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(A) According to Faraday's law of induction,a time-varying magnetic field induces an electric field.For a cylindrical region of radius $R$ with a magn

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$\frac{eR}{2m} \frac{dB}{dt}$ towards left

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(A) According to Faraday's law of induction,a time-varying magnetic field induces an electric field.For a cylindrical region of radius $R$ with a magnetic field $B$ changing at a rate $\frac{dB}{dt}$,the induced electric field $E$ at a distance $r=R$ from the center is given by:$\oint E \cdot dl = -\frac{d\Phi_B}{dt}$$E(2\pi R) = \pi R^2 \frac{dB}{dt}$$E = \frac{R}{2} \frac{dB}{dt}$Since the magnetic field is directed into the page and is increasing,by Lenz's law,the induced electric field will be in the counter-clockwise direction to oppose the change in flux.At point $P$ on the bottom periphery,the tangent to the counter-clockwise field points towards the left.The force on the proton is $F = eE$.The acceleration $a$ is given by $a = \frac{F}{m} = \frac{eE}{m}$.Substituting the value of $E$:$a = \frac{e}{m} \left( \frac{R}{2} \frac{dB}{dt} \right) = \frac{eR}{2m} \frac{dB}{dt}$ towards the left.

$A$ uniform magnetic field of induction $B$ is confined to a cylindrical region of radius $R$. The magnetic field is increasing at a constant rate of $\frac{dB}{dt} \text{ (T/s)}$. $A$ proton of charge $e$ and mass $m$ is placed at point $P$ on the periphery. Its acceleration is

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