Case: Read the passage and answer the question.
Consider a coil C1 connected to a galvanometer G as shown. A bar magnet is placed near the coil with its North pole facing the coil and its South pole away from it. The magnet is then moved towards the coil.
Answer the following question in this context.
The coil C1 has 100 turns. As the magnet is pushed in, the magnetic flux through each turn rises by \( 4\times10^{-4}\ \mathrm{Wb} \). The total resistance of the coil and the galvanometer is 20 Ω. The charge that flows through the galvanometer is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2\times10^{-3}\ \mathrm{C} \)
The moving magnet changes the flux through the coil, so an emf and a current are induced in it. The charge that flows is the current multiplied by the time for which it flows.
Given:
\( N=100 \)
\( \Delta\Phi_B=4\times10^{-4}\ \mathrm{Wb} \)
\( R=20\ \Omega \)
If the flux changes in a time \( \Delta t \), the average emf and current are
\( \begin{aligned} \varepsilon &= \frac{N\,\Delta\Phi_B}{\Delta t} \\ I &= \frac{\varepsilon}{R} \end{aligned} \)
\( \begin{aligned} q &= I\,\Delta t \\ &= \frac{N\,\Delta\Phi_B}{R} \\ &= \frac{100\times4\times10^{-4}}{20} \\ &= 2\times10^{-3}\ \mathrm{C} \end{aligned} \)
The time \( \Delta t \) does not appear in the result. So the same charge flows whether the magnet is pushed in quickly or slowly.
NCERT (2026-27):
Faraday’s Law of Induction – “In the case of a closely wound coil of N turns, change of flux associated with each turn, is the same. Therefore, the expression for the total induced emf is given by”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)
Summary – “Here ΦB is the flux linked with one turn of the coil. If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.”