"Electromagnetic Induction and Alternating Currents" — Practice

Question 1

A search coil of $500$ turns and area $4	imes10^{-4}$ m$^{2}$, with total circuit resistance $50\ \Omega$, is snatched out of a field of $0.8$ T. Find the charge that flows, and state how the answer changes if the snatch takes ten times as long.

Accepted Answer

$\Delta\Phi=BA=0.8	imes4	imes10^{-4}=3.2	imes10^{-4}$ Wb

$q=\dfrac{N\Delta\Phi}{R}=\dfrac{500	imes3.2	imes10^{-4}}{50}=3.2	imes10^{-3}$ C

A ten times slower snatch gives **exactly the same charge**. The current is ten times smaller and lasts ten times longer.

Question 2

A rod of length $L$ rotates at angular speed $\omega$ about an axis through its centre, perpendicular to a uniform field $B$. Find the emf between the centre and each end, and between the two ends.

Accepted Answer

Each half of length $L/2$ generates $\varepsilon=	frac12B\omega\left(\dfrac L2ight)^{2}=\dfrac{B\omega L^{2}}{8}$ between the centre and its end.

Both ends are at the **same** potential relative to the centre, because both halves sweep outward in the same sense.

The emf between the two ends is therefore **zero**, even though each end differs from the centre by $B\omega L^{2}/8$.

Question 3

A rod of mass $0.2$ kg and length $0.5$ m slides on frictionless horizontal rails in a vertical field of $0.4$ T, closed by a $2\ \Omega$ resistance. It is given a speed of $5$ m s$^{-1}$. Find the initial current, the initial retarding force and the total distance travelled.

Accepted Answer

$i=\dfrac{BLv}{R}=\dfrac{0.4	imes0.5	imes5}{2}=0.5$ A

$F=BiL=0.4	imes0.5	imes0.5=0.1$ N

$x=\dfrac{mv_0R}{B^{2}L^{2}}=\dfrac{0.2	imes5	imes2}{\left(0.16ight)\left(0.25ight)}=50$ m

The rod never formally stops, but the total distance converges because the velocity decays exponentially.

Question 4

Two coils have self-inductances $9$ mH and $4$ mH with a coupling coefficient of $0.8$. Find the mutual inductance and both series combinations.

Accepted Answer

$M=0.8\sqrt{9	imes4}=0.8	imes6=4.8$ mH

Series aiding: $L=9+4+2\left(4.8ight)=22.6$ mH

Series opposing: $L=9+4-2\left(4.8ight)=3.4$ mH

Both are positive, as they must be; a coupling coefficient above $1$ would give a negative inductance and is physically impossible.

Question 5

A coil of inductance $0.4$ H and resistance $8\ \Omega$ is connected to a $24$ V battery. Find the current after one time constant, and the energy stored in the steady state.

Accepted Answer

Final current $=\dfrac{24}{8}=3$ A, and $	au=\dfrac{0.4}{8}=0.05$ s.

After one time constant: $i=3\left(1-e^{-1}ight)=3	imes0.632=1.90$ A

$U=	frac12LI^{2}=	frac12\left(0.4ight)(9)=1.8$ J

At $t=0$ the inductor behaves as an open circuit, so no current flows at all in the first instant.

Question 6

A $5\ \mu$F capacitor charged to $100$ V is connected across a $20$ mH inductor of negligible resistance. Find the oscillation frequency, the maximum current, and the time at which the energy is equally shared.

Accepted Answer

$\omega=\dfrac{1}{\sqrt{\left(0.02ight)\left(5	imes10^{-6}ight)}}=\dfrac{1}{10^{-3.5}}=3162$ rad s$^{-1}$, so $f=503$ Hz.

$i_{max}=V\sqrt{\dfrac CL}=100\sqrt{\dfrac{5	imes10^{-6}}{0.02}}=1.58$ A

Energy is shared equally when $q=q_0/\sqrt2$, that is $\cos\omega t=1/\sqrt2$, giving $\omega t=\pi/4$:

$t=\dfrac{\pi}{4	imes3162}=2.5	imes10^{-4}$ s

Question 7

A series circuit of $R=40\ \Omega$, $L=0.2$ H and $C=30\ \mu$F is connected to $220$ V at $50$ Hz. Find the impedance, the current, the phase angle and whether the current leads or lags.

Accepted Answer

$X_L=2\pi\left(50ight)\left(0.2ight)=62.8\ \Omega$; $X_C=\dfrac{1}{2\pi\left(50ight)\left(3	imes10^{-5}ight)}=106.1\ \Omega$

$Z=\sqrt{1600+\left(-43.3ight)^{2}}=\sqrt{1600+1875}=58.9\ \Omega$

$I=\dfrac{220}{58.9}=3.73$ A

$	an\phi=\dfrac{-43.3}{40}=-1.083$, so $\phi=-47.3^{\circ}$: the circuit is capacitive and the current **leads**.

Question 8

In a series LCR circuit the voltages measured across $R$, $L$ and $C$ are $60$ V, $150$ V and $110$ V. Find the applied voltage and the power factor.

Accepted Answer

Net reactive voltage $=150-110=40$ V.

$V=\sqrt{60^{2}+40^{2}}=\sqrt{3600+1600}=72.1$ V

$\cos\phi=\dfrac{V_R}{V}=\dfrac{60}{72.1}=0.832$

Note that $V_L$ alone is more than twice the applied voltage, which the phasor triangle permits because $V_L$ and $V_C$ largely cancel.

Question 9

A series circuit with $R=5\ \Omega$ and $L=0.1$ H is to resonate at $200$ Hz. Find the required capacitance, the quality factor and the bandwidth.

Accepted Answer

$\omega_0=2\pi\left(200ight)=1257$ rad s$^{-1}$

$C=\dfrac{1}{\omega_0^{2}L}=\dfrac{1}{\left(1.58	imes10^{6}ight)\left(0.1ight)}=6.33	imes10^{-6}$ F

$Q=\dfrac{\omega_0L}{R}=\dfrac{1257	imes0.1}{5}=25.1$

$\Delta\omega=\dfrac RL=50$ rad s$^{-1}$, and $\dfrac{\omega_0}{\Delta\omega}=25.1$, confirming $Q$.

Question 10

An inductive load draws $8$ A at $240$ V with a power factor of $0.75$. Find the real power, the apparent power and the phase angle.

Accepted Answer

Apparent power $=240	imes8=1920$ VA

Real power $=1920	imes0.75=1440$ W

$\cos\phi=0.75\qquad\Rightarrow\qquad \phi=41.4^{\circ}$, with the current lagging.

The difference of $480$ VA is reactive: it flows back and forth without ever being consumed, yet it still heats the supply cables.

Question 11

A transformer steps $240$ V down to $12$ V and supplies $5$ A to a load, with an efficiency of $90\%$. Find the turns ratio, the primary current, and the power lost.

Accepted Answer

$\dfrac{N_s}{N_p}=\dfrac{12}{240}=\dfrac{1}{20}$

Output power $=12	imes5=60$ W, so input power $=\dfrac{60}{0.9}=66.7$ W.

$I_p=\dfrac{66.7}{240}=0.278$ A

Power lost $=6.7$ W, in copper resistance, hysteresis and eddy currents.

Question 12

A magnet is dropped down a long vertical copper pipe and reaches a steady speed. Explain why, and state what changes if the pipe is slit along its length.

Accepted Answer

The changing flux drives circular eddy currents in the pipe wall. By Lenz's law these oppose the magnet's approach below and its departure above, producing a retarding force that grows with speed.

A steady speed is reached when $mgv=i^{2}R$ summed over the pipe: all the gravitational energy released is being dissipated as heat.

A lengthwise slit breaks the circular current path, so almost no eddy current can flow and the magnet falls essentially freely.

Practice — "Electromagnetic Induction and Alternating Currents"

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