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Question

(B) The induced $EMF$ $(\varepsilon)$ in a conductor moving through a magnetic field is given by $\varepsilon = Bv\ell$. First, calculate the velocity

Vedclass · Accepted Answer

$20 \sqrt{10}$

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(B) The induced $EMF$ $(\varepsilon)$ in a conductor moving through a magnetic field is given by $\varepsilon = Bv\ell$. First, calculate the velocity $(v)$ of the wire after falling a distance $(h = 200 \ m)$ under gravity using the equation $v^2 = u^2 + 2gh$. Since the initial velocity $(u = 0)$, $v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 200} = \sqrt{4000} = 20\sqrt{10} \ m/s$. The magnetic field $(B)$ is $0.5 \ Gauss = 0.5 	imes 10^{-4} \ T$. The length of the wire $(\ell)$ is $20 \ m$. Substituting these values into the $EMF$ formula: $\varepsilon = (0.5 	imes 10^{-4} \ T) 	imes (20\sqrt{10} \ m/s) 	imes (20 \ m)$. $\varepsilon = 20\sqrt{10} 	imes 10^{-4} 	imes 10 = 20\sqrt{10} 	imes 10^{-3} \ V$. Since $1 \ V = 1000 \ mV$, the induced $EMF$ is $20\sqrt{10} \ mV$.

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