Question 166: Electromagnetic Induction

Question

A $20 \mathrm{~cm}$ long metallic rod is rotated with $210~ \mathrm{rpm}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field $0.2 \mathrm{~T}$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is ____________ $\mathrm{mV}$.

Take $\pi=\frac{22}{7}$

Accepted Answer

Given that the rod is rotating at 210 rpm, we first convert this to radians per second:
$\omega = 210 \cdot \frac{2\pi \mathrm{rad}}{60 \mathrm{s}} = 22 \mathrm{rad/s}$
Now, we can find the linear velocity $v$ of the tip of the rod:
$v = \omega r$
where $r$ is the length of the rod (0.2 m).
$v = 22 \mathrm{rad/s} \cdot 0.2 \mathrm{m} = 4.4 \mathrm{m/s}$
Now, we can find the emf developed between the center and the ring using the formula:
$\epsilon = \frac{1}{2} B\ell v$
where $B$ is the magnetic field (0.2 T), $\ell$ is the length of the rod (0.2 m), and $v$ is the linear velocity (4.4 m/s).
$\epsilon = \frac{1}{2} \cdot 0.2 \mathrm{T} \cdot 0.2 \mathrm{m} \cdot 4.4 \mathrm{m/s} = 0.088 \mathrm{V}$
To express this value in mV, we can simply multiply it by 1000:
$\epsilon = 0.088 \mathrm{V} \cdot 1000 = 88 \mathrm{mV}$
So the emf developed between the center and the ring is 88 mV.

Electromagnetic Induction

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