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# Derive an Expression for Self Inductance of a Long Solenoid of Length L, Cross-sectional Area a Having N Number of Turns. - Physics

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Derive an expression for self inductance of a long solenoid of length l, cross-sectional area A having N number of turns.

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#### Solution

Magnetic field B inside a solenoid carrying a current i is . mu_0 ni

B =mu_0ni

Let n be the number of turns per unit length.

Nphi_B = nlBA

Where,

N is total number of turns

l is the length of the solenoid

Inductance, L = (Nphi_B)/i

Substituting, we obtain

L = (nlBA)/i

Substituting the value of B, we obtain

L = (nLmu_0niA)/i

L = n^2lmu_0A

Inductance L of a solenoid is:

L =mu_0n^2lA

Concept: Inductance - Self-Inductance
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