Question

A long solenoid has \(400\) turns. When a current of \(100\,A\) is passed through it, the resulting magnetic flux linked with each turn of the solenoid is \(4\,mWb\). The self-inductance of the solenoid is

• A. \(1.6\,mH\)
• B. \(16\,mH\)
• C. \(16\,H\)
• D. \(0.16\,mH\)

Hint:

Self-inductance is \(L = \frac{N\phi}{I}\), where \(N\phi\) is total flux linkage.

Answer 1

The Correct Option is B

Step 1: Formula for self-inductance.
Self-inductance is given by:
\[ L = \frac{N\phi}{I} \]

Step 2: Identify given values.
\[ N = 400 \] \[ \phi = 4\,mWb = 4 \times 10^{-3}\,Wb \] \[ I = 100\,A \]

Step 3: Substitute values.
\[ L = \frac{400 \times 4 \times 10^{-3}}{100} \]

Step 4: Simplify numerator.
\[ 400 \times 4 \times 10^{-3} = 1.6 \]

Step 5: Calculate inductance.
\[ L = \frac{1.6}{100} \] \[ L = 0.016\,H \]

Step 6: Convert into \(mH\).
\[ 0.016\,H = 16\,mH \]

Step 7: Final conclusion.
\[ \boxed{16\,mH} \] Hence, correct answer is option (B).