Q 21. Consider a ring of radius \(R\), and charged with a linear charge density \(\lambda\), placed on a smooth frictionless surface, in a region of varying magnetic field that varies with time \(t\) as \(B = B_0e^{-\alpha t}\). Let its mass be \(m\). Find its angular velocity \(\omega\) as a function of time. Also find the tension \(T\) in the ring as a function of time. Neglect the interaction of the charges of the ring with itself.

Solution

From Faraday’s law, we can write:

\[\oint \vec{E}_{in} \cdot \vec{dl} = \frac{-d \phi_B }{dt}\]

Integrating over the closed loop, (\(\vec{E}_{in}\) is constant for the loop), we get,

\[\vec{E}_{in} = \frac{1}{2} \alpha B_0 R e^{-\alpha t}\]

\[\implies \ddot{\theta} = \frac{1}{m} \alpha B_0 \pi \lambda R e^{-\alpha t}\]

Integrating this to solve for \(\omega\), we get

\[\omega = \frac{\lambda B_0 \pi R}{m} \left( 1 - e^{-\alpha t} \right)\]

Tension in the ring is due to rotation of ring, and tension due to the charges in magnetic field. The net tension would be the vector sum of these. Since both are in the same direction(radially outwards), we can just add them, getting

\[T = \frac{B_0^2 \pi R^3 \lambda^2}{2m} ( 1 - e^{-2\alpha t} )\]