A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across its ends in such a way that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → BA/R
By Faraday’s law, the induced current is \( I = \frac{\varepsilon}{R} = \frac{1}{R}\,\frac{d\Phi}{dt} \) in magnitude. Since \( I = \frac{dq}{dt} \), the charge that flows depends only on the total change in flux, not on how fast the change happens.
\( \displaystyle q = \int I\,dt = \frac{\Delta\Phi}{R} \)
The ring is perpendicular to the field, so the initial flux is BA; after squeezing to zero area the flux is zero.
\( \displaystyle \Delta\Phi = BA - 0 = BA \)
\( \displaystyle q = \frac{BA}{R} \)