Question

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min, in a plane normal to earth's magnetic field at the place. If the magnitude of the field is 0.40 G, what is the induced e.m.f. between the axle and rim of the wheel?

Answer 1

Hint: Use concept of induced e.m.f. and use formula according to direction of magnetic field. Convert rev/min into rev/sec. convert gauss (G) into Tesla. To get a better imagination, take an example of a cycle.

Complete step by step answer:
 Before proceeding for a solution, let’s understand what the question wants to convey.
Calculate induced e.m.f between axle and rim of the wheel.
Given data:
Metallic spoke =N = 10,
Length of spoke (l) = 0.5m
Magnet field (B) = 0.4 gauss = $0.4*{{10}^{-4}}$T
Frequency = 120 rev/min
Convert it in to minute, we get
Frequency (f) = $\dfrac{120}{60}=2Hz$= 2 rev/sec
We know that one revolution is $2\pi $ radian.
1 rev = $2\pi $ radian
Form frequency we can calculate angular frequency.
$\begin{align}
  & \omega =2\pi f \\
 & \omega =2*\pi *2 \\
\end{align}$
$\omega $=4π radian/sec
We know that, velocity of wheel at rim = $\omega $r
And velocity of wheel at axel = 0
(Since wheel is rotate at axel therefore velocity will be zero and will be maximum at rim)
 Average velocity is given by
Avg.velocity =$\dfrac{0+\omega r}{2}=\dfrac{\omega r}{2}$
Formula of induced e.m.f (e) is
e = $Bvl$
$\begin{align}
  & e=(0.4\times {{10}^{-4}})\times (\dfrac{1}{2}\times 4\pi \times \dfrac{1}{2})\times \dfrac{1}{2} \\
 & e=0.4\times {{10}^{-4}}\times \dfrac{22}{7}\times 0.5 \\
 & e=6.28\times {{10}^{-5}}volt \\
\end{align}$
 Hence induced e.m.f calculated.
Note: It is assumed that all the spokes are parallel to each other, therefore induced e.m.f will be the same everywhere. Note that in question the magnetic field is perpendicular to the plane. Angle between B and plane leads to change the formula of e.m.f.