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

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