Question 260: Electromagnetic Induction

Question

A current of 0.5 A is passed through winding of a long solenoid having 400 turns. The magnetic flux linked with each turn is $3 	imes 10^{-3} \mathrm{~Wb}$. The self inductance of the solenoid is

Accepted Answer

The self-inductance $ L $ of a solenoid can be calculated using the formula for the flux linkage, which is given by the product of the number of turns $ N $ and the magnetic flux $ \Phi $ per turn:
$ L = \frac{N \Phi}{I} $
where:
$ N = 400 $ (the number of turns),
$ \Phi = 3 	imes 10^{-3} \, 	ext{Wb} $ (the magnetic flux per turn),
$ I = 0.5 \, 	ext{A} $ (the current).
Substitute these values into the formula:
$ L = \frac{400 	imes 3 	imes 10^{-3}}{0.5} $
Simplify the expression:
$ L = \frac{1.2}{0.5} = 2.4 \, 	ext{H} $
Therefore, the self-inductance of the solenoid is $ 2.4 \, 	ext{H} $.
Hence, the correct option is
Option A: 2.4 H
.

Answer

2.4 H

Answer

2.0 H

Answer

1.2 H

Answer

0.6 H

Electromagnetic Induction

A current of 0.5 A is passed through winding of a long solenoid having 400 turns. The magnetic flux linked with each turn is 3 \times 10^{-3} \mathrm{~Wb}. The self inductance of the solenoid is

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