Question
Physics Question on Electrostatic potential
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
1.0 henry
Explanation
Solution
The self-inductance of a solenoid can be calculated using the formula:
L = (N2 * Φ) / I
Given:
N = 500 turns
Φ = 4 x 10-3 Wb
I = 2A
Using the formula, we can substitute the given values:
L = (5002 * 4 x 10-3) / 2
L = (250000 * 4 x 10-3) / 2
L = (1000 x 10-3)
L = 1.0 Henry
Therefore, the self-inductance of the solenoid is 1.0 Henry. Hence, the correct option is (C) 1.0 henry.