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(D) The magnetic energy stored in a solenoid is given by $U_B = \frac{1}{2} LI^2$.For a solenoid,the self-inductance is $L = \mu_0 n^2 Al$ and the mag

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$\frac{B^2}{2 \mu_0}$

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(D) The magnetic energy stored in a solenoid is given by $U_B = \frac{1}{2} LI^2$.For a solenoid,the self-inductance is $L = \mu_0 n^2 Al$ and the magnetic field is $B = \mu_0 nI$,which implies $I = \frac{B}{\mu_0 n}$.Substituting these into the energy expression:$U_B = \frac{1}{2} (\mu_0 n^2 Al) \left( \frac{B}{\mu_0 n} \right)^2 = \frac{1}{2} (\mu_0 n^2 Al) \left( \frac{B^2}{\mu_0^2 n^2} \right) = \frac{B^2 Al}{2 \mu_0}$.The magnetic energy per unit volume $(u_B)$ is defined as the total energy divided by the volume $(V = Al)$:$u_B = \frac{U_B}{V} = \frac{B^2 Al / (2 \mu_0)}{Al} = \frac{B^2}{2 \mu_0}$.

The expression for the magnetic energy per unit volume stored in a solenoid in terms of magnetic field is . . . . . . .

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