"Magnetic Effects of Current and Magnetism" — Practice

Question 1

A wire carries current $I$ along a path made of two long straight tails and a quarter-circle arc of radius $R$ joining them. Find the field at the centre of the arc.

Accepted Answer

Both straight tails run along lines through the centre, so $d\vec l	imes\hat r=0$ and they contribute nothing.

Only the arc contributes, and for an angle $	heta=\pi/2$:

$B=\dfrac{\mu_0I	heta}{4\pi R}=\dfrac{\mu_0I}{8R}$

The direction is perpendicular to the plane of the arc, given by the right-hand rule.

Question 2

A square loop of side $a$ carries current $I$. Find the field at its centre.

Accepted Answer

Each side is at perpendicular distance $a/2$, with $	heta_1=	heta_2=45^{\circ}$:

$B_{side}=\dfrac{\mu_0I}{4\pi\left(a/2ight)}\left(2\sin45^{\circ}ight)=\dfrac{\mu_0I\sqrt2}{2\pi a}$

All four sides give fields in the same direction, so

$B=4	imes\dfrac{\mu_0I\sqrt2}{2\pi a}=\dfrac{2\sqrt2\,\mu_0I}{\pi a}$

Question 3

A cylindrical conductor of radius $R$ carries a uniform current density $J$. A cylindrical hole of radius $R/2$ is bored parallel to the axis, its centre at distance $R/2$ from the axis. Find the field at the centre of the hole.

Accepted Answer

Inside a uniform cylinder, $B=\dfrac{\mu_0Jr}{2}$ at distance $r$ from its own axis.

At the hole's centre, the negative cylinder contributes nothing (its own $r=0$), and the full cylinder gives

$B=\dfrac{\mu_0J}{2}	imes\dfrac{R}{2}=\dfrac{\mu_0JR}{4}$

The general result $\vec B=	frac12\mu_0\vec J	imes\vec d$ shows this same value holds at **every** point of the hole.

Question 4

A wire is bent into a semicircle of radius $R$ and carries current $I$ in a uniform field $B$ perpendicular to the plane of the semicircle. Find the net force on it.

Accepted Answer

The chord joining the ends of a semicircle is the diameter, of length $2R$.

$F = I\left(2Right)B$

directed perpendicular to both the diameter and $\vec B$.

Attempting to integrate around the arc gives the same answer after considerable work, which is exactly what the chord theorem saves.

Question 5

A rectangular loop of sides $a$ and $b$ carries current $I$ and lies in the plane of a long wire carrying $I_0$, with its sides of length $b$ parallel to the wire at distances $d$ and $d+a$. Find the net force on the loop.

Accepted Answer

The near side, at distance $d$, is attracted with $\dfrac{\mu_0I_0Ib}{2\pi d}$.

The far side, at $d+a$, is repelled with $\dfrac{\mu_0I_0Ib}{2\pi\left(d+aight)}$.

$F=\dfrac{\mu_0I_0Ib}{2\pi}\left(\dfrac1d-\dfrac{1}{d+a}ight)=\dfrac{\mu_0I_0Iab}{2\pi d\left(d+aight)}$

towards the wire. The perpendicular sides feel equal and opposite forces that cancel.

Question 6

A circular coil of $50$ turns and radius $0.1$ m carries $2$ A in a uniform field of $0.4$ T, with its plane parallel to the field. Find the magnetic moment, the torque, and the work needed to bring it into alignment.

Accepted Answer

$M=NIA=50	imes2	imes\pi\left(0.1ight)^{2}=3.14$ A m$^{2}$

With the plane parallel to $\vec B$, $\vec M$ is perpendicular to it, so

$	au=MB=3.14	imes0.4=1.26$ N m, the maximum value.

$W=U_{aligned}-U_{perpendicular}=-MB-0=-1.26$ J

The work is negative: the field turns the coil into alignment by itself.

Question 7

An electron enters a field of $2	imes10^{-3}$ T at $60^{\circ}$ to the field lines with speed $3	imes10^{6}$ m s$^{-1}$. Find the radius, period and pitch of its helix, taking $e/m=1.76	imes10^{11}$ C kg$^{-1}$.

Accepted Answer

$v_{\perp}=3	imes10^{6}\sin60^{\circ}=2.60	imes10^{6}$ m s$^{-1}$; $v_{\parallel}=1.5	imes10^{6}$ m s$^{-1}$.

$r=\dfrac{v_{\perp}}{\left(e/might)B}=\dfrac{2.60	imes10^{6}}{\left(1.76	imes10^{11}ight)\left(2	imes10^{-3}ight)}=7.4	imes10^{-3}$ m

$T=\dfrac{2\pi}{\left(1.76	imes10^{11}ight)\left(2	imes10^{-3}ight)}=1.79	imes10^{-8}$ s

Pitch $=v_{\parallel}T=1.5	imes10^{6}	imes1.79	imes10^{-8}=2.7	imes10^{-2}$ m

Question 8

A cyclotron accelerates deuterons in a field of $1.5$ T with dees of radius $0.6$ m. Find the operating frequency and the maximum kinetic energy, taking $q/m=4.8	imes10^{7}$ C kg$^{-1}$ and $q=1.6	imes10^{-19}$ C.

Accepted Answer

$f=\dfrac{qB}{2\pi m}=\dfrac{\left(4.8	imes10^{7}ight)\left(1.5ight)}{2\pi}=1.15	imes10^{7}$ Hz

$K=	frac12\left(\dfrac qmight)qB^{2}R^{2}=	frac12\left(4.8	imes10^{7}ight)\left(1.6	imes10^{-19}ight)\left(2.25ight)\left(0.36ight)$

$K=3.1	imes10^{-12}$ J, about $19$ MeV.

The frequency is fixed because the period does not depend on speed, which is what allows a single oscillator to drive the whole acceleration.

Question 9

A velocity selector uses a field of $0.05$ T crossed with an electric field of $2	imes10^{4}$ V m$^{-1}$. Find the selected speed, and explain why the answer does not depend on the particle's charge or mass.

Accepted Answer

$qE=qvB\qquad\Rightarrow\qquad v=\dfrac{E}{B}$

$v=\dfrac{2	imes10^{4}}{0.05}=4	imes10^{5}$ m s$^{-1}$

The charge cancels from both sides, and mass never entered because neither force depends on it. Only particles at exactly this speed pass undeflected, whatever they are made of.

Question 10

A uniformly charged solid sphere of charge $q$, mass $m$ and radius $R$ spins about a diameter at angular speed $\omega$. Find its magnetic moment and verify the gyromagnetic ratio.

Accepted Answer

$M=	frac15q\omega R^{2}$

$L=I\omega=	frac25mR^{2}\omega$

$\dfrac{M}{L}=\dfrac{q\omega R^{2}/5}{2m\omega R^{2}/5}=\dfrac{q}{2m}$

The same value follows for a ring and for a disc, even though their individual moments differ by factors of two and five.

Question 11

A galvanometer has $150$ turns of area $3	imes10^{-4}$ m$^{2}$ in a radial field of $0.25$ T, with torsional constant $1.5	imes10^{-8}$ N m per degree and coil resistance $80\ \Omega$. Find both sensitivities, and state the effect of doubling the number of turns.

Accepted Answer

$S_I=\dfrac{NAB}{k}=\dfrac{150	imes3	imes10^{-4}	imes0.25}{1.5	imes10^{-8}}=7.5	imes10^{5}$ degrees per ampere

$S_V=\dfrac{S_I}{R}=\dfrac{7.5	imes10^{5}}{80}=9.4	imes10^{3}$ degrees per volt

Doubling $N$ doubles $S_I$, but the coil length and hence $R$ also roughly double, so $S_V$ is essentially unchanged.

Question 12

At a certain place the total magnetic field of the Earth is $5.0	imes10^{-5}$ T and the dip is $60^{\circ}$. Find the horizontal and vertical components, and state what a compass needle does at the magnetic poles.

Accepted Answer

$B_H=B\cos\delta=5.0	imes10^{-5}	imes0.5=2.5	imes10^{-5}$ T

$B_V=B\sin\delta=5.0	imes10^{-5}	imes0.866=4.33	imes10^{-5}$ T

Check: $	an\delta=\dfrac{4.33}{2.5}=1.73=	an60^{\circ}$.

At a magnetic pole the dip is $90^{\circ}$, so $B_H=0$ and a horizontal compass needle has nothing to align with — it is useless there.

Practice — "Magnetic Effects of Current and Magnetism"

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