A metallic wire loop of side l = 0.1 text{ m} | vedclass

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(B) Given: Side length $l = 0.1 	ext{ m}$, Resistance of loop $R_{loop} = 1 \Omega$, Magnetic field $B = 2 	ext{ Wb m}^{-2}$, Current $I = 1 	ext{

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$2$

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(B) Given: Side length $l = 0.1 	ext{ m}$, Resistance of loop $R_{loop} = 1 \Omega$, Magnetic field $B = 2 	ext{ Wb m}^{-2}$, Current $I = 1 	ext{ mA} = 10^{-3} 	ext{ A}$.First, calculate the equivalent resistance of the resistor network connected to the loop. The network consists of four $3 \Omega$ resistors in a bridge-like configuration. The equivalent resistance of this network is $R_{net} = 3 \Omega$.The total resistance of the circuit is $R_{total} = R_{loop} + R_{net} = 1 \Omega + 3 \Omega = 4 \Omega$.The motional electromotive force (emf) induced in the loop is given by $E = Bvl$.Using Ohm's law, $I = \frac{E}{R_{total}} = \frac{Bvl}{R_{total}}$.Substituting the values: $10^{-3} = \frac{2 	imes v 	imes 0.1}{4}$.$10^{-3} = \frac{0.2v}{4} = 0.05v$.$v = \frac{10^{-3}}{0.05} = 0.02 	ext{ m s}^{-1} = 2 	ext{ cm s}^{-1}$.

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