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# 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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## 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$Step 2: Formula$\phi =BA$Step 3: CalculationThe magnetic flux is given by the formula$\phi =BA$$I=\frac{∆Q}{∆t}$----------(1)The induced emf is given by:$E=\frac{△\phi }{△t}$Here,$∆\phi$ is the change in magnetic flux.$∆t$ is the change in time.After the ring is suddenly squeezed to zero areas,$E=\frac{△\phi }{△t}$$∆\phi ={\phi }_{1}-{\phi }_{2}where{\phi }_{1}isinitial=0,{\phi }_{2}isfinal={\phi }_{}$ $=\frac{\phi -0}{t}\phantom{\rule{0ex}{0ex}}=\frac{\phi }{t}----\left(2\right)$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)$\frac{Q}{△t}=\frac{BA}{R△t}$ $Q=\frac{AB}{R}$Hence, Option(B) is correct.  Suggest Corrections  0      Explore more