Electromagnetic Induction and Alternating Currents Questions and Answers

Answer: A

The capacitive reactance is proportional to \(\dfrac{1}{f}\).

\(X_{C} = \frac{1}{2\pi fC}\). At higher frequencies the capacitor charges and discharges more often and passes more current, so it offers less opposition; for direct current (\(f = 0\)) it blocks completely.

Proportionality to \(f\) describes an inductor, \(X_{L} = 2\pi fL\). The squared and square-root forms do not occur for a single capacitor.

Answer: A

The average induced emf is \(4\pi\) volt.

By Faraday's law the average emf is the change in flux linkage divided by the time:

\[ \begin{aligned} \varepsilon &= \frac{NBA}{t} = \frac{200 \times 0.4 \times \pi(0.1)^{2}}{0.2} \\ &= \frac{0.8\pi}{0.2} = 4\pi\ \text{V} \end{aligned} \]

The radius must be in metres; leaving it in centimetres makes the answer far too large. \(40\pi\) is ten times the correct value.

Answer: D

The induced emf is 15 mV.

Initially the plane is perpendicular to the field, so the flux is \(BA\). When the plane lies along the field, the flux is zero. The average emf is the change in flux divided by the time:

\[ \begin{aligned} \Delta\phi &= 0.5 \times 150 \times 10^{-4} = 7.5 \times 10^{-3} \text{ Wb} \\ e &= \frac{7.5 \times 10^{-3}}{0.5} = 1.5 \times 10^{-2} \text{ V} \end{aligned} \]

That is 15 mV. 7.5 mV is the flux change in milliwebers, forgetting to divide by the 0.5 s.

Answer: C

The secondary voltage from B is 50 V.

For a transformer \(\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}\). With \(\frac{N_{p}}{N_{s}} = 2\), each transformer halves the voltage (step-down).

\[ \begin{aligned} \text{after } A &: 200 \div 2 = 100 \text{ V} \\ \text{after } B &: 100 \div 2 = 50 \text{ V} \end{aligned} \]

100 V is the output of A alone, and 400 V treats the transformers as step-up.

Answer: E

The emf is 0.2 V.

The induced emf has magnitude \(L\frac{\Delta I}{\Delta t}\).

\[ \begin{aligned} e &= 10 \times 10^{-3} \times \frac{5 - 1}{0.2} \\ &= 0.01 \times 20 = 0.2\ \text{V} \end{aligned} \]

Using the final current 5 A instead of the change 4 A gives 0.25 V, and dropping the milli prefix gives 200 V; neither is offered. The emf opposes the change in current (Lenz's law), so it acts against the increase while the current rises.

Answer: D

Lenz's law is a consequence of the conservation of energy.

Lenz's law says that the induced current flows in a direction that opposes the change producing it. When a magnet is pushed into a coil, the induced current makes the coil repel the magnet, so work has to be done to move it, and that work appears as electrical energy. If the induced current aided the change, energy would be created from nothing.

Conservation of charge, mass or momentum does not explain the direction of the induced current.

Answer: A

The total charge depends on the change in magnetic flux.

By Faraday's law the induced emf is the rate of change of flux. With circuit resistance R, the current is that emf over R, and the total charge is the current over the time of the change:

\[ \begin{aligned} q &= \int \frac{1}{R} \frac{d\Phi}{dt}\, dt \\ &= \frac{\Delta \Phi}{R} \end{aligned} \]

So the charge depends only on the total change in flux, not on how fast it happens. The rate of change decides the emf and current, which is the trap in the last option; the initial or final flux alone is not enough.

Answer: A

The induced emf remains the same.

Each spoke rotating in the field has the same emf \(\frac{1}{2}B\omega l^{2}\) between the axle and the rim, and all the spokes connect the axle to the rim side by side, that is, in parallel. Connecting identical emfs in parallel does not change the voltage, so the emf between axle and rim depends only on B, the angular speed and the spoke length.

Doubling the number of spokes therefore leaves the emf unchanged (it would only lower the internal resistance). The emf would double only if the spokes were in series, which they are not.

Answer: B

The values are \(4/\sqrt{2}\) A and 100 Hz.

Comparing \(I = 4\sin 200\pi t\) with \(I = I_0 \sin 2\pi f t\):

\[ \begin{aligned} I_{rms} &= \frac{I_0}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2.83\ \text{A} \\ 2\pi f &= 200\pi \Rightarrow f = 100\ \text{Hz} \end{aligned} \]

2 A is half the peak, not the r.m.s. value; \(4\sqrt{2}\) A multiplies by root 2 instead of dividing; and 50 Hz comes from taking 200 as \(4\pi f\).

Answer: B

The requirement is high initial permeability, low hysteresis loss and high specific resistance.

A transformer core must carry the magnetic flux easily, so its permeability should be high. It is magnetised and demagnetised fifty times a second, so the hysteresis loop must be narrow (low hysteresis loss). And to limit eddy currents induced in the core, its resistivity should be high, which is why silicon steel, laminated and insulated, is used.

Each of the other options gets one of these three properties the wrong way round.