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NEET PHYSICSELECTROMAGNETIC INDUCTIONEasy

Question

The magnetic potential energy stored in a certain inductor is 25 mJ25 \text{ mJ}, when the current in the inductor is 60 mA60 \text{ mA}. This inductor is of inductance:

A

0.138 H0.138 \text{ H}

B

138.88 H138.88 \text{ H}

C

1.389 H1.389 \text{ H}

D

13.89 H13.89 \text{ H}

Step-by-Step Solution

The magnetic potential energy (UU) stored in an inductor is given by the formula: U=12LI2U = \frac{1}{2} L I^2 Where: UU is the energy stored (25 mJ=25×103 J25 \text{ mJ} = 25 \times 10^{-3} \text{ J}) II is the current flowing (60 mA=60×103 A60 \text{ mA} = 60 \times 10^{-3} \text{ A})

  • LL is the self-inductance of the inductor.

Calculation: Rearranging the formula to solve for LL: L=2UI2L = \frac{2U}{I^2} L=2×(25×103 J)(60×103 A)2L = \frac{2 \times (25 \times 10^{-3} \text{ J})}{(60 \times 10^{-3} \text{ A})^2} L=50×1033600×106L = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}} L=50×1033.6×103L = \frac{50 \times 10^{-3}}{3.6 \times 10^{-3}} L=503.613.888... HL = \frac{50}{3.6} \approx 13.888... \text{ H}

Rounding to two decimal places, the inductance is 13.89 H13.89 \text{ H}.

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 INDUCTIONmagneticpotentialenergystoredcertain

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