A wheel with 20 metallic spokes,each 1 ,m lon | vedclass

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(B) Consider one spoke $(OP)$ as shown in the diagram.The induced emf across one spoke $(OP)$ is given by:$e = \frac{B \omega l^2}{2}$Given:$B = 0.4 \

Vedclass · Accepted Answer

$2.51 	imes 10^{-4} \;V$

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(B) Consider one spoke $(OP)$ as shown in the diagram.The induced emf across one spoke $(OP)$ is given by:$e = \frac{B \omega l^2}{2}$Given:$B = 0.4 \,G = 0.4 	imes 10^{-4} \,T$$l = 1 \,m$$f = 120 \,rpm = \frac{120}{60} \,rps = 2 \,rps$Angular velocity $\omega = 2 \pi f = 2 \pi 	imes 2 = 4 \pi \,rad/s$Substituting the values:$e = \frac{1}{2} 	imes (0.4 	imes 10^{-4}) 	imes (4 \pi) 	imes (1)^2$$e = 0.2 	imes 10^{-4} 	imes 4 	imes 3.14159$$e = 0.8 	imes 3.14159 	imes 10^{-4} \approx 2.51 	imes 10^{-4} \,V$Since all spokes are connected in parallel between the axle and the rim,the net induced emf is equal to the emf induced in a single spoke:$e_{	ext{Net}} = e = 2.51 	imes 10^{-4} \,V$

$A$ wheel with $20$ metallic spokes,each $1 \,m$ long,is rotated with a speed of $120 \,rpm$ in a plane perpendicular to a magnetic field of $0.4 \,G$. The induced emf between the axle and the rim of the wheel will be $\left(1 \;G = 10^{-4} \;T \right)$.

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