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Questions
Derive the expression for the self-inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.
Obtain an expression for self-inductance of a long solenoid of length l, area of cross-section A having N turns.
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Solution
The self-inductance of a solenoid is defined as the magnetic flux to current ratio via the solenoid. It is given by,
L = `phi/I`
Total number of turns in the solenoid,
N = nl
The magnetic field inside the long solenoid,
B = μ0ni
Flux through one turn,
`phi_1 = BA = mu_0niA`
Thus, total flux through N turns,
`phi_t = Nphi_1 = nl xx mu_0niA = mu_0n^2lAi`
Using `phi_t = Li`
where L is the self-inductance of the coil.
∴ `mu_0n^2lAi = Li`
L = `mu_0n^2Al`
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