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  • A non-conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field <img alt="\mathrm{B=4t^{2}}"

A non-conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field \mathrm{B=4t^{2}} is switched on at time t = 0. Mass of the ring is m and radius is R. The ring starts rotating after 2 s, the coefficient of friction between the ring and the table is:

Option: 1

\mathrm{\frac{4 \mathrm{qmR}}{\mathrm{g}}}


Option: 2

\mathrm{\frac{2 \mathrm{qmR}}{\mathrm{g}}}


Option: 3

\mathrm{\frac{8 \mathrm{qR}}{\mathrm{mg}}}


Option: 4

\mathrm{\frac{q R}{2 m g}}


Answers (1)

best_answer

\mathrm{As, \quad E l=\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}} or \quad E(2 \pi R)=\pi R^2 \cdot \frac{d B}{d t}=\pi R^2(8 t)}

\mathrm{\begin{aligned} \therefore \quad E= & 4 R t \\ & F=q E=4 q R t \quad \text { [tangential] } \\ & \tau_F=4 q R^2 t \\ \tau_r & =(\mu \mathrm{mgR}) \end{aligned}}

When, \mathrm{\tau_{\mathrm{F}}>\tau_{\mathrm{r}}} f ,ring will start rotating.

\mathrm{ \begin{aligned} & \mathrm{t}=2 \mathrm{~s}, \\ & 8 \mathrm{qR}^3=\mu \mathrm{mgR} \end{aligned} }

\mathrm{ Coefficient\: of\: friction, \mu=\frac{8 \mathrm{q} R}{\mathrm{mg}}}

 

Posted by

Pankaj Sanodiya

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