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(D) The $EMF$ induced in a rotating rod (spoke) of length $r$ in a magnetic field $B$ is given by $\varepsilon = \frac{1}{2} B_{\perp} \omega r^2$, wh

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$	an^{-1}\left(\frac{3}{4}ight)$

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(D) The $EMF$ induced in a rotating rod (spoke) of length $r$ in a magnetic field $B$ is given by $\varepsilon = \frac{1}{2} B_{\perp} \omega r^2$, where $B_{\perp}$ is the component of the magnetic field perpendicular to the plane of rotation.Since the plane of the wheel is vertical along East-West, the magnetic field component perpendicular to the plane is the horizontal component $B_{H}$.Given: $\omega = 100 	ext{ rpm} = \frac{100 	imes 2\pi}{60} = \frac{10\pi}{3} 	ext{ rad/s}$, $r = 0.3 \, m$, $\varepsilon = 3\pi 	imes 10^{-6} \, V$, $B_{V} = 15 	imes 10^{-6} \, T$.Substituting the values: $3\pi 	imes 10^{-6} = \frac{1}{2} 	imes B_{H} 	imes \frac{10\pi}{3} 	imes (0.3)^2$.$3\pi 	imes 10^{-6} = B_{H} 	imes \frac{5\pi}{3} 	imes 0.09 = B_{H} 	imes 0.15\pi$.$B_{H} = \frac{3\pi 	imes 10^{-6}}{0.15\pi} = 20 	imes 10^{-6} \, T = 20 \mu T$.The angle of dip $\delta$ is given by $	an \delta = \frac{B_{V}}{B_{H}} = \frac{15}{20} = \frac{3}{4}$.Therefore, $\delta = 	an^{-1}\left(\frac{3}{4}ight)$.

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