Electromagnetic Induction and Alternating Currents — Practice

Question 1

A coil sits in a very strong but perfectly steady magnetic field. What emf is induced in it?

Accepted Answer

Exactly zero. Induction responds only to the rate of change of flux, never to the flux itself, so a constant field of any magnitude induces nothing at all.

Question 2

State the phase relationship between current and voltage in a pure inductor and in a pure capacitor.

Accepted Answer

In an inductor the current lags the voltage by $\pi/2$; in a capacitor it leads by $\pi/2$. The mnemonic CIVIL captures it: in a C, I leads V, while V leads I in an L.

Question 3

The Indian mains is quoted as 220 V. Find the peak voltage and explain which value insulation must withstand.

Accepted Answer

$V_0 = 220\sqrt{2} = 311$ V. Quoted values and all meter readings are RMS, but insulation must be rated for the 311 V peak, which makes the distinction practical rather than academic.

Question 4

A 100-turn coil of resistance 20 $\Omega$ has its flux per turn fall from 0.06 to 0.02 Wb. Find the charge that flows.

Accepted Answer

The flux linkage change is $100	imes0.04 = 4$ Wb, so $q = N\Delta\Phi/R = 4/20 = 0.2$ C. No time information is needed, because the charge depends only on the total flux change and the resistance.

Question 5

A 0.5 m rod slides at 4 m s$^{-1}$ on rails in a 0.2 T perpendicular field with circuit resistance 0.1 $\Omega$. Find the emf, current, retarding force and mechanical power, and check the energy balance.

Accepted Answer

$\varepsilon = Bvl = 0.4$ V, giving $I = 4$ A. The retarding force $BIl = 0.4$ N, so the mechanical power $Fv = 1.6$ W. The electrical dissipation $I^{2}R$ is also 1.6 W, so the balance closes exactly as Lenz's law requires.

Question 6

A series circuit has $R = 30$, $X_L = 60$ and $X_C = 20\ \Omega$ across 200 V. Find $Z$, the current, the phase angle and the average power.

Accepted Answer

Net reactance is 40 $\Omega$, so $Z = \sqrt{900+1600} = 50\ \Omega$ and $I = 4$ A. $	an\phi = 40/30$ gives $\phi = 53^\circ$ with power factor 0.6, so $P = 480$ W. This checks against $I^{2}R = 480$ W, and the current lags because the circuit is inductive.

Question 7

A transformer steps 2200 V down to 220 V and delivers 8 A at 90 per cent efficiency. Find the turns ratio and the primary current.

Accepted Answer

The turns ratio is 10:1 step-down. Output power is 1760 W, so input is $1760/0.9 = 1956$ W, giving a primary current of 0.89 A. This exceeds the ideal 0.8 A because the extra current covers the 196 W lost to copper, eddy, hysteresis and leakage.

Question 8

A series circuit has $L = 2$ H, $C = 8\ \mu$F and $R = 10\ \Omega$ across 20 V. Find the resonant frequency, the quality factor and the voltage across the inductor at resonance.

Accepted Answer

$\omega_0 = 1/\sqrt{LC} = 250$ rad s$^{-1}$. At resonance $Z = R$, so $I = 2$ A, and $X_L = 500\ \Omega$ gives $V_L = 1000$ V. The quality factor is 50, so the magnification is exactly $Q$. Nothing is violated because $V_C$ is also 1000 V and exactly antiphase, so the two cancel.

Question 9

Explain why Lenz's law must have the sign it does, using an energy argument.

Accepted Answer

Suppose the induced current assisted the change instead. Pushing a magnet toward a coil would produce a current that pulled it in faster, increasing the rate of flux change, strengthening the current further. The magnet would accelerate indefinitely while also generating electrical energy — energy from nothing. The opposing sign forces you to do mechanical work, and that work is exactly the electrical energy produced.

Question 10

Why can a transformer not operate on direct current, and why did this decide the choice of AC for power distribution?

Accepted Answer

A steady current gives a steady flux, and a steady flux induces no secondary emf at all, so no energy transfers. Long-distance transmission needs very high voltage to keep $I^{2}R$ losses small, followed by a step down to safe voltages for use. Only alternating current can be stepped up and down efficiently, so the transformer settled the question and the world grid is AC.

Question 11

A transformer steps 220 V up to 4400 V and delivers 8 kW at the secondary with an efficiency of 80 per cent. Find the turns ratio, the secondary current and the primary current.

Accepted Answer

The turns ratio is $N_s/N_p = 4400/220 = 20$. The secondary current is $I_s = 8000/4400 = 1.82$ A. The primary must supply $P_p = 8000/0.8 = 10{,}000$ W, so $I_p = 10000/220 = 45.5$ A. The ideal relation would have given $I_p = 20 	imes 1.82 = 36.4$ A; the quoted efficiency raises the primary current above that figure, never below it.

Question 12

A load draws 5 A at 230 V, 50 Hz with a power factor of 0.7 lagging. Find the real power and the reactive power, and the capacitance that must be connected in parallel to bring the power factor to unity.

Accepted Answer

Apparent power is $S = (230)(5) = 1150$ VA and real power is $P = S\cos\phi = 805$ W. The reactive power is $Q = \sqrt{S^{2}-P^{2}} = \sqrt{1150^{2}-805^{2}} = 821$ VAR, lagging. To cancel it the capacitor must supply an equal leading reactive power, $Q_C = V^{2}\omega C$, so $C = 821/[(230)^{2}(314)] = 4.9	imes10^{-5}$ F, about 49 µF. The real power is untouched; only the current the cables carry falls, from 5 A to $805/230 = 3.5$ A.

Practice — Electromagnetic Induction and Alternating Currents

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities