A wheel with 10 metallic spokes,each 0.5 m l | vedclass

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Question

(C) Given: Length of each spoke $R = 0.5 \ m$,angular speed $\omega = 120 \ rev/min$,and magnetic field $B = H_E = 0.4 \ G = 0.4 	imes 10^{-4} \ T$.F

Vedclass · Accepted Answer

$6.28 	imes 10^{-2} \ mV$

Vedclass · Answer

(C) Given: Length of each spoke $R = 0.5 \ m$,angular speed $\omega = 120 \ rev/min$,and magnetic field $B = H_E = 0.4 \ G = 0.4 	imes 10^{-4} \ T$.First,convert the angular speed to $rad/s$:$\omega = \frac{120 	imes 2\pi}{60} = 4\pi \ rad/s$.The induced emf across the spokes of a rotating wheel in a magnetic field is given by the formula:$\varepsilon = \frac{1}{2} B \omega R^2$.Substituting the values:$\varepsilon = \frac{1}{2} 	imes (0.4 	imes 10^{-4} \ T) 	imes (4\pi \ rad/s) 	imes (0.5 \ m)^2$.$\varepsilon = 0.2 	imes 10^{-4} 	imes 4\pi 	imes 0.25$.$\varepsilon = 0.2 	imes 10^{-4} 	imes \pi$.$\varepsilon = 0.2 	imes 3.14159 	imes 10^{-4} \ V$.$\varepsilon = 0.6283 	imes 10^{-4} \ V = 6.283 	imes 10^{-5} \ V$.Converting to millivolts $(mV)$:$\varepsilon = 6.283 	imes 10^{-5} 	imes 10^3 \ mV = 6.283 	imes 10^{-2} \ mV$.Thus,the correct option is $C$.

$A$ wheel with $10$ metallic spokes,each $0.5 \ m$ long,is rotated with a speed of $120 \ rev/min$ in a plane normal to the horizontal component of Earth's magnetic field $H_E$ at a place. If $H_E = 0.4 \ G$ at the place,then the induced emf is: $(1 \ G = 10^{-4} \ T)$

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