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(B) $(1)$ Analyze the resistance network: The network is a Wheatstone bridge connected between points $B$ and $D$. The arms are $(1\, \Omega, 1\, \Ome

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$170$

Vedclass · Answer

(B) $(1)$ Analyze the resistance network: The network is a Wheatstone bridge connected between points $B$ and $D$. The arms are $(1\, \Omega, 1\, \Omega)$ in series and $(2\, \Omega, 2\, \Omega)$ in series. The central $3\, \Omega$ resistor is connected between $A$ and $C$. Since the bridge is balanced $(\frac{1}{1} = \frac{2}{2})$, the potential at $A$ equals the potential at $C$, so no current flows through the $3\, \Omega$ resistor. The equivalent resistance $R_{eq}$ is the parallel combination of the two branches: $R_{eq} = \frac{(1+1)(2+2)}{(1+1)+(2+2)} = \frac{2 	imes 4}{2+4} = \frac{8}{6} = \frac{4}{3}\, \Omega \approx 1.33\, \Omega$.$(2)$ Calculate the induced electromotive force $(EMF)$: The motional $EMF$ is $\varepsilon = B \ell v$. Here, $B = 1\, T$, $\ell = 5\, cm = 0.05\, m$, and $v = 1\, cm/s = 0.01\, m/s$. Thus, $\varepsilon = 1 	imes 0.05 	imes 0.01 = 5 	imes 10^{-4}\, V$.$(3)$ Calculate the total resistance: $R_{total} = R_{loop} + R_{eq} = 1.7\, \Omega + 1.33\, \Omega = 3.03\, \Omega$.$(4)$ Calculate the current: $i = \frac{\varepsilon}{R_{total}} = \frac{5 	imes 10^{-4}}{3.03} \approx 1.65 	imes 10^{-4}\, A = 165\, \mu A$. The closest option is $170\, \mu A$.

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