A wheel with 10 metallic spokes each 0.5 m long rotated with a speed of 120 rpm in a plane normal to the horizontal component of earth’s magnetic field \[{B_h}\] at a place. If \[{B_h} = 0.4G\] at the place. What is the induced emf between the axle and the rim of the wheel? (\[1G = {10^{ - 4}}T\])
(A) 0 V
(B) 0.628 mV
(C) 0.628 μV
(D) 62.8 μV

Hint: According to Faraday’s law of electromagnetic induction “When the flux of magnetic field through the area bounded by two consecutive spokes changes, an emf between the axle and the rim of the wheel will produce”.
The emf is given by
\[\varepsilon = - \dfrac{{d\phi }}{{dt}}\]
Where, \[\phi = \int {\vec B.d\vec s} \] is the flux of the magnetic field through the area.

Complete step by step answer:
Given
Number of spokes = 10
Length of each spoke or radius of the wheel (r) = 0.5 m
Angular speed of the wheel = 120 rev/min = \[4\pi {\text{ rad/s}}\]
Earth’s magnetic field (\[{B_h}\]) = 0.4 G =\[0.4 \times {10^4}T\]
Consider the (A) area covered by an angle is θ.
So, \[A = \pi {r^2}\dfrac{\theta }{{2\pi }}\]
\[\therefore A = {r^2}\dfrac{\theta }{2}\]
Now, The induced emf \[\varepsilon = - \dfrac{{d\phi }}{{dt}} = - \dfrac{{d(BA)}}{{dt}}\]
\[ \Rightarrow \varepsilon = - \dfrac{{Bd(A)}}{{dt}}\]……………….. (ii)
Substitute the given values in the equation (ii), we get
\[\Rightarrow \varepsilon = \dfrac{{2\pi \times 2 \times 0.4 \times {{10}^{ - 4}} \times {{(0.5)}^2}}}{2} = 6.28 \times {10^{ - 5}}V\]
\[\therefore \varepsilon = 0.628mV\]

Hence, Option (B) is the correct answer.

Note: Magnetic flux changes by change in:-
(i) Magnitude of magnetic field
(ii) Crossing Area and
(iii) Angle between magnetic field vector and area vector.
The direction of the induced magnetic field in a loop can be obtained by using an electromagnetic induction equation. If the flux increases with time, \[\dfrac{{d\phi }}{{dt}}\]is positive and ε will be negative similarly when \[\dfrac{{d\phi }}{{dt}}\]is negative ε will be positive.