Question

A long solenoid has 500 turns, when a current of 2A is passed through it, the resulting flux linked with each turns of solenoid is 4 x 10^-3 Wb, then the self induction of solenoid is

• A. 2.0 henry
• B. 4.0 henry
• C. 1.0 henry
• D. 2.5 henry

Answer 1

The Correct Option is C

The self-inductance of a solenoid can be calculated using the formula:

\( L = \frac{N^2 \cdot \Phi}{I}\)

Given:

• \( N = 500 \) turns
• \( \Phi = 4 \times 10^{-3} \, \text{Wb} \)
• \( I = 2 \, \text{A} \)

Using the formula, we substitute the given values:

\(L = \frac{500^2 \cdot 4 \times 10^{-3}}{2}\)

\(L = \frac{250000 \cdot 4 \times 10^{-3}}{2}\)

\(L = \frac{1000 \times 10^{-3}}{2}\)

\(L = 1.0 \, \text{Henry}\)

Therefore, the self-inductance of the solenoid is 1.0 Henry. Hence, the correct option is (C) 1.0 Henry.

Answer 2

The self-inductance (L) of a solenoid is related to the total flux linkage (\(\Phi\)) and the current (I) by the formula:

\(L = \frac{N\phi}{I}\)

Where:

• \(L\) is the self-inductance in Henrys (H)
• \(N\) is the number of turns
• \(\phi\) is the magnetic flux linked with each turn in Webers (Wb)
• \(I\) is the current in Amperes (A)

Given:

• \(N = 500\) turns
• \(I = 2 \, \text{A}\)
• \(\phi = 4 \times 10^{-3} \, \text{Wb}\)

Substitute the given values into the formula:

\(L = \frac{500 \times 4 \times 10^{-3}}{2}\)

\(L = \frac{2000 \times 10^{-3}}{2}\)

\(L = \frac{2}{2}\)

\(L = 1 \, \text{H}\)

Therefore, the self-inductance of the solenoid is 1.0 henry.