Question

Define self inductance. Derive an expression for self inductance of a long solenoid

Answer 1

Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.

Consider a solenoid having N number of turns and the current flowing through that be I and solenoid is having length of L and radius of R.

The magnetic field within the solenoid is given by

B=u0*N*I

The total flux through the solenoid wire which has having NL turns is

Flux=NL*B*pi*R*R

The expression for inductance is =flux/current

Hence the self inductance of solenoid is =u0*N^2*pi*R^2*

hope it helps you

Answer 2

Explanation:

$\mathtt \pink{ \xcancel{d B = \frac { M _ { 0 } n d x I a ^ { 2 } } { 2 [ ( r - x ) ^ { 2 } + a ^ { 2 } ] ^ { 3 / 2 } }}}$