The magnetic energy stored in a solenoid of cross sectional area A and length l is
(Symbols have their usual meaning)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\mathrm{B}^{2}\mathrm{Al}}{2\mu_o} \)
The energy stored in a solenoid carrying current I is \( \frac{1}{2}LI^{2} \). For a long solenoid with n turns per unit length, the self-inductance and the field inside are
\( L = \mu_0 n^{2}Al \)
\( B = \mu_0 nI \)
Eliminate I using \( I = B/\mu_0 n \):
\( \displaystyle U = \frac{1}{2}\,\mu_0 n^{2}Al \left(\frac{B}{\mu_0 n}\right)^{2} \)
\( \displaystyle U = \frac{B^{2}Al}{2\mu_0} \)
This is the magnetic energy density \( B^{2}/2\mu_0 \) multiplied by the volume Al of the solenoid.
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
\( \displaystyle U_B = \frac{1}{2}LI^2 = \frac{1}{2}L\left(\frac{B}{\mu_0 n}\right)^2 \qquad \left(\text{since } B = \mu_0 nI \text{ for a solenoid}\right) \)
\( \displaystyle = \frac{1}{2}\left(\mu_0 n^2 A l\right)\left(\frac{B}{\mu_0 n}\right)^2 \qquad [\text{from Eq. (6.15)}] \)
\( \displaystyle = \frac{1}{2\mu_0}B^2 A l \)”