A bicycle wheel of radius 0.4, m has 20 spoke | vedclass

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(C) The induced $emf$ across the ends of each spoke is given by the formula:$e = \frac{1}{2} B l^2 \omega$where $B$ is the magnetic field, $l$ is the

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$6 	imes 10^{-5} \,V$

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(C) The induced $emf$ across the ends of each spoke is given by the formula:$e = \frac{1}{2} B l^2 \omega$where $B$ is the magnetic field, $l$ is the length of each spoke (which is equal to the radius of the wheel), and $\omega$ is the angular velocity.Given:Radius $l = 0.4\, m$Magnetic field $B = 0.4 	imes 10^{-4}\, T$Frequency $f = 180\, 	ext{rpm} = \frac{180}{60} = 3\, 	ext{rev/s}$Angular velocity $\omega = 2 \pi f = 2 \pi 	imes 3 = 6 \pi\, 	ext{rad/s}$Since all spokes are connected in parallel, the total induced $emf$ between the rim and the centre is the same as that induced in a single spoke.$e = \frac{1}{2} 	imes (0.4 	imes 10^{-4}) 	imes (0.4)^2 	imes (2 \pi 	imes 3)$$e = \frac{1}{2} 	imes 0.4 	imes 10^{-4} 	imes 0.16 	imes 6 \pi$$e = 0.2 	imes 10^{-4} 	imes 0.16 	imes 6 	imes 3.14$$e \approx 6.03 	imes 10^{-5}\, V$Rounding to the nearest option, we get $6 	imes 10^{-5}\, V$.

$A$ bicycle wheel of radius $0.4\, m$ has $20$ spokes. It is rotating at the rate of $180$ revolutions per minute, perpendicular to the horizontal component of the Earth's magnetic field of $0.4 \times 10^{-4}\, T$. The $emf$ induced between the rim and the centre of the wheel will be

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