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A magnetic field in a certain region is given by B=B_{0}\cos \left ( \omega t \right ) and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (figure). Find the magnitude and the direction of the current at (a, 0, 0) at

t=\frac{\pi }{2\omega },t=\frac{\pi }{\omega }\; and\; t=\frac{3\pi }{2\omega }

Answers (1)

\phi _{m}=\vec{B}.\vec{A}=BA\; \cos \theta

And as we know both \vec{A} (area vector ) and \vec{B}  (magnetic field vector) are directed along z-axis. So, angle between them is 0.

So, \cos \theta = 1 \left ( \because \theta =0 \right )

\Rightarrow \phi _{m}=BA

Area of coil of radius  a=\pi a ^{2}

\varepsilon =B_{0}(\pi a^{2})\; \cos \omega t

By Faraday's law of electromagnetic induction, Magnitude of induced emf is given by

\varepsilon =B_{0}(\pi a^{2})\; \omega \sin \omega t

This causes flow of induced current, which is given by

I=\frac{B_{0}(\pi a^{2})\; \omega \sin \omega t}{R}

Now, the value of current at different instants,

(i) t=\frac{\pi }{2\omega }

I=\frac{B_{0}(\pi a^{2})\omega }{R} along\; \hat{j}

Because \sin \omega t=\sin \left ( \omega \frac{\pi }{2\omega } \right )=\sin \frac{\pi }{2}=1

(ii) t=\frac{\pi }{\omega },I=\frac{B(\pi a^{2})\omega }{R} =0

because, \sin \omega t=\sin \left ( \omega \frac{\pi }{2\omega } \right )=\sin \pi =0

(iii) t=\frac{3}{2}\frac{\pi }{\omega }

I = \frac{B(\pi a^{2})\omega }{R}\; along -\hat{j}

\sin \omega t=\sin \left ( \omega .\frac{3\pi }{2\omega } \right )=\sin \frac{3\pi }{2}=-1

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