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Q.

A long solenoid is placed in a uniform magnetic field of strength B with its axis parallel to the field direction. The self inductance of the solenoid is L and it's area of cross section is A. It is not carrying any current. If the total magnetic flux linked with the solenoid is ϕ, then energy of the magnetic field stored in the solenoid is.

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a

LϕL

b

ϕ2L

c

ϕ22L

d

zero

answer is C.

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Detailed Solution

Flux linked with each turn of the solenoid = BATotal number of turns = nℓ where ℓ is the length of the solenoidTotal flux linked ϕ=BAnℓ∴B=ϕAnℓEnergy per unit volume = =B22μ0∴ total energy of the magnetic field stored in the solenoid=B22μ0Aℓ=12μ0ϕ2A2n2ℓ2Aℓ=12⋅ϕ2μ0An2⁡ℓ=ϕ22L ∵L=μ0An2ℓ

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A long solenoid is placed in a uniform magnetic field of strength B with its axis parallel to the field direction. The self inductance of the solenoid is L and it's area of cross section is A. It is not carrying any current. If the total magnetic flux linked with the solenoid is ϕ, then energy of the magnetic field stored in the solenoid is.