Question

A $1.0m$ long metallic rod is rotated with an angular frequency of $400\text{rad/s}$ 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 of $0.5T$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Answer 1

As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force. Thus, the resulting separation of charges produces an emf across the ends of the rod.
At a certain value of emf, there is no more flow of electrons, and a steady state is reached. The magnitude of the emf $\left(ϵ\right)$ generated, across a length 𝑑𝑟 of the rod as it moves at right angles to the magnetic field, is given by
$dϵ=Bvdr$
Hence, $ϵ=\int dϵ={\int }_0^{R}Bvdr$
$ϵ={\int }_0^{R}B\omega rdr$
$ϵ=\frac{B\omega R^2}{2}=\frac{1}{2}B\omega l^2$
Where,
$\omega$ = angular velocity of rotation,
l = length of the rod
B = Magnetic field
Putting the given values.
Length of the rod, " $l=1m$
Angular frequency, $\omega =400{\text{rads}}^{-1}$
Magnetic field strength, $B=0.5T$
We get,
$ϵ=\frac{1}{2}\times 0.5\times 400\times 1^2$
$ϵ=100V$
Hence, the emf developed between the centre and the ring is $100V$.
Final answer: $ϵ=100V$