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(B) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$.The induced $e.m.

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$1$

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(B) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$.The induced $e.m.f.$ in a moving conductor is given by $e = \int B v \, dr$.For the left arm at distance $x_1 = 10\, cm = 0.1\, m$,the magnetic field is $B_1 = \frac{\mu_0 I}{2\pi x_1} = \frac{2 	imes 10^{-7} 	imes 1}{0.1} = 2 	imes 10^{-6}\, T$.The induced $e.m.f.$ in the left arm is $e_1 = B_1 l v = (2 	imes 10^{-6}) 	imes (0.1) 	imes (10) = 2 	imes 10^{-6}\, V = 2\,\mu V$.For the right arm at distance $x_2 = x_1 + l = 0.1 + 0.1 = 0.2\, m$,the magnetic field is $B_2 = \frac{\mu_0 I}{2\pi x_2} = \frac{2 	imes 10^{-7} 	imes 1}{0.2} = 1 	imes 10^{-6}\, T$.The induced $e.m.f.$ in the right arm is $e_2 = B_2 l v = (1 	imes 10^{-6}) 	imes (0.1) 	imes (10) = 1 	imes 10^{-6}\, V = 1\,\mu V$.The net induced $e.m.f.$ is $e = e_1 - e_2 = 2\,\mu V - 1\,\mu V = 1\,\mu V$.

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