If the magnetic flux through each turn is \( \Phi \), the self-inductance of an N-turn coil carrying a current I is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \frac{N\Phi}{I} \)
In a coil of \( N \) turns the same flux \( \Phi \) passes through every turn, so the flux linkage of the coil is \( N\Phi \). It is proportional to the current in the coil, so \( N\Phi=LI \). The constant of proportionality \( L \) is the self-inductance.
Option (1) \( N\Phi I \) — Incorrect. Here the flux linkage is multiplied by the current. In \( L \) it must be divided by the current. This product would also grow as the square of the current, since \( \Phi \) itself is proportional to \( I \).
Option (2) \( \frac{N\Phi}{I} \) — Correct. From \( N\Phi=LI \), the self-inductance \( L=N\Phi/I \) is the flux linkage per unit current.
Option (3) \( \frac{N}{\Phi I} \) — Incorrect. With the flux in the denominator, a coil that links more flux for the same current would have a smaller inductance, the reverse of \( N\Phi=LI \).
Option (4) \( \frac{\Phi I}{N} \) — Incorrect. It places the current in the numerator and the turns in the denominator. But \( L \) rises with the number of turns linking the flux, and it is the linkage divided by the current.
NCERT (2026-27):
Self-inductance – “… In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as
\( N\Phi_B\propto I \)
\( N\Phi_B=L\,I \)\( \qquad (6.13) \)
where constant of proportionality L is called self-inductance of the coil.”