A horizontal telegraph wire of length 30 m s | vedclass

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(A) $1$. The wire falls freely under gravity from a height $h = 20 \ m$. The velocity $v$ of the wire just before hitting the ground is given by $v =

Vedclass · Accepted Answer

$0.3 \ mA$

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(A) $1$. The wire falls freely under gravity from a height $h = 20 \ m$. The velocity $v$ of the wire just before hitting the ground is given by $v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 20} = \sqrt{400} = 20 \ m \ s^{-1}$.$2$. The induced electromotive force $(EMF)$ in a conductor moving through a magnetic field is given by $\epsilon = Bvl$,where $B$ is the magnetic field component perpendicular to the velocity and length,$v$ is the velocity,and $l$ is the length of the wire.$3$. Here,$B = 2 	imes 10^{-5} \ T$,$v = 20 \ m \ s^{-1}$,and $l = 30 \ m$.$4$. The induced $EMF$ is $\epsilon = (2 	imes 10^{-5}) 	imes 20 	imes 30 = 1200 	imes 10^{-5} = 1.2 	imes 10^{-2} \ V$.$5$. The induced current $I$ is given by $I = \frac{\epsilon}{R}$,where $R = 40 \ \Omega$.$6$. $I = \frac{1.2 	imes 10^{-2}}{40} = 0.03 	imes 10^{-2} \ A = 3 	imes 10^{-4} \ A = 0.3 \ mA$.

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