Electromagnetic Induction and Alternating Currents Questions and Answers

Answer: A

The induced emf is +5 V.

By Faraday's law, \(\varepsilon = -\frac{d\phi}{dt}\):

\[ \begin{aligned} \frac{d\phi}{dt} &= 10t - 25 = 20 - 25 = -5 \\ \varepsilon &= -(-5) = +5\ \text{V} \end{aligned} \]

-5 V forgets the minus sign of Lenz's law, and the constant term 150 plays no part, since it does not change with time.

The positive sign means the induced emf opposes the decrease in flux that is happening at that instant (the flux is decreasing at 5 Wb s^-1).

Answer: E

The self-inductance is 0.50 H.

Self-inductance is the total flux linkage per unit current: \(L = \frac{N\Phi}{I}\).

\[ \begin{aligned} L &= \frac{1000 \times 10^{-3}}{2} \\ &= \frac{1}{2} \\ &= 0.50\ \text{H} \end{aligned} \]

The flux of \(10^{-3}\) Wb is per turn, so it must be multiplied by the 1000 turns. Leaving out the division by the 2 A current would give 1 H, which is not offered.

Answer: A

The core is laminated to minimize eddy current loss.

The changing magnetic flux in the core induces circulating eddy currents in it, which waste energy as heat. Building the core from thin sheets insulated from each other breaks up the paths of these currents and greatly reduces the loss.

Copper loss is the heating of the windings (\(I^{2}R\)) and is reduced by thicker conductors. Hysteresis loss is reduced by choosing a soft magnetic material such as silicon steel, not by laminating. Laminations do not reduce cost or flux leakage.

Answer: D

The induced emf is 12 V.

\[ \begin{aligned} e &= M\frac{dI}{dt} = 3 \times \frac{2 - 0}{0.5} \\ &= 3 \times 4 = 12\ \text{V} \end{aligned} \]

6 V is M times the change in current, without dividing by the 0.5 s over which it changes.

The emf in the second coil depends on the rate of change of current in the first, not on the current itself, so once the current stops changing at 2 A the induced emf falls to zero. The 12 V is its value while the current is rising at a steady 4 A/s.

Answer: E

The key gives the current as 2 A, its rms value.

The inductive reactance is

\[ X_{L} = \omega L = 100\pi \times 0.5 = 50\pi\ \Omega \]

The peak voltage is \(100\sqrt{2}\pi\) V, so

\[ \begin{aligned} I_{0} &= \frac{100\sqrt{2}\pi}{50\pi} = 2\sqrt{2}\ \text{A} \\ I_{rms} &= \frac{I_{0}}{\sqrt{2}} = 2\ \text{A} \end{aligned} \]

An alternating current is normally quoted by its rms value, which is what an ammeter reads; \(2\sqrt{2}\) A is the peak value.

Answer: E

The change of flux is 12 Wb.

By Faraday's law, the magnitude of the induced emf equals the rate of change of flux, \(e = \frac{\Delta\phi}{\Delta t}\).

\[ \Delta\phi = e \times \Delta t = 2 \times 6 = 12 \text{ Wb} \]

Here the emf is taken as steady over the 6 s, so the flux changes at a constant 2 Wb per second; one weber per second is one volt. 3 Wb comes from dividing 6 by 2 instead of multiplying, and the other values do not follow from the law.

Answer: E

The self-inductance becomes 0.75 H.

A solenoid's self-inductance is \(L = \mu_r\mu_0 n^{2}Al\), so filling it with a core multiplies \(L\) by the relative permeability:

\[ \begin{aligned} L &= 1500 \times 0.5\ \text{mH} \\ &= 750\ \text{mH} = 0.75\ \text{H} \end{aligned} \]

0.5 H keeps the number 0.5 but switches the unit, which is not what the core does; 1.5 H and 1.25 H are not equal to 1500 times 0.5 mH.

Answer: A

The r.m.s. current is 5 A.

The two terms combine into one sinusoid with peak value

\[ \begin{aligned} I_0 &= \sqrt{(4\sqrt{2})^{2} + (3\sqrt{2})^{2}} = \sqrt{32 + 18} \\ &= \sqrt{50} = 5\sqrt{2}\ \text{A} \\ I_{rms} &= \frac{I_0}{\sqrt{2}} = 5\ \text{A} \end{aligned} \]

\(5\sqrt{2}\) A is the peak value, not the r.m.s. value, and \(7\sqrt{2}\) A simply adds the two amplitudes.

Answer: B

The impedance is \(2\sqrt{2}\,\pi^{2}\ \Omega\).

First find the inductive reactance, then combine it with R.

\[ \begin{aligned} X_L &= 2\pi f L = 2\pi \times 50 \times 0.02\pi \\ &= 2\pi^{2}\ \Omega \\ Z &= \sqrt{R^{2} + X_L^{2}} = \sqrt{2}\times 2\pi^{2} \end{aligned} \]

R and \(X_L\) are equal, so the impedance is \(\sqrt{2}\) times either. \(2\pi^{2}\ \Omega\) is R or \(X_L\) alone, and the other options drop a factor of \(\pi\).

Answer: D

The current is 0.8 A.

Connected in series (aiding), the emfs add and so do the internal resistances:

\[ I = \frac{2 + 6}{8 + 1 + 1} = \frac{8}{10} = 0.8\ \text{A} \]

Ignoring the internal resistances gives 1 A. If the cells opposed each other the net emf would be 4 V and the current 0.4 A, which is not offered.