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(A) The magnetic field due to a long wire at a distance $y$ is $B = \frac{\mu_0 I}{2 \pi y}$.The motional electromotive force $(EMF)$ induced in the l

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$4$

Vedclass · Answer

(A) The magnetic field due to a long wire at a distance $y$ is $B = \frac{\mu_0 I}{2 \pi y}$.The motional electromotive force $(EMF)$ induced in the loop is due to the motion of the vertical sides of the loop in the $y$-direction.The $EMF$ induced in the side at distance $d$ is $\varepsilon_1 = B_1 l v_y = \left( \frac{\mu_0 I}{2 \pi d} ight) l v_y$.The $EMF$ induced in the side at distance $d+a$ is $\varepsilon_2 = B_2 l v_y = \left( \frac{\mu_0 I}{2 \pi (d+a)} ight) l v_y$.The net $EMF$ is $\varepsilon = \varepsilon_1 - \varepsilon_2 = \frac{\mu_0 I l v_y}{2 \pi} \left( \frac{1}{d} - \frac{1}{d+a} ight)$.Given $I = 10 \ A$,$l = 4 \ cm = 0.04 \ m$,$a = 2 \ cm = 0.02 \ m$,$d = 4 \ cm = 0.04 \ m$,$R = 0.1 \ \Omega$,and $i = 10 \ \mu A = 10^{-5} \ A$.$i = \frac{\varepsilon}{R} \Rightarrow 10^{-5} = \frac{4 \pi 	imes 10^{-7} 	imes 10 	imes 0.04 	imes v_y}{2 \pi 	imes 0.1} \left( \frac{1}{0.04} - \frac{1}{0.06} ight)$.$10^{-5} = 2 	imes 10^{-6} 	imes 0.4 	imes v_y 	imes \left( \frac{0.06 - 0.04}{0.0024} ight) = 8 	imes 10^{-7} 	imes v_y 	imes \left( \frac{0.02}{0.0024} ight) = 8 	imes 10^{-7} 	imes v_y 	imes \frac{25}{3}$.$v_y = \frac{10^{-5} 	imes 3}{8 	imes 10^{-7} 	imes 25} = \frac{300}{200} = 1.5 \ m/s$.Wait,re-evaluating with given dimensions: $l=4cm, a=2cm$. The velocity vector is $\vec{v} = v(\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y}) = v_x \hat{x} + v_y \hat{y}$.Thus $v_y = v/2$ and $v_x = v\sqrt{3}/2$.Using the formula $\varepsilon = \frac{\mu_0 I l v_y}{2 \pi} \left( \frac{1}{d} - \frac{1}{d+a} ight)$:$10^{-5} = \frac{2 	imes 10^{-7} 	imes 10 	imes 0.04 	imes (v/2)}{0.1} \left( \frac{1}{0.04} - \frac{1}{0.06} ight) = 4 	imes 10^{-6} 	imes v 	imes \frac{0.02}{0.0024} = 4 	imes 10^{-6} 	imes v 	imes \frac{25}{3}$.$v = \frac{3 	imes 10^{-5}}{100 	imes 10^{-6}} = 0.3 \ m/s$. Re-checking calculation: $i = 10 \mu A$,$R=0.1 \Omega$,$\varepsilon = 10^{-6} V$.$10^{-6} = \frac{2 	imes 10^{-7} 	imes 10 	imes 0.04 	imes v_y}{0.1} 	imes (25 - 16.66) \approx 2 	imes 10^{-6} 	imes 0.4 	imes v_y 	imes 8.33 \Rightarrow v_y = 0.15$. The provided solution suggests $v=4$.

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