A player with a 3 text{ m} long iron rod runs | vedclass

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(B) The induced electromotive force (emf) in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $e = Bvl \sin 	he

Vedclass · Accepted Answer

$1 	imes 10^{-3} 	ext{ V}$ in the vertical position and zero in the horizontal position

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(B) The induced electromotive force (emf) in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $e = Bvl \sin 	heta$,where $	heta$ is the angle between the velocity vector and the magnetic field vector. The Earth's horizontal magnetic field $B_H$ acts from South to North.$1$. When the rod is held vertically,the rod is perpendicular to both the velocity (East) and the magnetic field (North). Thus,the rod cuts the magnetic field lines. The induced emf is:$v = 30 	ext{ km/hr} = 30 	imes \frac{5}{18} 	ext{ m/s} = \frac{25}{3} 	ext{ m/s}$$e = B_H \cdot v \cdot l = (4 	imes 10^{-5} 	ext{ Wb/m}^2) 	imes (\frac{25}{3} 	ext{ m/s}) 	imes (3 	ext{ m}) = 1 	imes 10^{-3} 	ext{ V}$.$2$. When the rod is held horizontally and parallel to the direction of motion (East),the rod is parallel to the velocity vector. Since the rod is not cutting the magnetic field lines,the induced emf is $e = 0$.

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