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(C) The induced $emf$ across a rotating rod in a magnetic field is given by $\epsilon = \frac{1}{2} B \omega R^2$.Here,$B = H_E = 0.4 \; G = 0.4 	ime

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The induced $emf$ across a rotating rod in a magnetic field is given by $\epsilon = \frac{1}{2} B \omega R^2$.Here,$B = H_E = 0.4 \; G = 0.4 	imes 10^{-4} \; T$.The angular velocity $\omega = 2 \pi 
u$,where $
u = 120 \; rev/min = \frac{120}{60} \; rev/s = 2 \; rev/s$.So,$\omega = 2 \pi 	imes 2 = 4 \pi \; rad/s$.The length of the spoke $R = 0.5 \; m$.Substituting the values: $\epsilon = \frac{1}{2} 	imes (0.4 	imes 10^{-4}) 	imes (4 \pi) 	imes (0.5)^2$.$\epsilon = 0.2 	imes 10^{-4} 	imes 4 \pi 	imes 0.25 = 0.2 	imes 10^{-4} 	imes \pi = 0.2 	imes 3.14159 	imes 10^{-4} \approx 6.28 	imes 10^{-5} \; V$.The number of spokes is immaterial because the $emf$s across the spokes are in parallel.

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