A wheel with ten metallic spokes,each 0.50 m | vedclass

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(D) The induced $e.m.f.$ $(e)$ across a rotating spoke is given by the formula $e = \frac{1}{2} B \omega l^2$.Given:$B = 0.4 \ G = 0.4 	imes 10^{-4}

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$6.28 	imes 10^{-5} \ V$

Vedclass · Answer

(D) The induced $e.m.f.$ $(e)$ across a rotating spoke is given by the formula $e = \frac{1}{2} B \omega l^2$.Given:$B = 0.4 \ G = 0.4 	imes 10^{-4} \ T$$l = 0.50 \ m$Frequency $
u = 120 \ rev/min = \frac{120}{60} \ rev/s = 2 \ rev/s$Angular velocity $\omega = 2 \pi 
u = 2 	imes 3.14 	imes 2 = 12.56 \ rad/s$Substituting the values:$e = \frac{1}{2} 	imes (0.4 	imes 10^{-4}) 	imes (12.56) 	imes (0.5)^2$$e = 0.2 	imes 10^{-4} 	imes 12.56 	imes 0.25$$e = 0.05 	imes 12.56 	imes 10^{-4}$$e = 0.628 	imes 10^{-4} \ V = 6.28 	imes 10^{-5} \ V$.

$A$ wheel with ten metallic spokes,each $0.50 \ m$ long,is rotated with a speed of $120 \ rev/min$ in a plane normal to the earth's magnetic field at the place. If the magnitude of the field is $0.4 \ G$,the induced $e.m.f.$ between the axle and the rim of the wheel is equal to:

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