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Karnataka PUC II Electromagnetic Induction — practice questions

Explore 9 free Electromagnetic Induction questions in Karnataka PUC II 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 1–9 of 9.

Magnetic flux through a flat loop in a uniform field B isRead the answer and worked solution →Self-inductance L of a coil is defined byRead the answer and worked solution →Motional emf induced in a straight conductor of length l moving with velocity v perpendicular to a uniform B isRead the answer and worked solution →Mutual inductance M between two coils is defined such that the emf induced in coil 2 due to changing current in coil 1 isRead the answer and worked solution →Eddy currents areRead the answer and worked solution →In a transformer, the ratio of secondary to primary voltage equalsRead the answer and worked solution →Energy stored in an inductor carrying current I isRead the answer and worked solution →A horizontal conducting rod 0.5 m long moves at 10 m/s perpendicular to a 0.4 T magnetic field. The induced emf isRead the answer and worked solution →Match each quantity to its unitRead the answer and worked solution →

Worked practice questions

Magnetic flux through a flat loop in a uniform field B is

  1. Φ_B = B · A cos θ, where θ is the angle between B and the area normal
  2. Φ_B = B · A sin θ
  3. Φ_B = B + A
  4. Φ_B = B / A
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Correct answer: A. Φ_B = B · A cos θ, where θ is the angle between B and the area normal

We have Φ_B = B · A. So Φ_B = BA cos θ. Maximum when normal is parallel to B (θ = 0), zero when perpendicular (θ = π/2).

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Self-inductance L of a coil is defined by

  1. L = I / Φ_B
  2. L = ε · I
  3. L = R · t
  4. L = Φ_B / I
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Correct answer: D. L = Φ_B / I

L = N Φ_B / I — the flux linkage per unit current. SI unit is the henry (H) = V·s/A. The induced emf is ε = −L dI/dt.

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Motional emf induced in a straight conductor of length l moving with velocity v perpendicular to a uniform B is

  1. ε = Bl/v
  2. ε = Bv²l
  3. ε = Blv
  4. ε = B/(lv)
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Correct answer: C. ε = Blv

ε = (v × B) · l, magnitude Bvl when all three are mutually perpendicular. This is the basis of generators.

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More Physics topics

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