Question

A long solenoid of n turns has a self inductance l and area of cross section a. When a current i flows through the solenoid, the magnetic field inside it has magnitude b. The current i is equal to

Answer 1

Answer:

Current in the solenoid is given as

$i = \frac{NaB}{L}$

Explanation:

As we know that the magnetic field inside solenoid is given as

$B = \frac{\mu_0 N i}{l}$

also we know that the self inductance of the solenoid is given as

$L = \frac{\mu_o N^2 a}{l}$

now we have

$\frac{B}{L} = \frac{i}{Na}$

so we have

$i = \frac{NaB}{L}$

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Topic : Solenoid

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Answer 2

Explanation:

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