The magnetic energy stored in an induction coil (choose the correct statement).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Depends on squared value of magnetic field
The energy stored in the magnetic field of a coil of self inductance L carrying current I is \( U = \frac{1}{2}LI^{2} \). The same energy can be written through the field: the magnetic energy per unit volume is \( u_B = \frac{B^{2}}{2\mu_0} \), so for a solenoid of cross-section A and length l:
\( \displaystyle U = u_B\times(Al) = \frac{B^{2}}{2\mu_0}\,Al \)
Each statement is now judged.
Option (1) Does not depend on magnitude of current — Incorrect. \( U = \frac{1}{2}LI^{2} \) depends on the square of the current.
Option (2) Depends on squared value of magnetic field — Correct. \( U\propto B^{2} \) for a given coil, as the expression above shows.
Option (3) Does not depend on self inductance of the coil — Incorrect. L appears directly in \( U = \frac{1}{2}LI^{2} \).
Option (4) Does not depend on area of coil — Incorrect. The volume \( Al \) containing the field appears in \( U = \frac{B^{2}}{2\mu_0}Al \), and L itself is proportional to A.
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. …
\( \displaystyle U_B = \frac{1}{2\mu_0} B^2 A l \)
… In both the cases energy is proportional to the square of the field strength.”