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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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