JEE Main 2023 — Electromagnetic Induction Question with Solution

Given that the rod is rotating at 210 rpm, we first convert this to radians per second:

$\omega =210\cdot \frac{2\pi \mathrm{rad}}{60\mathrm{s}}=22rad/s$

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).

$v=22rad/s\cdot 0.2\mathrm{m}=4.4m/s$

Now, we can find the emf developed between the center and the ring using the formula:

$ϵ=\frac{1}{2}B\ell v$

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).

$ϵ=\frac{1}{2}\cdot 0.2\mathrm{T}\cdot 0.2\mathrm{m}\cdot 4.4m/s=0.088\mathrm{V}$

To express this value in mV, we can simply multiply it by 1000:

$ϵ=0.088\mathrm{V}\cdot 1000=88\mathrm{mV}$

So the emf developed between the center and the ring is 88 mV.