When an ac source v is connected to a pure inductor, Kirchhoff’s loop rule gives \( v-L\frac{di}{dt}=0 \). Which law explains the negative sign?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Lenz’s law
In \( v-L\,di/dt=0 \), the term \( -L\,di/dt \) is the self-induced emf of the inductor. It appears whenever the current through the inductor changes.
Option (1) Lenz’s law — Correct. By Lenz’s law the induced emf opposes the change in current that produces it, so it acts against the source when the current grows. This opposition is shown by the negative sign.
Option (2) Ampere’s circuital law — Incorrect. It relates the line integral of the magnetic field around a closed loop to the current through the loop. It says nothing about the direction of an induced emf.
Option (3) Biot-Savart law — Incorrect. It gives the magnetic field produced by a current element. It does not deal with the emf induced by a changing current.
Option (4) Ohm’s law — Incorrect. It relates the voltage across a resistor to the current through it. There is no resistor in this circuit, and the law gives no sign for an induced emf.
NCERT (2026-27):
AC Voltage Applied to an Inductor – “… Using the Kirchhoff’s loop rule, \( \sum\varepsilon(t)=0 \), and since there is no resistor in the circuit,
\( \displaystyle v-L\frac{di}{dt}=0 \qquad (7.10) \)
where the second term is the self-induced Faraday emf in the inductor; and L is the self-inductance of the inductor. The negative sign follows from Lenz’s law (Chapter 6).”