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AP Intermediate 2nd Year Electromagnetic Induction — practice questions

Explore 10 free Electromagnetic Induction questions in AP Intermediate 2nd Year Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electromagnetic Induction questions and exercises with answers

Showing questions 1–10 of 10.

A current of 4 mA in a coil of self-inductance 10 mH and 200 turns produces total flux:Read the answer and worked solution →When two inductors $L_1$ and $L_2$ are connected in **series** (no mutual inductance), the equivalent inductance is:Read the answer and worked solution →Two inductors of 4 H and 12 H are connected in **parallel** (no mutual inductance). The equivalent inductance is:Read the answer and worked solution →If the current in a coil of inductance 5 H changes from 2 A to 5 A in 0.1 s, the magnitude of induced EMF is:Read the answer and worked solution →If the current in a primary coil changes at 30 A/s and induces 0.6 V in a secondary coil, the mutual inductance is:Read the answer and worked solution →The energy stored per unit volume in a magnetic field $B = 0.1$ T in vacuum (take $\mu_0 = 4\pi \times 10^{-7}$ N/A²) is approximately:Read the answer and worked solution →A coil of self-inductance 2 H carries a current of 4 A. The energy stored in its magnetic field is:Read the answer and worked solution →Faraday's law of electromagnetic induction states EMF induced equals:Read the answer and worked solution →Lenz's law states the induced current direction is such that it:Read the answer and worked solution →SI unit of magnetic flux is:Read the answer and worked solution →

Worked practice questions

A current of 4 mA in a coil of self-inductance 10 mH and 200 turns produces total flux:

  1. $8 \times 10^{-3}$ Wb total
  2. $4 \times 10^{-5}$ Wb total (i.e. $2 \times 10^{-7}$ Wb per turn)
  3. $2 \times 10^{-5}$ Wb total
  4. $2 \times 10^{-7}$ Wb per turn
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Correct answer: B. $4 \times 10^{-5}$ Wb total (i.e. $2 \times 10^{-7}$ Wb per turn)

**Total flux linkage:** $\phi = L i = (10 \times 10^{-3}) \times (4 \times 10^{-3}) = 4 \times 10^{-5}$ Wb. Flux per turn = $\phi/N = (4 \times 10^{-5})/200 = 2 \times 10^{-7}$ Wb.

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When two inductors $L_1$ and $L_2$ are connected in **series** (no mutual inductance), the equivalent inductance is:

  1. $L_1 + L_2$
  2. $L_1 L_2/(L_1 + L_2)$
  3. $1/L_1 + 1/L_2$
  4. $\sqrt{L_1 L_2}$
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Correct answer: A. $L_1 + L_2$

**Inductors in series add directly:** $L_{eq} = L_1 + L_2$ (analogous to resistors in series). Parallel combination uses the reciprocal sum.

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Two inductors of 4 H and 12 H are connected in **parallel** (no mutual inductance). The equivalent inductance is:

  1. 8 H
  2. 16 H
  3. 4 H
  4. 3 H
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Correct answer: D. 3 H

Parallel: $1/L_{eq} = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3$ ⇒ $L_{eq} = 3$ H. Always less than the smaller value.

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