Electromagnetic Induction and Alternating Currents Questions and Answers

Answer: C

The mismatched pair is Induced emf : Biot - Savart law.

Induced emf is governed by Faraday's law of electromagnetic induction (with Lenz's law for its direction). The Biot-Savart law gives the magnetic field produced by a current, a different phenomenon.

The other pairs are correct: induction furnaces heat metal with eddy currents, transformer cores are laminated to reduce eddy currents, an ac generator works by electromagnetic induction, and coaxial coils are the standard example of mutual inductance.

Answer: D

The power is 100 W.

Power in an ac circuit uses the rms voltage:

\[ \begin{aligned} V_{rms} &= \frac{282.8}{\sqrt{2}} = 200\ \text{V} \\ P &= \frac{V_{rms}^{2}}{R} = \frac{40000}{400} = 100\ \text{W} \end{aligned} \]

Using the peak voltage gives 200 W, the usual slip; 150, 250 and 300 W do not follow.

For a sinusoidal supply the rms value is \(\frac{1}{\sqrt{2}}\) of the peak, and 282.8 V was chosen as \(200\sqrt{2}\) so that \(V_{rms}\) is exactly 200 V.

Answer: C

The flux linked with the coil is 2 Wb.

With the plane perpendicular to the field, the field is along the normal and the angle between them is 0. The flux linked with all N turns is

\[ \begin{aligned} N\phi &= NBA\cos 0^{\circ} \\ &= 100 \times 1 \times 0.02 = 2\ \text{Wb} \end{aligned} \]

Zero would apply if the plane were parallel to the field. The flux through a single turn is 0.02 Wb; the 100 turns raise the linkage to 2 Wb.

Answer: E

The current is 5 A.

The energy stored in an inductor is \(U = \frac{1}{2}LI^{2}\).

\[ \begin{aligned} 25 &= \tfrac{1}{2}(2)I^{2} \\ I^{2} &= 25 \Rightarrow I = 5\ \text{A} \end{aligned} \]

25 A is \(I^{2}\), not I. 10 A doubles the answer, and 2 A confuses the current with the inductance.

Answer: A

The self-inductance is 0.4 H.

The magnitude of the self-induced emf is \(e = L\frac{dI}{dt}\).

\[ L = \frac{e}{dI/dt} = \frac{12}{30} = 0.4\ \text{H} \]

The other values are arithmetic slips; multiplying 12 by 30 instead of dividing would give 360 H, far outside the options.

Answer: C

The induced emf is 0.3 V.

For a rod moving perpendicular to the field with its length perpendicular to both the field and its velocity,

\[ \varepsilon = Blv = 0.2 \times 0.5 \times 3 = 0.3\ \text{V} \]

0.6 V doubles the result, and 0.1 or 0.2 V leave out one of the three factors.

Answer: B

The rms value is \(\frac{200}{\sqrt{2}}\) V, about 141 V.

For a sinusoidal alternating voltage, the root mean square value is the peak value divided by \(\sqrt{2}\):

\[ \begin{aligned} V_{\text{rms}} &= \frac{V_{0}}{\sqrt{2}} = \frac{200}{\sqrt{2}} \\ &\approx 141\ \text{V} \end{aligned} \]

100 V simply halves the peak value; it is neither the rms value nor the half-cycle average, \(\frac{2V_{0}}{\pi} \approx 127\) V. \(200\sqrt{2}\) V multiplies instead of dividing, and \(\frac{200}{\sqrt{3}}\) V belongs to a triangular wave, not a sine wave.

Answer: B

The power is proportional to \(v^{2}\).

For a rod of length l moved at speed v through a field B, the induced emf is \(\varepsilon = Blv\) and the current in a circuit of resistance R is \(I = \frac{Blv}{R}\). The magnetic force on the rod opposes the motion, so an equal force must push it:

\[ \begin{aligned} F &= IlB = \frac{B^{2}l^{2}v}{R} \\ P &= Fv = \frac{B^{2}l^{2}v^{2}}{R} \end{aligned} \]

This power appears as Joule heat in the circuit. Power proportional to v alone would hold for a constant force, but here the opposing force itself grows with v.

Answer: D

The angle is 60°.

Magnetic flux through a flat coil is \(\phi = BA\cos\theta\), where \(\theta\) is the angle between \(\vec{B}\) and the area vector:

\[ \begin{aligned} 4 &= 16 \times 0.5 \times \cos\theta \\ \cos\theta &= \frac{4}{8} \\ &= \frac{1}{2}, \quad \theta \\ &= 60^{\circ} \end{aligned} \]

30° uses \(\sin\theta\), which is correct only if \(\theta\) is measured from the plane of the coil. 0° would give the maximum flux of 8 Wb, and 90° zero flux.

Answer: D

Self-inductance does not depend on the current through it.

For a long solenoid, \(L = \frac{\mu N^{2}A}{l}\): it depends on the number of turns, the area of cross-section (and so the radius), the length and the permeability of the core. These are all properties of the coil's geometry and material.

The current sets the flux \(\phi = LI\) and the stored energy \(\frac{1}{2}LI^{2}\), but \(L\) itself is the ratio \(\frac{\phi}{I}\), which stays the same as the current changes (for a core without saturation).