A long solenoid has 1000 turns. When a current of 4 A flows through it the magnetic flux linked with each turn of the solenoid is 4 × 10 - 3 Wb. The self-inductance of the solenoid is:

Question

Aicharya · Accepted Answer

In the case of self-inductance of the solenoid, magnetic flux $(ϕ)$ is directly proportional to the current $(i)$ flowing in the coil. The flux through each turn is equal to the flux linkage.

The relation is given as $N ϕ = L i$ , here $N$ is the number of turns and $L$ is inductance.

So putting the given values, we get

$1000 \times 4 \times 10^{- 3} = L \times 4$

Thus, the self inductance is $1 H$ .

A long solenoid has 1000 turns. When a current of 4 A flows through it the magnetic flux linked with each…

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