Question

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

• A. 3 H
• B. 2 H
• C. 1 H
• D. 4 H

Answer 1

The Correct Option is C

To find the self-inductance of the solenoid, we can use the relationship between the self-inductance \( L \), the magnetic flux \( \Phi \), and the current \( I \). The formula for self-inductance is given by:

L = \frac{N \Phi}{I}

where:

• L is the self-inductance in Henry (H),
• N is the number of turns of the solenoid,
• \Phi is the magnetic flux linked with each turn in Weber (Wb),
• I is the current in Amperes (A).

Given:

• N = 1000 turns
• \Phi = 4 \times 10^{-3} \, \text{Wb}
• I = 4 \, \text{A}

Substituting these values into the formula for self-inductance:

L = \frac{1000 \times 4 \times 10^{-3}}{4}

Calculating:

L = \frac{4000 \times 10^{-3}}{4} = \frac{4}{4} = 1\, \text{H}

Therefore, the self-inductance of the solenoid is 1 H.

Thus, the correct answer is:

1 H