Self induction due to solenoid and mutual induction due to solenoid
Self induction due to solenoid and mutual induction due to solenoid
Sabreen , 7 Years ago
Grade 12
Electromagnetic Induction> Self induction due to solenoid and mutual...
1 Answers
Arun
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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