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Q.

A square wire loop with side L = 1 .0 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field as shown in figure. The resistance of the loop is 35 Ω and the loop contains an ideal battery with emf ε = 6.0 V. If the magnitude of the field varies with time according to B = 5.0 - 2.0t, with B in tesla and t in seconds, what is current (in A) around the loop?

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Detailed Solution

Let magnetic field at any time be B. Let upward direction of area vector as positive, the magnetic flux associated with the circuit isΦB=BL22 and the induced emf is εi=−dΦBdt=−L22dBdt.  As, B=5.0−2.0t, hence, dBdt=−2.0T/s It means the induced emf εi=−(1.0)22(−2.0)=1.0V. As εi is positive, so the induced emf is  in the same direction as the emf of the battery. The total emf isεtotal =ε+εi=6.0V+1.0V=7.0VThe current is in the sense of the total emf (countercloclcvise).The magnitude of the current in the loopi=εtotal R=7.035.0=0.2A
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