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प्रश्न
Obtain an expression for the self-inductance of a solenoid.
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उत्तर
- Consider a current I established in the windings (turns) of a long solenoid. The current produces a magnetic flux `phi_"B"` through the central region.
- The inductance of the solenoid is given by,
L = `("N"phi_"B")/"I",`
where N = the number of turns,
ΦB = magnetic flux linkage. - The flux linkage for a length l near the middle of the solenoid is,
NΦB = (nl) `(vec"B". vec"A")` = nlBA, (for θ = 0°),
where n = the number of turns per unit length,
B = magnetic field
A = the cross-sectional area of the solenoid. - The magnetic field inside the solenoid is given as, B = `mu_0"ni"`
- Hence, L = `("N"phi_"B")/"i"`
= `(("nl")"BA")/"i"`
= `("nl"(mu_0"ni")"A")/"i"`
∴ L = μ0n2lA
where, Al is the interior volume of solenoid.
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| AC Source | ||
| S.No. | V (volts) | I (A) |
| 1 | 3.0 | 0.5 |
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| DC Source | ||
| S.No. | V (volts) | I (A) |
| 1 | 4.0 | 1.0 |
| 2 | 6.0 | 1.5 |
| 3 | 8.0 | 2.0 |
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