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?

Question

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?

Infinity Learn · Accepted Answer

Let magnetic field at any time be B. Let upward direction of area vector as positive, the magnetic flux associated with the circuit isΦ$$B=BL22$$ and the induced emf is ε$$i=$$−dΦ$$Bdt=$$−$$L22dBdt$$.  As, $$B=5.0$$−$$2.0t$$, hence, $$dBdt=$$−$$2.0T/s$$ It means the induced emf ε$$i=$$−(1.0)22($$−$$2.0)=1.0V$$. As εi is positive, so the induced emf is  in the same direction as the emf of the battery. The total emf isεtotal $$=$$ε$$+$$ε$$i=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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