Electromagnetic Induction and Alternating Currents Questions and Answers

Answer: A

There are momentary galvanometer deflections in opposite directions that die out shortly.

This is electromagnetic induction. When the key is inserted, the current in circuit B rises from zero, so the magnetic field it produces through circuit A changes and induces a brief current in A, and the galvanometer deflects. Once the current in B is steady, the field stops changing and the deflection dies away.

When the key is removed, the current in B falls to zero, the field through A changes the other way, and the induced current flows in the opposite direction. So the deflections are in opposite directions, not the same direction, and there is an effect on removing the key as well as on inserting it.

Answer: A

The EMF induced in a conductor moving in a magnetic field is \(e = Blv\sin\theta\).

\[ \begin{aligned} e = 1.2 \times 1.5 \times 60 \times \sin 30^\circ \\ &= 108 \times 0.5 = 54\ \text{V} \end{aligned} \]

The length must be changed to 1.5 m. The value 60 V would result from taking \(\sin\theta = 0.555\), 45 V from a length of 1.25 m and 52 V has no basis. The answer is 54 V, option A.

Answer: C

The capacitive reactance will decrease by a factor of four.

Capacitive reactance is inversely proportional to frequency:

\[ \begin{aligned} X_C &= \frac{1}{2\pi f C} \\ \frac{X_{200}}{X_{50}} &= \frac{50}{200} = \frac{1}{4} \end{aligned} \]

The capacitance (30 µF) and supply voltage do not affect the ratio. 'Increase by four' would be true of an inductor, whose reactance \(X_L = 2\pi f L\) grows with frequency. The factor-of-two options would apply to a doubling of frequency, not a fourfold rise.

Answer: D

The correct statement is that the maximum voltage across the capacitor equals the maximum voltage across the resistor.

\[ \begin{aligned} X_L &= \omega L = 400 \times 0.175 = 70\ \Omega \\ X_C &= \frac{1}{\omega C} = \frac{1}{400 \times 62.5 \times 10^{-6}} = 40\ \Omega \\ Z &= \sqrt{40^{2} + (70 - 40)^{2}} = 50\ \Omega \end{aligned} \]

The current amplitude is \(I_0 = 50/50 = 1\) A, so the voltage amplitudes are \(V_R = 40\) V, \(V_L = 70\) V and \(V_C = 40\) V. The capacitor and resistor both carry 40 V.

The inductor's 70 V exceeds the 50 V source, which is possible because \(V_L\) and \(V_C\) are out of phase and partly cancel. The capacitor's 40 V is less than 60 V, and the resistor carries 40 V, not the full 50 V, because the circuit is not at resonance.

Answer: C

The resonant frequency is \(\frac{2}{\sqrt{3\,LC}}\).

The smaller coil is a closed loop with no resistance, so the magnetic flux through it cannot change: it carries whatever current cancels the larger coil's field inside it. With the outer field \(\mu_{0}NI\) and the inner coil's field \(\mu_{0}(2N)i\), cancellation needs \(i = -\frac{I}{2}\), and the field inside the small coil's cross-section is zero.

Now count the flux linkage of the larger coil. Half its length (\(Nd/2\) turns) surrounds the small coil, where only the area \(S - \frac{S}{2}\) carries field; the other half sees the full area \(S\):

\[ \begin{aligned} \Phi &= \tfrac{Nd}{2}\,\mu_{0}NI\tfrac{S}{2} + \tfrac{Nd}{2}\,\mu_{0}NIS \\ &= \tfrac{3}{4}\,\mu_{0}N^{2}SdI = \tfrac{3}{4}LI \end{aligned} \]

So the effective inductance is \(\frac{3L}{4}\), and

\[ \begin{aligned} \omega_{0} &= \frac{1}{\sqrt{\frac{3}{4}LC}} \\ &= \frac{2}{\sqrt{3\,LC}} \end{aligned} \]

The other options would need an effective inductance of \(\frac{15L}{16}\), \(\frac{5L}{36}\) or \(\frac{3L}{2}\); none of these follows from the shorted coil, which can only reduce the inductance, and here by exactly a quarter.

Answer: C

The correct match is P → 3, Q → 2, R → 1, S → 4.

The emf is proportional to the rate of change of the flux through the part of the loop lying in \(x \gt 0\), and the current follows its sign. The loop turns clockwise through \(\omega t\).

(P) A half-disc moves into the field region at a steady rate: the field-filled area grows linearly during the first half period and shrinks linearly during the second. The emf is constant and reverses at \(\frac{T}{2}\): graph (3).

(R) A single 60° sector enters the field during the first sixth of the period, stays fully inside (constant flux, no current), then leaves between \(\frac{T}{2}\) and \(\frac{2T}{3}\): one positive pulse, a gap, then one negative pulse: graph (1).

(S) The two sectors lie opposite each other, so one leaves the field exactly as the other enters, and the total flux stays constant: no current, graph (4).

(Q) The two sectors enter and leave at different times, giving two pulses of each sign: graph (2), the only one left.

Answer: D

The self-inductance becomes \(4L\).

For a solenoid, \(L = \frac{\mu_{0}N^{2}A}{l}\). The field inside is proportional to \(N\), and the flux links all \(N\) turns, so \(L \propto N^{2}\). Doubling \(N\) with the same length and area multiplies \(L\) by \(2^{2} = 4\).

\(2L\) assumes \(L \propto N\), counting either the field or the linkage but not both. \(8L\) would be a cube law.

Answer: E

The constant current is 10A.

The heating effect of an alternating current equals that of a direct current equal to its rms value:

\[ \begin{aligned} I_{rms} &= \frac{I_{0}}{\sqrt{2}} \\ &= \frac{10\sqrt{2}}{\sqrt{2}} = 10\ \text{A} \end{aligned} \]

\(10\sqrt{2}\) A is the peak, which is reached only momentarily, so it would overheat the wire. 14 A is the peak rounded, and 5 A or 7 A halve the correct values.

Answer: A

The acceleration of the magnet is less than g.

As the magnet falls through the ring, the magnetic flux through the ring changes, inducing currents (eddy currents) in the conducting iron ring. By Lenz's law these currents oppose the change that causes them: the ring repels the approaching magnet and attracts the receding one, so in both phases there is an upward force on the magnet and its acceleration is less than g.

It would equal g only if no current could be induced, and it cannot exceed g because the induced force always opposes the motion.

Answer: E

Resonance needs an inductor and capacitor.

Resonance in an a.c. circuit occurs at the frequency where the inductive reactance equals the capacitive reactance, \(X_{L} = X_{C}\), so that the two cancel: \(\omega_{0} = 1/\sqrt{LC}\). At that frequency the current in a series LCR circuit is a maximum. Both an inductor and a capacitor are needed, because the effect comes from energy passing back and forth between the magnetic field of one and the electric field of the other.

A circuit with only an inductor, only a capacitor, or either one with a resistor has a single kind of reactance and nothing to cancel it, so it cannot resonate. A resistor is not needed for resonance; it only limits the peak current and sets the sharpness of the resonance.