Questions
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 changes
with time according to the relation
B = (0.01 - 2 t) tesla.
The total e.m.f. of the battery will be
detailed solution
Correct option is C
See fig. (3). Induced e.m.f. is given bye=−dΦdt=−ddt(BA)=−AdBdt=−l22ddt(0⋅01−2t)=(1)22×2=1 voltSimilar Questions
A conductor rod AB moves parallel to X-axis in a uniform magnetic field, pointing in the positive X-direction. The end A of the rod gets
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