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प्रश्न
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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उत्तर
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`
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संबंधित प्रश्न
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(c) magnetic field energy if the same current goes through them
(d) time constant if one solenoid is connected to one battery and the other is connected to another battery.
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