Question

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 ${B_h} = 0.4\;{{G}}$at the place. What is the induced emf between the axle and the rim of the wheel? ($1\;{{G = 1}}{{{0}}^{ - 4}}\;{{T}}$ )
(A) $0\;{{V}}$
(B) $0.628\;{{mV}}$
(C) $0.628\;{{\mu V}}$
(D) $62.8\;{{\mu V}}$

Answer 1

Hint: The relative motion of the magnet will produce an induced current because of the induced electromotive force. The change of magnetic flux per unit time is the induced emf. Where, the magnetic flux is the product of the magnetic field and the area covered. The change in the magnetic flux only brings the induced emf between the axle and the rim of the wheel.

Complete step by step solution:
Given data:-
Length of the metallic spoke,$r = 0.5\;{{m}}$
 Linear speed of the wheel, $v = 120\;{{rev/min = }}\dfrac{{120}}{{60}}{{rev/s = 2}}\;{{rev/s}}$
Earth’s magnetic field , ${B_h} = 0.4\;{{G = 0}}{{.4}} \times {{1}}{{{0}}^{ - 4}}\;{{T}}$
The area covered by an angle $\theta $, where $r$ is the radius is given as,
$
  A = \pi {r^2}\dfrac{\theta }{{2\pi }} \\
   = \dfrac{{{r^2}}}{2}\theta \\
$
The expression for the induced emf is given as,
$E = \dfrac{{d\phi }}{{dt}}$
Where $\phi $is the magnetic flux. Magnetic flux is the product of magnetic fields and the area covered.
Therefore, Magnetic flux $\phi = BA$
Where B is the magnetic field and A is the area covered.
Substituting above expression for induced emf,
$
  E = \dfrac{{d\phi }}{{dt}} \\
   = \dfrac{{d(BA)}}{{dt}} \\
$
Substituting in the above expression. We get,
$
  E = \dfrac{{d(B\dfrac{{{r^2}}}{2}\theta )}}{{dt}} \\
   = B\dfrac{{{r^2}}}{2}\dfrac{{d\theta }}{{dt}} \\
$
The angle covered in unit time is the angular speed $\omega $ .
$E = B\dfrac{{{r^2}}}{2}\omega $
Therefore the induced emf is given as,
$E = \dfrac{1}{2}B{r^2}\omega $
The angular speed $\omega = 2\pi v$
Substituting the value of linear speed,
\[\omega = 2\pi \times {{2}}\;{{rev/s}} = 4\pi \;{{rev/s}}\]
Substituting the values in the above expression,
$
  E = \dfrac{1}{2} \times {{0}}{{.4}} \times {{1}}{{{0}}^{ - 4}}\;{{T}} \times {\left( {0.5\;{{m}}} \right)^2} \times 4\pi \;{{rev/s}} \\
  {{ = 6}}{{.28}} \times {{1}}{{{0}}^{ - 5}}\;{{V}} \\
  {{ = 0}}{{.628}}\;{{\mu V}} \\
$
Thus the induced emf between the axle and the rim of the wheel is ${{0}}{{.628}}\;{{\mu V}}$.

Thus option C is correct.

Note:
We can find the induced emf between the axle and the rim of the wheel by following the method We have the number of spokes ,$N = 10$ . The length of each spoke ,$l = 0.5\;{{m}}$ ,magnetic field $B = {{0}}{{.4}} \times {{1}}{{{0}}^{ - 4}}\;{{T}}$and the linear speed $v = {{2}}\;{{rev/s}}$.