A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of $50$ revolutions per minute. If the length of each spoke is $30.0cm$ and the horizontal component of the earth’s magnetic field is $4 \times {10^{ - 5}}T$, find the emf induced between the axis and the outer end of a spoke. Neglect centripetal force acting on the free electrons of the spoke.

Hint: When charged particles move with uniform acceleration inside a uniform magnetic field, an electric field is formed. That electric field then induces an emf. Any particle performing circular motion is always under uniform acceleration, because even though it’s angular velocity remains constant, its direction keeps changing.

Complete step by step answer:
Angular speed of the wheel $\left( \omega \right)$ is given as,
$
  \omega = 50\dfrac{{rev}}{{\min }} \\
   \Rightarrow \omega = 100\pi \dfrac{{rad}}{{\min }} \\
   \Rightarrow \omega = \dfrac{{100\pi }}{{60}}\dfrac{{rad}}{{\sec }} \\
   \Rightarrow \omega = \dfrac{{5\pi }}{3}\dfrac{{rad}}{{\sec }} $
The emf $\left( \varepsilon \right)$ induced by the moving charged particles can be calculated by,
$\Rightarrow \varepsilon = \dfrac{{\omega B{l^2}}}{2}$
Where, $B = $Horizontal component of earth’s magnetic field,
$l = $ Length of the spoke
$\Rightarrow \varepsilon = \dfrac{{5\pi \times 4 \times {{10}^{ - 5}} \times 0.3 \times 0.3}}{{3 \times 2}}$
$\Rightarrow \varepsilon = 9.42\mu V$

Hence the emf induced is $9.42\mu V$.

Note: Any charge at rest always emits uniform electric field. When a charge moves at a uniform velocity creates a magnetic field. When a charge moves with uniform acceleration it produces an electromagnetic field. In that electromagnetic field, electric fields and magnetic fields oscillate perpendicular to each other.