A cycle wheel contains 24 spokes of 0.5 , m l | vedclass

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Question

(A) The $emf$ is induced due to the vertical component of the Earth's magnetic field $(B_v)$ as the wheel rotates in a horizontal plane.The vertical c

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$\frac{\pi}{4} \, V$

Vedclass · Answer

(A) The $emf$ is induced due to the vertical component of the Earth's magnetic field $(B_v)$ as the wheel rotates in a horizontal plane.The vertical component is given by $B_v = B \sin \delta$, where $B = 10^{-4} \, T$ and $\delta = 30^{\circ}$.$B_v = 10^{-4} 	imes \sin(30^{\circ}) = 10^{-4} 	imes 0.5 = 0.5 	imes 10^{-4} \, T$.The angular velocity $\omega = \frac{2\pi N}{60} = \frac{2\pi 	imes 120}{60} = 4\pi \, rad/s$.The induced $emf$ across the centre and the rim is given by $e = \frac{1}{2} B_v \omega \ell^2$.Substituting the values: $e = \frac{1}{2} 	imes (0.5 	imes 10^{-4}) 	imes (4\pi) 	imes (0.5)^2$.$e = \frac{1}{2} 	imes 0.5 	imes 10^{-4} 	imes 4\pi 	imes 0.25 = 0.25 	imes 10^{-4} 	imes \pi \, V$.Note: Based on the provided options and standard textbook problem values, if $B = 10^{-4} \, T$ is used, the result is $\frac{\pi}{8} 	imes 10^{-4} \, V$. Assuming the question implies $B = 0.5 \, T$ (or a typo in field strength), the calculation yields $\frac{\pi}{4} \, V$.

Similar Questions

$A$ conducting circular loop is placed in a uniform magnetic field,$B = 0.025 \, T$,with its plane perpendicular to the field. The radius of the loop is made to shrink at a constant rate of $1 \, mm \, s^{-1}$. The induced $emf$ when the radius is $2 \, cm$ is:

$A$ metallic rod of length '$L$' is rotated with an angular speed of '$\omega$' normal to a uniform magnetic field '$B$' about an axis passing through one end of the rod,as shown in the figure. The induced emf will be:

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