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  • The magnetic field between the plate of a capacitor when  is given byOption:

The magnetic field between the plate of a capacitor when r<R is given by

Option: 1

\frac{\mu_0 i_D r}{2 \pi R^2}


Option: 2

\frac{\mu_0 i_D}{2 \pi R}


Option: 3

\frac{\mu_0 i_D}{2 \pi r}


Option: 4

Zero


Answers (1)

best_answer

Consider a loop of radius r<(R) between the two circular plates, placed coaxially with them. The area of the \text { loop }=\pi r^2
Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely. Final momentum
\mathrm{P_f=\frac{E}{c} \quad \text { (negative value) }}
\mathrm{\therefore} Change in momentum
\mathrm{ \begin{aligned} \Delta p & =p_f-p_i \\ & =-\frac{E}{c}-\frac{E}{c}=-\frac{2 E}{c} \end{aligned} }
Thus, momentum transferred to the surface is
\mathrm{\Delta p^{\prime}=|\Delta p|=\frac{2 E}{c}}
 

Posted by

Suraj Bhandari

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