Electromagnetic Induction
Maharashtra Board · Class 12 · Physics
Practice quiz for Electromagnetic Induction — Maharashtra Board Class 12 Physics. MCQs and questions with answers to test your preparation.
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Quick Quiz: Electromagnetic Induction
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According to Faraday's first law of electromagnetic induction, what condition is necessary for an EMF to be induced in a circuit?
A coil of 100 turns has an area of 0.01 m². If a uniform magnetic field of 0.5 T perpendicular to the coil is reduced to zero in 0.1 s, calculate the induced EMF.
According to Lenz's law, the direction of induced current is such that it:
A conducting rod of length 0.2 m moves with velocity 5 m/s perpendicular to a magnetic field of 0.8 T. Calculate the motional EMF induced.
Sample Questions
Which of the following factors affect the self-inductance of a solenoid?
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Number of turns, Cross-sectional area, Length of solenoid, Permeability of core material
Self-inductance L = μ₀n²Al = μ₀N²A/l, where N is number of turns, A is area, l is length, and μ₀ is permeability. Current does not affect inductance as it's a property of the geometry and material.
The unit of inductance is:
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Henry
The unit of inductance is Henry (H), named after Joseph Henry. 1 H = 1 V·s/A = 1 Ω·s. Weber is the unit of magnetic flux, Tesla is for magnetic field, and Farad is for capacitance.
A solenoid has 500 turns, length 0.5 m, and cross-sectional area 0.01 m². Calculate its self-inductance in air.
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6.28 mH
For a solenoid: L = μ₀N²A/l. Given: N = 500, A = 0.01 m², l = 0.5 m, μ₀ = 4π × 10⁻⁷ H/m. L = (4π × 10⁻⁷ × 500² × 0.01)/0.5 = 6.28 × 10⁻³ H = 6.28 mH
The energy stored in an inductor of 2 H carrying current of 3 A is:
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9 J
Energy stored in an inductor: U = ½LI². Given: L = 2 H, I = 3 A. Therefore, U = ½ × 2 × 3² = ½ × 2 × 9 = 9 J
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