A square loop of side $$1$$ m is placed in a perpendicular magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of e.m.l $$10$$ V and negligible internal resistance is connected in the loop. [Fig. (3)]. The magnetic field changeswith time according to the relationB $$=$$ (0.01$$ $$-$$ $$2$$ $$t) tesla.The total e.m.f. of the battery will be

Question

Infinity Learn · Accepted Answer

See fig. (3). Induced e.m.f. is given $$bye=$$−dΦ$$dt=$$−$$ddt(BA)=$$−$$AdBdt=$$−$$l22ddt(0$$⋅$$01$$−$$2t)=(1)22$$×$$2=1$$ volt

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