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(A) According to Faraday's law of electromagnetic induction,the induced electric field $E$ in a circular path of radius $r$ $(r > R)$ is given by $\oi

Vedclass · Accepted Answer

$\frac{q B_0 R^2}{2mr}$

Vedclass · Answer

(A) According to Faraday's law of electromagnetic induction,the induced electric field $E$ in a circular path of radius $r$ $(r > R)$ is given by $\oint E \cdot dl = -\frac{d\phi}{dt}$.Since the magnetic field is $B(t) = B_0 + B_0 t$,the rate of change of the magnetic field is $\frac{dB}{dt} = B_0$.The magnetic flux $\phi$ through the circular region of radius $R$ is $\phi = B \cdot A = B(\pi R^2)$.Therefore,the magnitude of the induced electromotive force $(EMF)$ is $\left| \frac{d\phi}{dt} \right| = \left| \frac{d}{dt} (B \pi R^2) \right| = \pi R^2 \frac{dB}{dt} = \pi R^2 B_0$.The induced electric field $E$ at distance $r$ satisfies $E(2\pi r) = \pi R^2 B_0$,which gives $E = \frac{R^2 B_0}{2r}$.The force on the charged particle is $F = qE$,and the acceleration is $a = \frac{F}{m} = \frac{qE}{m}$.Substituting the expression for $E$,we get $a = \frac{q}{m} \left( \frac{R^2 B_0}{2r} \right) = \frac{q B_0 R^2}{2mr}$.

$A$ circular region of radius $R$ has a uniform magnetic field $B = B_0 + B_0 t(-\hat{k})$. At $t = 0$,what is the acceleration of a charged particle of mass $m$ and charge $q$ placed at a distance $r$ $(r > R)$ from the center?

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