A square metallic wire loop of side 0.1 , m a | vedclass

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(B) $1$. The network of resistances is a balanced Wheatstone bridge. The equivalent resistance of the bridge between points $P$ and $Q$ is $3 \, \Omeg

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$2$

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(B) $1$. The network of resistances is a balanced Wheatstone bridge. The equivalent resistance of the bridge between points $P$ and $Q$ is $3 \, \Omega$.$2$. The total resistance of the circuit is $R_{total} = R_{loop} + R_{bridge} = 1 \, \Omega + 3 \, \Omega = 4 \, \Omega$.$3$. The motional electromotive force (emf) induced in the loop is given by $e = Bvl$, where $B = 2 \, Wb/m^2$, $l = 0.1 \, m$, and $v$ is the velocity.$4$. The induced current is $i = \frac{e}{R_{total}} = \frac{Bvl}{R_{total}}$.$5$. Given $i = 1 \, mA = 10^{-3} \, A$, we have $10^{-3} = \frac{2 	imes v 	imes 0.1}{4}$.$6$. Solving for $v$: $10^{-3} = \frac{0.2v}{4} \Rightarrow 10^{-3} = 0.05v \Rightarrow v = \frac{10^{-3}}{0.05} = 0.02 \, m/s$.$7$. Converting to $cm/s$: $v = 0.02 	imes 100 \, cm/s = 2 \, cm/s$.

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