Case: Read the passage and answer the question.
The number of magnetic field lines crossing a surface is called the magnetic flux. An electric current can be induced in a coil by a flux change produced by another coil in its vicinity or by a flux change produced by the same coil. The former is called mutual induction and the latter is called self-induction.
A steady current of 2 A in a coil gives a flux linkage \( N\Phi_B \) of 0.4 Wb. The current is then changed at the rate of \( 5\ \mathrm{A\,s^{-1}} \). The magnitude of the emf induced in the coil is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 V
The flux linkage of a coil is proportional to the current in it, \( N\Phi_B=LI \). Here \( L \) is the self-inductance of the coil. When the current changes, the emf induced in the coil has the magnitude \( L\,dI/dt \).
Given:
\( N\Phi_B=0.4\ \mathrm{Wb} \)
\( I=2\ \mathrm{A} \)
\( dI/dt=5\ \mathrm{A\,s^{-1}} \)
\( \begin{aligned} L &= \frac{N\Phi_B}{I} \\ &= \frac{0.4}{2} \\ &= 0.2\ \mathrm{H} \end{aligned} \)
\( \begin{aligned} |\varepsilon| &= L\frac{dI}{dt} \\ &= 0.2\times5 \\ &= 1\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Summary – “5. Inductance is the ratio of the flux-linkage to current. It is equal to NΦ/I.”
Self-inductance – “When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using … the induced emf is given by”
\( \displaystyle \varepsilon=-\frac{d(N\Phi_B)}{dt} \)
\( \displaystyle \varepsilon=-L\frac{dI}{dt} \)\( \qquad (6.14) \)