A coil is rewound with wire of a different non-magnetic material, so that its resistance becomes three times, while the number of turns, shape and size of the coil stay the same. How does its self-inductance change?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → It is unaffected.
The self-inductance of a coil depends only on its geometry (number of turns, shape and size) and on the permeability of the medium. For a long solenoid, \( L=\mu_0 n^{2}Al \), and the resistance of the wire does not enter it.
Option (1) It becomes 3 times. — Incorrect. The resistance depends on the resistivity of the wire, which has no place in \( L \). The flux linked per unit current, \( N\Phi_B/I \), is set by the turns and the shape and size of the coil.
Option (2) It becomes 9 times. — Incorrect. Since \( L\propto n^{2} \), a ninefold rise would need three times as many turns per unit length, but the number of turns is unchanged.
Option (3) It becomes 1/9 times. — Incorrect. \( L \) could fall only through fewer turns, a smaller coil or a medium of lower permeability. The new wire is non-magnetic, so the medium inside the coil is as before.
Option (4) It is unaffected. — Correct. The number of turns, the shape and the size are the same. The non-magnetic wire leaves the permeability unchanged. So the self-inductance is unaffected, and only the resistance of the coil changes.
NCERT (2026-27):
Self-inductance – “… the self-inductance is,
\( \displaystyle L=\frac{N\Phi_B}{I}=\mu_0 n^{2}Al \qquad (6.15) \)
… The self-inductance of the coil depends on its geometry and on the permeability of the medium.”