Question

Derive an expression for the self inductance of a solenoid

Answer 1

Answer: let one be the length of the Solenoid with air core and N be the number of turn if current I flowing though it, the magnetic field it is given

B=uo n/1 Ian

Flus b =uo n square /1 iA

Induced e. M. F is the result of varying current,

=-d/dt(uo N square /1 iA)

=-uo N square/1 A di /dt

=-L di/dt

Similarly, L = uo N2A/1

L=uon21A

n=number of turns per unit length.

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