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 $\Omega$ and the loop contains an ideal battery with emf $\epsilon$ = 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?

Let magnetic field at any time be B. Let upward direction of area vector as positive, the magnetic flux associated with the circuit is

$\begin{array}{l}{\mathrm{\Phi }}_B=\frac{B{L}^2}{2}\text{ and the induced emf is }{\epsilon }_i=-\frac{d{\mathrm{\Phi }}_B}{dt}=-\frac{L^2}{2}\frac{dB}{dt}\text{. }\\ \text{ As, }B=5.0-2.0t\text{, hence, }\frac{dB}{dt}=-2.0\mathrm{T}/\mathrm{s}\\ \text{ It means the induced emf }{\epsilon }_i=-\frac{(1.0{)}^2}{2}(-2.0)=1.0\mathrm{V}\text{. }\end{array}$

As ${\epsilon }_i$ is positive, so the induced emf is  in the same direction as the emf of the battery. The total emf is

${\epsilon }_{\text{total }}=\epsilon +{\epsilon }_i=6.0\mathrm{V}+1.0\mathrm{V}=7.0\mathrm{V}$

The current is in the sense of the total emf (countercloclcvise).
The magnitude of the current in the loop

$i=\frac{{\epsilon }_{\text{total }}}{R}=\frac{7.0}{35.0}=0.2\mathrm{A}$