A uniform magnetic field of induction B is co | vedclass

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Question

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

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$\frac{1}{2} \frac{eR}{m} \frac{dB}{dt}$ towards the 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 $R$,the induced electric field $E$ at a point on the periphery $(r = R)$ is given by:$E = \frac{R}{2} \frac{dB}{dt}$The force $F$ on an electron of charge $e$ at point $P$ is $F = eE = \frac{eR}{2} \frac{dB}{dt}$.The acceleration $a$ of the electron is given by $a = \frac{F}{m} = \frac{eR}{2m} \frac{dB}{dt}$.Since the magnetic field is increasing,the induced electric field lines form concentric circles in the direction opposite to the induced current (Lenz's law). For an electron (negative charge),the force is directed opposite to the induced electric field,which points tangentially. Given the geometry,the acceleration is directed towards the center or tangentially depending on the orientation,but among the options provided,the magnitude is $\frac{1}{2} \frac{eR}{m} \frac{dB}{dt}$.

$A$ 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 $\frac{dB}{dt} \ (T/s)$. An electron is placed at the point $P$ on the periphery of the field. The acceleration experienced by the electron is:

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