First slide

Inductance

Question

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

Moderate

A

2.5 H
B

2.0 H
C

1.0 H
D

4.0 H

Solution

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, i.e.,
$ϕ \propto I$
$or ϕ = LI$ …..(i)
Where I is called the self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through solenoid,
$ϕ = 500 \times 4 \times 10^{- 3} = 2 Wb$
[after putting values in Eq. (i)]
or L = 1 H

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