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NEET PHYSICSELECTROMAGNETIC INDUCTIONEasy

Question

A long solenoid has 1000 turns. When a current of 4 A4 \text{ A} flows through it, the magnetic flux linked with each turn of the solenoid is 4×103 Wb4 \times 10^{-3} \text{ Wb}. The self-inductance of the solenoid is:

A

3 H

B

2 H

C

1 H

D

4 H

Step-by-Step Solution

The self-inductance (LL) of a coil is defined by the relationship between the total magnetic flux linkage (NΦBN\Phi_B) and the current (II) flowing through it: NΦB=LIN\Phi_B = LI .

Given: Number of turns (NN) = 1000 Current (II) = 4 A4 \text{ A}

  • Magnetic flux per turn (ΦB\Phi_B) = 4×103 Wb4 \times 10^{-3} \text{ Wb}

Calculation:

  1. Calculate Total Flux Linkage: Total Flux=N×ΦB\text{Total Flux} = N \times \Phi_B Total Flux=1000×(4×103) Wb=4 Wb\text{Total Flux} = 1000 \times (4 \times 10^{-3}) \text{ Wb} = 4 \text{ Wb}

  2. Calculate Self-Inductance (LL): Using the formula LI=NΦBLI = N\Phi_B: L=NΦBIL = \frac{N\Phi_B}{I} L=4 Wb4 AL = \frac{4 \text{ Wb}}{4 \text{ A}} L=1 HL = 1 \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 INDUCTIONsolenoidcurrentthroughmagneticlinked

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