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(B) The induced electromotive force $(EMF)$ in the moving connector is given by $e = B l v$.Given: $B = 0.1\, T$,$l = 10\, cm = 0.1\, m$,and $v = 1\,

Vedclass · Accepted Answer

$\frac{1}{220}\, A$

Vedclass · Answer

(B) The induced electromotive force $(EMF)$ in the moving connector is given by $e = B l v$.Given: $B = 0.1\, T$,$l = 10\, cm = 0.1\, m$,and $v = 1\, m/s$.$e = 0.1 	imes 0.1 	imes 1 = 0.01\, V$.The connector acts as a battery with internal resistance $r = 1\, \Omega$. The two sides of the loop with resistances $R_1 = 2\, \Omega$ and $R_2 = 3\, \Omega$ are connected in parallel to this battery.The equivalent resistance of the external circuit is $R_{eq} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{2 	imes 3}{2 + 3} = \frac{6}{5} = 1.2\, \Omega$.The total resistance of the circuit is $R_{total} = r + R_{eq} = 1 + 1.2 = 2.2\, \Omega$.The current in the connector is $I = \frac{e}{R_{total}} = \frac{0.01}{2.2} = \frac{1}{220}\, A$.

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