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प्रश्न
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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उत्तर
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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Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency `(200/pi)` Hz and a DC source are applied across the coil.
| AC Source | ||
| S.No. | V (volts) | I (A) |
| 1 | 3.0 | 0.5 |
| 2 | 6.0 | 1.0 |
| 3 | 9.0 | 1.5 |
| DC Source | ||
| S.No. | V (volts) | I (A) |
| 1 | 4.0 | 1.0 |
| 2 | 6.0 | 1.5 |
| 3 | 8.0 | 2.0 |
