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Self induction due to solenoid and mutual induction due to solenoid

Self induction  due to solenoid and mutual induction due to solenoid 
 
 

Grade:12

1 Answers

Arun
25758 Points
3 years ago

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*L.

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