QJEEM23Ph66c9: JEE Main 2023 Physics - Class 12 JEEM23Ph66c9 Solution: A long metallic rod is rotated with about an axis normal to the …

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.