Question

# A thin circular ring of the area $A$is held perpendicular to a uniform magnetic field $B$of induction. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is $R$. When the ring is suddenly squeezed to zero areas, the charge flowing through the galvanometer is: $\frac{BR}{A}$$\frac{AB}{R}$$ABR$None

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Solution

## The correct option is B $\frac{AB}{R}$The explanation for the correct option:Step 1: Given data$\text{Area}=A$$\text{Resistance}=R$$\text{Magneticfield}=B$Step 2: Formula$\mathrm{Ï†}=BA$Step 3: CalculationThe magnetic flux is given by the formula$\mathrm{Ï†}=BA$$I=\frac{âˆ†Q}{âˆ†t}$----------(1)The induced emf is given by:$E=\frac{â–³\mathrm{Ï†}}{â–³t}$Here,$âˆ†\mathrm{Ï†}$ is the change in magnetic flux.$âˆ†t$ is the change in time.After the ring is suddenly squeezed to zero areas,$E=\frac{â–³\mathrm{Ï†}}{â–³t}$$âˆ†\mathrm{Ï†}={\mathrm{Ï†}}_{1}-{\mathrm{Ï†}}_{2}where{\mathrm{Ï†}}_{1}isinitial=0,{\mathrm{Ï†}}_{2}isfinal={\mathrm{Ï†}}_{}$ $=\frac{\mathrm{Ï†}-0}{t}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{Ï†}}{t}----\left(2\right)$Induced emf is produced when the magnetic flux changes.The current is given by$I=\frac{E}{R}$Here, $I$ is the current.$t$ is time.$E$ is the emf.Now substitute the value of current and induced emf in the above equation ( equating equation 1and 2)$\frac{Q}{â–³t}=\frac{BA}{Râ–³t}$ $Q=\frac{AB}{R}$Hence, Option(B) is correct.

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