Question

A ring of mass, radius r having q uniformly distribute over it and free to rotate sbout its own axis place in a region having a magnetic field B parallel to its axis.If the magnetic field is suddenly Switched off the angular velocity accquired by the ring is

• A. $\frac{qB}{m}$
• B. $\frac{2qB}{m}$
• C. $\frac{qB}{2m}$
• D. None of these

Answer 1

The correct option is C $\frac{qB}{2m}$

$\begin{array}{cc}\underset{–}{B}y\text{Lenz's law}:& \text{Induced emf}\left(e\right)=-\frac{d\varphi }{dt}\text{(where}\varphi \to \text{magnetic}flux)\\ & =\pi r^2\cdot \frac{dB}{dt}\end{array}$

$\text{Here, linear charge density}\left(\lambda \right)=\frac{q}{2\pi r}\text{, of the given ring.}$

$\begin{array}{c}\text{so, torque}\left(\tau \right)=\int \lambda \cdot Edl\\ \text{Thus, angular momentum}=I\omega =\int \tau \cdot dt\end{array}$

$\begin{array}{c}\Rightarrow I\omega =R\lambda {\pi }^2\cdot B\\ \Rightarrow \left(m{r}^2\right)\omega =B\times r\times \frac{q}{2\pi r}\times {\pi }^2.\\ \therefore \omega =\frac{qB}{2m}\end{array}$