
when the currents are unequal it feels as though the larger one ought to push harder, and the formula is read as belonging to one wire rather than to the pair
the integral sign suggests a sum of positive things, and the habit from ordinary integrals of positive functions is that a zero total means a zero integrand
wrong$$\oint\vec B\cdot d\vec l = 0 \Rightarrow \vec B = \vec 0$$
right$$\oint\vec B\cdot d\vec l = 0 \ \text{only constrains the signed total around the path}$$
⚠ Counting a current that does not thread the loop
a nearby wire visibly dominates the field on the path, and it feels wrong to discard the largest number in the problem
right$$I_{\rm enc} = \sum_{\text{wires through the loop}} \pm I_i$$
⚠ Adding the enclosed currents without their signs
the sign depends on a walking direction that was chosen silently and then never written down, so there is nothing on the page to check the sign against
⚠ Using the whole current at a point inside the metal
the current in the wire is the number printed in the problem, and the idea that only part of it counts requires noticing that the Amperian circle is inside the conductor
⚠ Putting the total number of turns into the solenoid formula
$N$ is the number printed in the question and $n$ has to be computed, so the eye reaches for the one that is already there
wrong$$B = \mu_0 N I$$
right$$B = \mu_0 n I = \mu_0\frac{N}{L}I$$
⚠ Using the toroid formula as though the field were uniform
a toroid looks like a solenoid bent into a ring, and the solenoid field is uniform, so the property is carried over with the picture
wrong$$B_{\rm toroid} = \mu_0 n I \ \text{everywhere inside}$$
right$$B_{\rm toroid} = \frac{\mu_0 NI}{2\pi r},\qquad \text{different at every radius}$$
⚠ Using the solenoid formula near an end
the formula carries no position in it, so it reads as though it applies everywhere inside the coil
wrong$$B(\text{at the mouth of the coil}) = \mu_0 nI$$
right$$B(\text{at the mouth of the coil}) \approx \tfrac12\mu_0 nI,\qquad B = \mu_0 nI\ \text{only well inside}$$
⚠ Using the straight wire formula at the centre of a loop
both formulas are $\mu_0 I$ divided by a length with a small number in front, and $R$ appears in both, so they are easy to interchange while writing quickly
wrong$$B_{\rm centre} = \frac{\mu_0 I}{2\pi R}$$
right$$B_{\rm centre} = \frac{\mu_0 I}{2R}$$
⚠ Putting degrees into the arc formula
the angle of an arc is naturally spoken of in degrees, and the formula does not look like one that cares
order of magnitude values, for checking rather than for quoting
Check yourself
Close the page and write out from memory: the field of a long straight wire, including which power of the distance appears and how the direction is found; the force per unit length between two parallel currents and which way it acts; the statement of Ampere's law, what the enclosed current means and what zero circulation does and does not imply; the field inside and outside a uniform solid conductor and where the maximum sits; the field inside a solenoid and inside a toroid, and which of the two is uniform; the Biot and Savart law, the field at the centre of a loop and of an arc, and which pieces of a bent wire contribute nothing; and what a core does to the field of a coil for each of the three classes of material.
Write down the field of a long straight wire without hesitating over whether it is $1/r$ or $1/r^{2}$, get its direction with your right hand, and add the fields of two wires at a point that is not on the line joining them?
c-wire-field
State which way two parallel currents push each other, and set up the levitation problem by equating a force per unit length to a weight per unit length rather than to a weight?
c-parallel-wires
Say in one sentence why a nearby wire that misses the loop changes the field on it but not the circulation, and count a set of enclosed currents with the right signs after choosing a walking direction?
c-ampere-law
Produce the field at any radius in a thick wire or a coaxial cable, using an area fraction for the enclosed current, and say where the field is largest without differentiating anything?
c-ampere-apply
Convert a total number of turns into turns per metre before using the solenoid formula, and say what stretching a coil, or bending it into a ring, does to the field inside?
c-solenoid
Look at a bent wire and say immediately which pieces contribute nothing at the field point, then treat the remaining arc as a fraction of a full loop?
c-biot-savart
Compute the field in an iron core, and then reject your own answer if it comes out above about two tesla, explaining what has gone wrong in terms of the permeability rather than the arithmetic?
c-materials
Glossary (22 terms)
permeability of free spaceboşluğun manyetik geçirgenliği
The constant $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$ that fixes how much magnetic field a given current produces. It plays the role in magnetism that the electric constant plays in electrostatics, except that it multiplies rather than divides.
sağ el kuralı
The rule that fixes the direction of the field around a current: point the thumb of the right hand along the current and the fingers curl the way the field circles the wire.
superposition of fieldsalanların üst üste binmesi
The rule that the field produced by several sources at a point is the vector sum of the fields each would produce there alone.
ampereamper
The SI unit of electric current. It is now defined by counting elementary charges per second instead.
circulationdolanım
The closed path integral of a field along its own direction, written with the ring integral sign.
Ampere's lawAmpere yasası
The statement that the circulation of the magnetic field around any closed path equals the permeability of free space times the net current passing through that path, for steady currents.
enclosed currentçevrelenen akım
The net current passing through any surface bounded by a chosen closed path, counted with a sign fixed by curling the right hand along the walking direction.
Amperian loopAmpere çevrimi
The imaginary closed path chosen when applying Ampere's law, of any shape and not a physical object; the skill lies in choosing one on which the field is constant in magnitude and fixed in orientation relative to the path.
coaxial cableeş eksenli kablo
A cable in which a central conductor carries current one way and a surrounding sleeve carries the same current back.
solenoidsolenoid
A wire wound into a long closely spaced helix, inside which the field is uniform, points along the axis, and equals the permeability of free space times the turns per metre times the current.
turns per unit lengthbirim uzunluktaki sarım sayısı
The number of turns of a winding divided by the length over which they are spread, in turns per metre.
toroidtoroid
A solenoid bent round into a closed ring, whose field is confined inside the winding, is zero in the hole and outside, and falls off as one over the distance from the centre of the ring, so it is not uniform.
Biot and Savart lawBiot Savart yasası
The rule giving the small field contributed by a short piece of current carrying wire: proportional to the current and the length of the piece, inversely proportional to the square of the distance, and perpendicular to both the piece and the line to the field point.
current elementakım elemanı
A short piece of a current carrying wire, treated as a vector of length equal to that piece and direction along the current in it.
magnetic dipole momentmanyetik dipol momenti
The quantity equal to the number of turns times the current times the area of a coil, pointing perpendicular to its plane.
relative permeabilitybağıl geçirgenlik
The dimensionless factor by which a material multiplies the field a coil would otherwise produce.
magnetic susceptibilitymanyetik alınganlık
The relative permeability minus one, so a small positive number for paramagnetic materials and a small negative number for diamagnetic ones.
ferromagnetismferromanyetizma
The behaviour of iron, nickel, cobalt and their alloys, in which regions of the material called domains align with an applied field and multiply it by a factor in the hundreds or thousands.
paramanyetizma
A weak strengthening of the field by a material, with a susceptibility of order plus ten to the minus five, so that a sample is drawn very slightly into a strong field.
diyamanyetizma
A weak weakening of the field by a material, with a negative susceptibility, so that a sample is pushed very slightly out of a strong field.
saturationdoyum
The condition of a ferromagnet in which essentially all of its domains are already aligned, so that further increases in the applied field produce almost no further increase in the field inside it.
hysteresishisterezis
The property of a ferromagnet that the field inside it depends not only on the applied field but on the history of what has been applied before, so that raising and lowering the applied field trace different curves.
What comes next
§12 · Electromagnetic Induction and Faraday's Law
Every current on this page was steady, and every field it made sat still.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook, whose treatment of the sources of the magnetic field covers the same ground as this section. Its end of chapter problems on coaxial cables and bent wire geometries run a level harder than the ones here.
Course syllabus, week 11 line The scope of this section comes from the week line, which reads Sources of Magnetic Field, and from nowhere else.
Course syllabus, catalogue description and assessment weights The catalogue description of the course lists magnetic field and Ampere's law among its topics, which is what places this material inside it.
SI definitions and constants The permeability of free space is taken as $4\pi\times10^{-7}\ \mathrm{T\,m/A}$ throughout, which is exact to well beyond the three figures used here even though the SI redefinition of the ampere made it a measured quantity.