Magnetic Effects of Current and Magnetism

A wire is bent into an elaborate zig-zag and carries current $I$ from point $A$ to point $B$ through a uniform field $\vec{B}$. What force does it feel?

Exactly the same force as a straight wire from $A$ to $B$ carrying the same current. The shape in between is irrelevant.

The reason is that the force on each element is $Id\vec{l}\times \vec{B}$, and with $\vec{B}$ constant it can be taken outside the sum:

$\vec{F}=I\left(\int d\vec{l}\right)\times \vec{B}=I{\vec{L}}_{AB}\times \vec{B}$

where ${\vec{L}}_{AB}$ is the straight vector from start to finish. Every wiggle cancels against its neighbours.

The corollary is immediate and just as useful: for a closed loop, $\oint d\vec{l}=0$, so a closed loop in a uniform field feels no net force at all — only a torque.

That is the tone of the chapter. Main asks for the field of an infinite wire and the force on a straight rod. Advanced asks for the field of a finite arc, the field inside a bored-out cylinder, the torque on a rotating charge distribution, and the pitch of a helix.

1. Biot-Savart by integration

The law itself is

$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2}$

and the four results that follow cover most of what Advanced asks.

A finite straight wire at perpendicular distance $r$, with its ends subtending ${\theta }_1$ and ${\theta }_2$ at the foot of the perpendicular:

$B=\frac{{\mu }_{0}I}{4\pi r}\left(\sin {\theta }_1+\sin {\theta }_2\right)$

Both angles at $90^{\circ }$ gives the infinite wire, ${\mu }_{0}I/2\pi r$. An arc of angle $\theta$ at its centre gives $B=\frac{{\mu }_{0}I\theta }{4\pi R}$, the full ring being the case $\theta =2\pi$. A ring on its axis gives $\frac{{\mu }_{0}I{R}^2}{2{\left(R^2+x^2\right)}^{3/2}}$. A long solenoid has ${\mu }_{0}nI$ inside and exactly half that at each end.

Illustration 1

A wire is bent into a semicircle of radius $R$ with two long straight tails along the same diameter. Find the field at the centre.

The straight portions lie along the line through the centre, so $d\vec{l}\times \hat{r}=0$ for every element: they contribute nothing.

The semicircle contributes $\frac{{\mu }_{0}I\pi }{4\pi R}=\frac{{\mu }_{0}I}{4R}$.

Half a ring gives half a ring's field, which is obvious once the tails are dismissed — and dismissing them correctly is the whole test.

Illustration 2

Find the field at the end of a long solenoid, and explain the factor of one half.

At the middle, contributions arrive from turns on both sides and $B={\mu }_{0}nI$. At the end, only one side exists, so

$B_{end}=\frac{1}{2}{\mu }_{0}nI$

The general result is $B=\frac{1}{2}{\mu }_{0}nI\left(\cos {\alpha }_1-\cos {\alpha }_2\right)$, and setting one angle to $90^{\circ }$ recovers the end value. The half is not a fudge; it is the limit of the exact expression.

2. Ampère's law past the standard cases

Ampère's law is $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{enc}$, and it is useful whenever symmetry keeps $B$ constant along a chosen loop.

A thick wire with non-uniform current density still has cylindrical symmetry, so only $I_{enc}$ changes:

$I_{enc}={\int }_0^{r}J\left(r^{\prime }\right)2\pi r^{\prime }d{r}^{\prime },B=\frac{{\mu }_0{I}_{enc}}{2\pi r}$

A cylindrical cavity bored off-axis in a uniformly current-carrying cylinder gives a uniform field inside:

${\vec{B}}_{cavity}=\frac{{\mu }_0}{2}\vec{J}\times \vec{d}$

the exact magnetic twin of the electrostatic and gravitational cavity results, and for the same reason: the field inside a uniform cylinder is linear in the position vector, so superposing a negative current density cancels the position dependence.

Two further statements complete the toolkit. A moving point charge produces

$\vec{B}=\frac{{\mu }_0}{4\pi }\frac{q\vec{v}\times \hat{r}}{r^2}$

and Gauss's law for magnetism, $\oint \vec{B}\cdot d\vec{A}=0$, says there are no magnetic monopoles: every field line that leaves a region comes back, so magnetic flux through any closed surface vanishes.

Illustration 3

A cylinder of radius $R$ carries current with density $J=J_0\frac{r}{R}$. Find $B$ inside and check it against the outside field at $r=R$.

$I_{enc}={\int }_0^r\frac{J_0{r}^{\prime }}{R}2\pi r^{\prime }d{r}^{\prime }=\frac{2\pi J_0{r}^3}{3R}$

$B_{in}=\frac{{\mu }_0{J}_0{r}^2}{3R}$, growing as $r^2$ rather than linearly.

Total current is $\frac{2\pi J_0{R}^2}{3}$, so $B_{out}=\frac{{\mu }_0{J}_0{R}^2}{3r}$, and at $r=R$ both give $\frac{{\mu }_0{J}_{0}R}{3}$.

The continuity check costs one line and catches most algebra errors, because a discontinuity in $B$ would require a surface current that the problem never mentioned.

3. Forces on wires, and torque on loops

The shape-independence result from the hook is the first tool. The second is that a closed loop is a magnetic dipole of moment $\vec{M}=I\vec{A}$, so in a uniform field

$\vec{\tau }=\vec{M}\times \vec{B},U=-\vec{M}\cdot \vec{B}$

and in a non-uniform field there is a net force $F=M\frac{dB}{dx}$ as well. Two parallel wires carry the familiar force per unit length $\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$, attracting when the currents are parallel.

Illustration 4

A square loop of side $a$ carrying current $I$ lies in the plane of a long straight wire carrying $I_0$, with its nearest side at distance $d$ and parallel to the wire. Find the net force on the loop.

The two sides parallel to the wire feel opposite forces, at distances $d$ and $d+a$:

$F=\frac{{\mu }_0{I}_{0}Ia}{2\pi }\left(\frac{1}{d}-\frac{1}{d+a}\right)=\frac{{\mu }_0{I}_{0}I{a}^2}{2\pi d\left(d+a\right)}$

The two perpendicular sides feel equal and opposite forces that cancel.

The net force is towards the wire when the near side's current is parallel to $I_0$. It is non-zero only because the field is non-uniform, exactly as the closed-loop theorem requires.

Illustration 5

A circular loop of radius $R$ carrying $I$ is placed in a uniform field $B$ with its plane parallel to $\vec{B}$. Find the torque and the work needed to rotate it until its plane is perpendicular to $\vec{B}$.

$M=I\pi R^2$, and with the plane parallel to $\vec{B}$, $\vec{M}$ is perpendicular to $\vec{B}$:

$\tau =MB=\pi R^{2}IB$, the maximum possible.

$W=U_{final}-U_{initial}=-MB-0=-\pi R^{2}IB$

The work is negative, meaning the field does the work: the loop turns itself into alignment, which is precisely how an electric motor's rotor is driven.

4. Charged particles: helices, cyclotrons and selectors

Resolve the velocity into components along and across $\vec{B}$. The parallel component is untouched; the perpendicular component circles. The result is a helix:

$r=\frac{m{v}_{\perp }}{qB},T=\frac{2\pi m}{qB},\text{pitch}=v_{\parallel }T=\frac{2\pi m{v}_{\parallel }}{qB}$

The period is independent of speed, which is what makes a cyclotron possible: the accelerating voltage can be reversed at a fixed frequency $f=qB/2\pi m$ however fast the particle has become. Its exit energy is

$K_{max}=\frac{q^2{B}^2{R}^2}{2m}$

A velocity selector crosses $\vec{E}$ and $\vec{B}$ so that only particles with $v=E/B$ pass undeflected, whatever their charge or mass.

Illustration 6

A proton enters a field of $0.5$ T at $30^{\circ }$ to the field lines with speed $4\times 10^6$ m s${}^{-1}$. Find the radius and pitch of its helix. Take $m/q=1.04\times 10^{-8}$ kg C${}^{-1}$.

$v_{\perp }=4\times 10^6\sin 30^{\circ }=2\times 10^6$ m s${}^{-1}$, and $v_{\parallel }=3.46\times 10^6$ m s${}^{-1}$.

$r=\frac{m{v}_{\perp }}{qB}=\frac{\left(1.04\times 10^{-8}\right)\left(2\times 10^6\right)}{0.5}=0.042$ m

$T=\frac{2\pi \left(1.04\times 10^{-8}\right)}{0.5}=1.31\times 10^{-7}$ s, so pitch $=3.46\times 10^6\times 1.31\times 10^{-7}=0.45$ m

The pitch is ten times the radius here, because the particle entered close to the field direction. A particle entering perpendicular has zero pitch and simply circles.

Illustration 7

A cyclotron with dees of radius $0.5$ m operates at $1.2$ T. Find the operating frequency and the exit energy for a proton, with $q/m=9.6\times 10^7$ C kg${}^{-1}$.

$f=\frac{qB}{2\pi m}=\frac{\left(9.6\times 10^7\right)\left(1.2\right)}{2\pi }=1.83\times 10^7$ Hz

$K=\frac{1}{2}m{\left(\frac{qBR}{m}\right)}^2=\frac{1}{2}\left(\frac{q}{m}\right)q{B}^2{R}^2$, giving

$K=\frac{1}{2}\left(9.6\times 10^7\right)\left(1.6\times 10^{-19}\right)\left(1.44\right)\left(0.25\right)=2.8\times 10^{-12}$ J, about $17$ MeV.

The energy limit is set by the dee radius, not by the voltage. Raising the accelerating voltage only reduces the number of turns needed to reach the same exit energy.

5. Magnetic moment and the gyromagnetic ratio

Any rotating charge distribution is a current loop. Integrating over the rings that make it up gives

$M_{ring}=\frac{1}{2}q\omega R^2,M_{disc}=\frac{1}{4}q\omega R^2,M_{sphere}=\frac{1}{5}q\omega R^2$

for a uniformly charged ring, disc and solid sphere. But the striking result is what happens when the moment is compared with the angular momentum:

$\frac{M}{L}=\frac{q}{2m}$

for every one of them. The geometry cancels completely, provided the charge and mass are distributed in the same way. This ratio is why atomic magnetism is expressed in Bohr magnetons and why the electron's anomalous value of $g\approx 2$ was such a decisive clue.

Illustration 8

A disc of mass $m$, charge $q$ and radius $R$ spins about its axis at $\omega$. Verify the gyromagnetic ratio.

$M=\frac{1}{4}q\omega R^2$ and $L=I\omega =\frac{1}{2}m{R}^2\omega$

$\frac{M}{L}=\frac{q\omega R^2/4}{m\omega R^2/2}=\frac{q}{2m}$

Neither $R$ nor $\omega$ survives. Repeating with a ring gives $\frac{q\omega R^2/2}{m{R}^2\omega }$, the same value — which is the general theorem in two examples.

Illustration 9

A bar magnet of moment $M$ is placed at $60^{\circ }$ to a uniform field $B$. Find the torque, the potential energy, and the work needed to reverse it completely.

$\tau =MB\sin 60^{\circ }=0.866MB$

$U=-MB\cos 60^{\circ }=-0.5MB$

To reverse to $180^{\circ }$: $W=\left(+MB\right)-\left(-0.5MB\right)=1.5MB$

Full reversal from perfect alignment would cost $2MB$, the largest work any orientation change can require.

6. The moving-coil galvanometer

A rectangular coil of $N$ turns and area $A$ hangs in a magnetic field between shaped pole pieces, with a soft-iron cylinder at its centre. The purpose of that geometry is to make the field radial, so that the plane of the coil always contains $\vec{B}$ and the torque is

$\tau =NIAB$

with no $\sin \theta$ factor to spoil it. Balanced against a restoring couple $k\varphi$ from the suspension,

$I=\frac{k}{NAB}\varphi$

so the scale is linear — which a uniform field would not give.

Two sensitivities follow, and the distinction between them is a favourite:

$S_I=\frac{\varphi }{I}=\frac{NAB}{k},S_V=\frac{\varphi }{V}=\frac{NAB}{kR}$

Adding turns raises $N$ but also raises the coil's own resistance $R$, so current sensitivity always improves while voltage sensitivity may not.

Illustration 10

A galvanometer has $200$ turns of area $2\times 10^{-4}$ m${}^2$ in a field of $0.2$ T, with a torsional constant of $10^{-8}$ N m per degree and coil resistance $50\Omega$. Find both sensitivities.

$S_I=\frac{NAB}{k}=\frac{200\times 2\times 10^{-4}\times 0.2}{10^{-8}}=8\times 10^5$ degrees per ampere

$S_V=\frac{S_I}{R}=\frac{8\times 10^5}{50}=1.6\times 10^4$ degrees per volt

Doubling the number of turns doubles $S_I$, but it also roughly doubles $R$, so $S_V$ barely moves. Improving voltage sensitivity needs a weaker suspension or a stronger field instead.

7. Magnetism of materials

Inside matter the field has two sources — the free currents you control and the magnetisation of the material:

$\vec{B}={\mu }_0\left(\vec{H}+\vec{M}\right),\vec{M}=\chi \vec{H},{\mu }_r=1+\chi$

Diamagnets have small negative $\chi$ and are weakly repelled; paramagnets have small positive $\chi$ obeying Curie's law $\chi =C/T$; ferromagnets have $\chi$ in the thousands, retain magnetisation, and lose it above the Curie temperature.

The area of the hysteresis loop is the energy dissipated per unit volume per cycle. Permanent magnets want a wide loop; transformer cores want a narrow one, which is why the two use quite different alloys.

Illustration 11

A paramagnetic sample has susceptibility $2.4\times 10^{-4}$ at $300$ K. Find it at $150$ K and at $600$ K.

Curie's law gives $\chi \propto 1/T$:

At $150$ K: $\chi =4.8\times 10^{-4}$

At $600$ K: $\chi =1.2\times 10^{-4}$

Cooling helps because thermal agitation is what randomises the dipoles. A diamagnet, whose response is induced rather than aligned, shows almost no temperature dependence at all.

8. The Earth's field, and measuring with it

The Earth's field at a place is described by three elements: the declination (the angle between geographic and magnetic north), the dip $\delta$ (the angle below the horizontal), and the horizontal component $B_H$. They are related by

$\tan \delta =\frac{B_V}{B_H},B=\sqrt{B_H^2+B_V^2}$

A tangent galvanometer uses this directly. Its coil is set in the magnetic meridian so that its own field is perpendicular to $B_H$, and the needle settles where

$B_{coil}=B_H\tan \theta \Rightarrow I=\frac{2R{B}_H\tan \theta }{{\mu }_{0}N}$

Illustration 12

At a place the dip is $45^{\circ }$ and the horizontal component is $3.2\times 10^{-5}$ T. Find the total field and the vertical component.

$B_V=B_H\tan 45^{\circ }=3.2\times 10^{-5}$ T

$B=\sqrt{2}\times 3.2\times 10^{-5}=4.5\times 10^{-5}$ T

At the magnetic equator the dip is zero and the field is entirely horizontal; at the poles it is $90^{\circ }$ and a compass needle stands on end, which is why compasses are useless at high latitudes.

Illustration 13

A tangent galvanometer of $50$ turns and radius $0.1$ m gives a deflection of $45^{\circ }$. Find the current, taking $B_H=3.2\times 10^{-5}$ T.

$I=\frac{2R{B}_H\tan \theta }{{\mu }_{0}N}=\frac{2\left(0.1\right)\left(3.2\times 10^{-5}\right)\left(1\right)}{\left(4\pi \times 10^{-7}\right)\left(50\right)}$

$I=\frac{6.4\times 10^{-6}}{6.28\times 10^{-5}}=0.102$ A

Sensitivity is greatest near $45^{\circ }$, because $\tan \theta$ changes fastest per degree there — which is why the instrument is always adjusted to deflect near that value.

Summary

• A wire of any shape in a uniform field feels the force of the straight chord joining its ends; a closed loop feels zero net force, only torque.
• Finite wire: $B=\frac{{\mu }_{0}I}{4\pi r}\left(\sin {\theta }_1+\sin {\theta }_2\right)$; arc: $\frac{{\mu }_{0}I\theta }{4\pi R}$; ring axis: $\frac{{\mu }_{0}I{R}^2}{2{\left(R^2+x^2\right)}^{3/2}}$.
• Straight sections through the field point contribute nothing, since $d\vec{l}\times \hat{r}=0$.
• Solenoid: ${\mu }_{0}nI$ inside, exactly half that at each end.
• Non-uniform $J(r)$ keeps cylindrical symmetry; only $I_{enc}$ changes, and $B$ must be continuous at the surface.
• Cavity in a current-carrying cylinder: $\vec{B}=\frac{{\mu }_0}{2}\vec{J}\times \vec{d}$, uniform throughout the cavity.
• Loop as dipole: $\vec{M}=I\vec{A}$, $\vec{\tau }=\vec{M}\times \vec{B}$, $U=-\vec{M}\cdot \vec{B}$, and $F=MdB/dx$ in a gradient.
• Parallel wires: $\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$ per unit length, attracting for parallel currents.
• Helix: $r=\frac{m{v}_{\perp }}{qB}$, $T=\frac{2\pi m}{qB}$ independent of speed, pitch $=v_{\parallel }T$.
• Cyclotron: $f=\frac{qB}{2\pi m}$, $K_{max}=\frac{q^2{B}^2{R}^2}{2m}$; velocity selector passes $v=E/B$.
• Rotating charge: $M=\frac{1}{2}q\omega R^2$ (ring), $\frac{1}{4}q\omega R^2$ (disc), $\frac{1}{5}q\omega R^2$ (sphere) — but always $\frac{M}{L}=\frac{q}{2m}$.
• Galvanometer: the radial field removes $\sin \theta$, giving $\tau =NIAB$ and a linear scale; $S_I=NAB/k$, $S_V=S_I/R$.
• $\oint \vec{B}\cdot d\vec{A}=0$ — no magnetic monopoles; a moving charge gives $\vec{B}=\frac{{\mu }_0}{4\pi }\frac{q\vec{v}\times \hat{r}}{r^2}$.
• $\vec{B}={\mu }_0\left(\vec{H}+\vec{M}\right)$, ${\mu }_r=1+\chi$; paramagnets obey $\chi =C/T$, ferromagnets lose order above the Curie point.
• Hysteresis loop area is the energy lost per cycle: wide for permanent magnets, narrow for transformer cores.
• Earth: $\tan \delta =B_V/B_H$; a tangent galvanometer reads $I=\frac{2R{B}_H\tan \theta }{{\mu }_{0}N}$, most sensitive near $45^{\circ }$.