Question
A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self -inductance of the solenoid is
-
2.5 H
-
2.0 H
-
1.0 H
-
4.0 H
Solution
C.
1.0 H
The inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of 1 As-1. If I is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to I, ie,
where L is called the self-inductance or coefficients of self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through the solenoid.
Φ = 500 x 4 x 10-3 = 2 Wb
Or 2 = L x 2 [after putting value in eq. or L = 1H]
