The current flowing through an inductor is halved. How does its self-inductance change?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → It remains unchanged.
The self-inductance of a coil is defined by \( N\Phi_B=LI \). It depends only on the geometry of the coil and the permeability of the medium. The current through the coil does not change it.
Option (1) It is halved. — Incorrect. Halving the current halves the flux linkage \( N\Phi_B \) as well, so the ratio \( N\Phi_B/I \), which is \( L \), stays the same.
Option (2) It is doubled. — Incorrect. The current appears in the denominator of \( L=N\Phi_B/I \), but the flux in the numerator falls in the same proportion. \( L \) would change only if the turns, the size or shape of the coil, or its core changed.
Option (3) It becomes four times. — Incorrect. Nothing in \( L \) varies as the square of the current. It is the stored energy \( LI^{2}/2 \) that depends on \( I^{2} \), and it falls to one-quarter.
Option (4) It remains unchanged. — Correct. The coil and its core are the same, so the self-inductance is unchanged. For a long solenoid, \( L=\mu_0 n^{2}Al \) has no current term in it.
NCERT (2026-27):
Self-inductance – “The self-inductance of the coil depends on its geometry and on the permeability of the medium.”