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NEET PHYSICSELECTROMAGNETIC INDUCTIONEasy

Question

The current in an inductor of self-inductance 4 H4 \text{ H} changes from 4 A4 \text{ A} to 2 A2 \text{ A} in 1 s1 \text{ s}. The emf induced in the coil is:

A

-2 V

B

2 V

C

-4 V

D

8 V

Step-by-Step Solution

The induced electromotive force (emf) E\mathcal{E} in an inductor is given by Faraday's law of induction: E=LdIdt\mathcal{E} = -L \frac{dI}{dt}.

  1. Identify Given Values: Self-inductance (LL) = 4 H4 \text{ H} Initial Current (IiI_i) = 4 A4 \text{ A} Final Current (IfI_f) = 2 A2 \text{ A} Time interval (Δt\Delta t) = 1 s1 \text{ s}

  2. Calculate Rate of Change of Current: dIdt=IfIiΔt=241=2 A/s\frac{dI}{dt} = \frac{I_f - I_i}{\Delta t} = \frac{2 - 4}{1} = -2 \text{ A/s}

  3. Calculate Induced EMF: E=(4 H)×(2 A/s)\mathcal{E} = - (4 \text{ H}) \times (-2 \text{ A/s}) E=8 V\mathcal{E} = 8 \text{ V}.

The positive sign indicates the direction opposes the change in current (Lenz's Law), but the magnitude is 8 V.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC INDUCTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROMAGNETIC INDUCTIONcurrentinductorselfinductancechangesinduced

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