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(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi_B}{dt}$.Fo

Vedclass · Accepted Answer

$\frac{a^2}{2r} \left( \frac{dB}{dt} \right)$

Vedclass · Answer

(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi_B}{dt}$.For a point $P$ at a distance $r > a$ from the center,we consider a circular path of radius $r$ centered at $O$.The magnetic flux $\phi_B$ through this circular path is limited to the region of radius $a$,where the magnetic field $B(t)$ exists.Thus,$\phi_B = B(t) \cdot \pi a^2$.The induced electric field $E$ along the circular path is related to the $EMF$ by $\varepsilon = \oint E \cdot dl = E(2\pi r)$.Equating the magnitudes: $E(2\pi r) = \left| \frac{d}{dt} (B(t) \cdot \pi a^2) \right| = \pi a^2 \left| \frac{dB}{dt} \right|$.Solving for $E$: $E = \frac{\pi a^2}{2\pi r} \left| \frac{dB}{dt} \right| = \frac{a^2}{2r} \left| \frac{dB}{dt} \right|$.Therefore,the correct option is $A$.

$A$ time-varying magnetic field $B(t)$ exists in a circular region of radius '$a$' and is directed into the plane of the paper,as shown in the figure. The magnitude of the induced electric field at a point '$P$' at a distance '$r$' $(r > a)$ from the center of the circular region is:

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