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  • A metallic rod of mass m length R is pivoted at O. It is free to rotate about a horizontal axis passing through O. An inward magnetic field is applied as shown in the figure. The ratio of

A metallic rod of mass m length R is pivoted at O. It is free to rotate about a horizontal axis passing through O. An inward magnetic field is applied as shown in the figure. The ratio of maximum electric field inside the metallic rod to the applied magnetic field induction is:

Option: 1

\sqrt{\mathrm{gR}}


Option: 2

\sqrt{\mathrm{2gR}}


Option: 3

\sqrt{\mathrm{3gR}}


Option: 4

\sqrt{\mathrm{5gR}}


Answers (1)

best_answer

The induced emf between 0 and P   

\begin{aligned} & =\varepsilon=-\mathrm{\frac{d \phi}{d t}=-\frac{d(B A)}{d t}=-B \frac{d A}{d t}} \\ & =\varepsilon=\mathrm{-B \frac{d}{d t}\left[\frac{r^2}{2} \theta\right]=-\frac{B r^2 \omega}{2} \quad\left(\because \omega=\frac{d \theta}{d t}\right)} \end{aligned}

The potential gradient   =\frac{\mathrm{d} \varepsilon}{\mathrm{dr}}=\left(\frac{-\mathrm{B} \omega}{2}\right) \frac{\mathrm{d}}{\mathrm{r}}\left(\mathrm{r}^2\right)=-\mathrm{B} \omega \mathrm{r}

\Rightarrow the corresponding electric field =\mathrm E =\mathrm{B} \omega \mathrm{r}

\Rightarrow \mathrm E_{\max }=\mathrm{B} \omega \mathrm{R} .

Where, \omega = maximum angular speed of the rod, that is possible
when it becomes vertical.

Conservation of mechanical energy between horizontal and vertical
yield,

      

 

\begin{aligned} & \mathrm {\Delta P E_{g r}+\Delta K E_r=0 }\\ & \mathrm {-m g \frac{R}{2}+\frac{1}{2} I_0 \omega^2=0 }\\ & \mathrm {\Rightarrow \omega=\sqrt{\frac{3 g}{R}}, \text { since } l_o=\frac{m R^2}{3}} \end{aligned}

Putting the value of \omega_o we obtain

\begin{aligned} & \mathrm {E_{\max }=B \sqrt{\frac{3 g}{R}} \cdot R} \\ & \mathrm {\Rightarrow \quad \frac{E_{\max }}{B}=\sqrt{3 g R}} \end{aligned}

Posted by

jitender.kumar

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