The expression of the magnetic energy stored in a solenoid in terms of magnetic field \( B \), cross-sectional area \( A \) and length \( l \) of the solenoid, is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( U_B = \displaystyle\frac{1}{2\mu_0}B^{2}Al \)
The energy stored in an inductor is \( U = \tfrac{1}{2}LI^{2} \). For a long solenoid with \( n \) turns per unit length, the self-inductance is \( L = \mu_0 n^{2}Al \) and the field is \( B = \mu_0 n I \), so \( I = B/(\mu_0 n) \).
\( \displaystyle U_B = \frac{1}{2}\,(\mu_0 n^{2} A l)\left(\frac{B}{\mu_0 n}\right)^{2} \)
\( \displaystyle U_B = \frac{B^{2}}{2\mu_0}\,A l \)
The product \( Al \) is the volume of the solenoid, so the energy per unit volume is \( B^{2}/2\mu_0 \).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Self-Inductance – “… Example 6.9 (a) 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.”
Self-Inductance – “… Solution (a) From Eq. (6.17), the magnetic energy is
\( \displaystyle U_B = \frac{1}{2}LI^2 = \frac{1}{2}L\left(\frac{B}{\mu_0 n}\right)^2 = \frac{1}{2}\left(\mu_0 n^2 A l\right)\left(\frac{B}{\mu_0 n}\right)^2 = \frac{1}{2\mu_0}B^2 A l \)”