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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$\frac{L ϕ}{L}$

$\frac{ϕ^{2}}{L}$

$\frac{ϕ^{2}}{2 L}$

zero

answer is C.

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Detailed Solution

Flux linked with each turn of the solenoid = BA

Total number of turns = $n ℓ$ where $ℓ$ is the length of the solenoid

Total flux linked $ϕ = BA n ℓ ∴ B = \frac{ϕ}{An ℓ}$

Energy per unit volume = $= \frac{B^{2}}{2 μ_{0}}$

$∴$ total energy of the magnetic field stored in the solenoid

$\begin{array}{l} = \frac{B^{2}}{2 μ_{0}} Aℓ = \frac{1}{2 μ_{0}} (\frac{ϕ^{2}}{A^{2} n^{2} ℓ^{2}}) Aℓ \\ = \frac{1}{2} \cdot \frac{ϕ^{2}}{μ_{0} {An}^{2} ℓ} = \frac{ϕ^{2}}{2 L} [∵ L = μ_{0} {An}^{2} ℓ] \end{array}$

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