The magnetic energy stored in a current carrying solenoid is (if B = magnetic field, l = length of solenoid, A = area of cross section of solenoid) :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{B^2 A l}{2 \mu_0} \)
The energy stored in an inductor is \( U = \frac{1}{2}LI^{2} \). For a long solenoid with \( n \) turns per unit length, the self-inductance and the field inside are
\( \displaystyle L = \mu_0 n^{2} A l, \qquad B = \mu_0 n I \;\Rightarrow\; I = \frac{B}{\mu_0 n} \)
Substituting,
\( \displaystyle U = \frac{1}{2}\,\mu_0 n^{2} A l \times \frac{B^{2}}{\mu_0^{2} n^{2}} = \frac{B^{2} A l}{2\mu_0} \)
So the energy density is \( B^{2}/2\mu_0 \), and the field fills the volume \( A l \) 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 – …
