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Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field $$B,$$ area $$A$$ and length $$l$$ of the solenoid.
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## QuestionPhysicsClass 12

Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field $$B,$$ area $$A$$ and length $$l$$ of the solenoid.

$$U_m=\frac{B^2\mathit{Al}}{2\mu _0}$$
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## Solution

Magnetic energy stored in the solenoid is given by:
$$U_m=\frac 1 2LI^2$$
where$$L$$is the self-inductance and$$I$$is the current flowing
But, magnetic field inside the solenoid:
$$B=\mu _0nI$$
where$$n$$is the number of turns per unit length
$$⟹I=\frac B{\mu _0n}$$
Substitute the value of $$I$$ in $$U_m$$ , we get
$$U_m=\frac 1 2L(\frac B{\mu _0n})^2$$  ...(i)
And, the self-inductance for the solenoid:
$$L=\mu _0n^2\mathit{Al}$$
where$$A$$is the area of cross-section of the solenoid
Substitute the value of$$L$$in (i).
Hence, the magnetic energy stored in the solenoid is given by:
$$U_m=\frac 1 2(\mu _0n^2\mathit{Al})\left(\frac B{\mu _0n}\right)^2$$
$$⟹U_m=\frac{B^2\mathit{Al}}{2\mu _0}$$