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

Question

Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon:

Answer 1

To determine the self-inductance of the solenoid, we can use the relationship between the magnetic flux linked with each turn of the solenoid, the current, and the self-inductance.

The formula relating magnetic flux, self-inductance, and current is:

$\Phi = L \cdot I$

where:

$\Phi$ is the magnetic flux linked with each turn.
L is the self-inductance of the solenoid.
I is the current flowing through the solenoid.

From the question, we know:

The magnetic flux per turn, $\Phi = 4 \times 10^{-3}$ Wb.
The current, I = 2 A.
The number of turns, N = 500.

The total magnetic flux linked with the entire solenoid is the flux per turn multiplied by the number of turns:

$\Phi_{\text{total}} = N \cdot \Phi = 500 \times 4 \times 10^{-3} = 2$ Wb

The self-inductance L can now be calculated using $\Phi_{\text{total}} = L \cdot I$:

L = \frac{\Phi_{\text{total}}}{I} = \frac{2}{2} = 1.0 henry

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

Let's examine the given options:

Option 1: 1.0 henry - Correct
Option 2: 4.0 henry - Incorrect
Option 3: 2.5 henry - Incorrect
Option 4: 2.0 henry - Incorrect

Therefore, the correct answer is 1.0 henry.

Answer 2