Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A long solenoid has 500 turns. When a current of $2 \text{ A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \text{ Wb}$ . The self-inductance of the solenoid is:

2.5 H

2.0 H

1.0 H

4.0 H

Step-by-Step Solution

The self-inductance ( $L$ ) of a coil is defined by the relationship between the total magnetic flux linkage ( $N\phi_B$ ) and the current ( $I$ ) flowing through it.

Formula: $N\phi_B = LI$ Where: $N$ = Total number of turns $\phi_B$ = Magnetic flux linked with each turn $I$ = Current $L$ = Self-inductance
Given Values: $N = 500$ $I = 2 \text{ A}$

$\phi_B = 4 \times 10^{-3} \text{ Wb}$

Calculation: Rearranging the formula to solve for $L$ : $L = \frac{N\phi_B}{I}$ $L = \frac{500 \times (4 \times 10^{-3})}{2}$ $L = \frac{2000 \times 10^{-3}}{2}$ $L = \frac{2.0}{2}$ $L = 1.0 \text{ H}$

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