The figure shows a square loop L of side 5, c | vedclass

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(D) $1$. Analyze the resistance network: The network is a Wheatstone bridge with arms $1\, \Omega, 2\, \Omega, 1\, \Omega, 2\, \Omega$ and a central $

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) $1$. Analyze the resistance network: The network is a Wheatstone bridge with arms $1\, \Omega, 2\, \Omega, 1\, \Omega, 2\, \Omega$ and a central $3\, \Omega$ resistor. Since $\frac{1}{1} = \frac{2}{2}$,it is a balanced Wheatstone bridge. The central $3\, \Omega$ resistor carries no current. The equivalent resistance of the bridge is $R_{eq} = \frac{(1+2)(1+2)}{(1+2)+(1+2)} = \frac{3 	imes 3}{3+3} = 1.5\, \Omega$.$2$. Calculate the induced electromotive force $(EMF)$: The motional $EMF$ is given by $\varepsilon = B \ell v$. 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: The total resistance $R_{total} = R_{loop} + R_{eq} = 1.7\, \Omega + 1.5\, \Omega = 3.2\, \Omega$.$4$. Calculate the current: The current $i = \frac{\varepsilon}{R_{total}} = \frac{5 	imes 10^{-4}}{3.2} \approx 1.5625 	imes 10^{-4}\, A = 156.25\, \mu A$. The closest option is $150\, \mu A$ or $170\, \mu A$. Re-evaluating the bridge: $R_{eq} = \frac{3 	imes 3}{3+3} = 1.5\, \Omega$. Total $R = 1.7 + 1.5 = 3.2\, \Omega$. $i = 156.25\, \mu A$. Given the options,$150\, \mu A$ is the closest value.

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