Q150 / 160 · TS EAMCET 2023 (13 May Shift 1) · Physics
The expression for the magnetic energy stored in a solenoid of length $L$, in terms of magnetic field $B$ and area $\mathrm{A}$ is
- Correct answerA$\frac{1}{2 \mu_0} \mathrm{~B}^2 \mathrm{AL}$
- B$\frac{1}{2 \epsilon_0} \mathrm{~B}^2 \mathrm{AL}$
- C$\frac{1}{2 \mu_0} \mathrm{BA}^2 \mathrm{~L}$
- D$\frac{1}{2 \epsilon_0} \mathrm{BA}^2 \mathrm{~L}$
Solution
The expression for the magnetic energy stored in a solenoid is given by
$$
\begin{aligned}
& \frac{U}{V}=\frac{1}{2} \epsilon_0 E^2 \\
& \frac{U}{A L}=\frac{1}{2} \frac{1}{C^2 \mu_0}(B c)^2\left(\because c^2=\frac{1}{\mu_0 \epsilon_0}, c=\frac{E}{B}\right) \\
& U=\frac{A L}{2 c^2 \mu_0} \times B^2 c^2=\frac{1}{2 \mu_o} B^2 A L
\end{aligned}
$$
Hence, the correct answer is option: 'A'.