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  • A square loop of side 1m is placed in a perpendicular magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of e.m.f. 10 V and negligible internal resistane is conn

A square loop of side 1m is placed in a perpendicular magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of e.m.f. 10 V and negligible internal resistane is connected in the loop. The magnetic field changes with time according to the relation.
B = (0.1 – 2t) Tesla.
The total e.m.f. of the battery will be

Option: 1

1V


Option: 2

11 V


Option: 3

9V


Option: 4

10V


Answers (1)

best_answer

Induced e.mf. is given

\mathrm{\begin{aligned} & =-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \\ & =-\frac{\mathrm{l}^2}{2} \frac{\mathrm{d}}{\mathrm{dt}}(0.01-2 \mathrm{t}) \\ & =\frac{(1)^2}{2} \times 2=1 \text { volt } \\ & \therefore \text { Resultant e.m.f. }=10-1=9 \text { volt } \end{aligned} }

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Sanket Gandhi

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