Magnetic Effects of Current and Magnetism — Practice

Question 1

Why is the net magnetic flux through any closed surface always zero?

Accepted Answer

Because there are no magnetic monopoles, so no closed surface can enclose a net magnetic charge. Every field line that enters must leave again, since lines have nowhere to terminate. This is exactly why magnetic field lines are always closed loops, unlike electric field lines which begin and end on charges.

Question 2

Can a uniform magnetic field increase the kinetic energy of a charged particle?

Accepted Answer

No. The magnetic force is always perpendicular to the velocity, so it does no work at any instant and the speed is constant. It changes the direction of motion only, which is why the resulting path is circular or helical rather than accelerating.

Question 3

Two long parallel wires carry currents in the same direction. Do they attract or repel, and how does this compare with two like charges?

Accepted Answer

They attract. This is the reverse of the electrostatic case, where two like charges repel, and it is worth memorising precisely because the analogy misleads — there is no magnetic charge for it to rest on. This attraction is what defines the ampere.

Question 4

A wire has two radial straight segments and a $60^\circ$ arc of radius $R$ carrying current $I$. Find the field at the centre.

Accepted Answer

The radial segments have $d\vec{l}	imes\hat{r} = 0$, so they contribute nothing. The arc gives $\mu_0 I	heta/4\pi R$ with $	heta = \pi/3$, which simplifies to $B = \mu_0 I/12R$. Recognising that radial segments contribute nothing turns this into a ten-second problem.

Question 5

A proton and an alpha particle enter the same field perpendicular to it at the same speed. Compare their radii and periods, then compare radii at equal kinetic energy instead.

Accepted Answer

Both $r = mv/qB$ and $T = 2\pi m/qB$ depend on $m/q$, which doubles for the alpha particle, so both are twice the proton's — and neither period depends on speed. At equal kinetic energy, $r = \sqrt{2mK}/qB$, giving a ratio $\sqrt{4}/2 = 1$, so the radii are equal instead.

Question 6

Two long parallel wires 10 cm apart carry 5 A and 8 A in the same direction. Find the force per metre.

Accepted Answer

Using $F/L = \mu_0 I_1I_2/2\pi d$, the $\pi$ terms cancel to leave $(2	imes10^{-7})(40)/0.1 = 8	imes10^{-5}$ N m$^{-1}$. Since the currents are parallel, the force is attractive.

Question 7

A galvanometer of resistance 50 $\Omega$ gives full-scale deflection at 2 mA. Convert it to a 5 A ammeter and to a 10 V voltmeter.

Accepted Answer

Ammeter: the shunt is $S = I_gG/(I-I_g) = 0.1/4.998 \approx 0.02\ \Omega$ in parallel, making the instrument low-resistance. Voltmeter: the series resistance is $V/I_g - G = 5000 - 50 = 4950\ \Omega$, making it high-resistance. Each conversion pushes the meter toward its ideal form.

Question 8

An electron enters a 0.01 T field at $30^\circ$ to it with speed $4	imes10^{6}$ m s$^{-1}$. Find the radius and pitch, taking $m/e = 5.7	imes10^{-12}$ kg C$^{-1}$.

Accepted Answer

The perpendicular component $v\sin30^\circ = 2	imes10^{6}$ sets the radius at $r = mv_\perp/eB = 1.14$ mm, while the parallel component $3.46	imes10^{6}$ carries it along the axis. The period $2\pi m/eB = 3.58$ ns, so the pitch is $v_\parallel T \approx 1.24$ cm, roughly eleven times the radius. The two components never interact.

Question 9

A 100-turn coil of radius 5 cm carries 2 A in a 0.4 T field. Find the maximum torque and the work to rotate it from aligned to antiparallel.

Accepted Answer

$m = NIA = 1.57$ A m$^{2}$, so the maximum torque $mB = 0.628$ N m. The energy $U = -mB\cos	heta$ runs from $-0.628$ J to $+0.628$ J, so the work is 1.256 J. Note that torque peaks at $90^\circ$ where the energy is zero, while energy peaks at $180^\circ$ where the torque is zero.

Question 10

Explain why increasing the number of turns in a galvanometer coil may fail to improve its voltage sensitivity.

Accepted Answer

Current sensitivity $NAB/k$ rises directly with $N$. Voltage sensitivity is $NAB/kR$, but adding turns also lengthens the wire and raises the coil's resistance $R$ roughly in proportion to $N$. The two effects cancel, leaving voltage sensitivity essentially unchanged — so the two sensitivities cannot be improved independently by winding more turns.

Question 11

A galvanometer of resistance 100 Ω gives full-scale deflection at 2 mA. Find the shunt needed to convert it into an ammeter reading to 1 A, and the series resistance needed to convert it into a voltmeter reading to 5 V.

Accepted Answer

For the ammeter, $S = I_gG/(I-I_g) = (0.002)(100)/(1-0.002) = 0.200$ Ω. For the voltmeter, $R = V/I_g - G = 5/0.002 - 100 = 2400$ Ω. The two differ by four orders of magnitude, and in the required directions: an ammeter must present almost no resistance to the branch it joins, a voltmeter almost infinite resistance to the element it straddles.

Question 12

A wire is bent into a square of side $a$ and carries current $I$. Find the magnetic field at the centre of the square.

Accepted Answer

Each side is a finite straight wire at perpendicular distance $a/2$ from the centre, subtending 45 degrees on each side, so it contributes $B = \dfrac{\mu_0 I}{4\pi(a/2)}(2\sin 45^\circ) = \dfrac{\sqrt{2}\mu_0 I}{2\pi a}$. All four contributions point the same way, so the total is $B = \dfrac{2\sqrt{2}\mu_0 I}{\pi a}$. As a check, this exceeds the field at the centre of the inscribed circle of radius $a/2$, which is $\mu_0 I/a$, as it must, since the square's sides lie closer to the centre on average than that circle does.

Practice — Magnetic Effects of Current and Magnetism

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