Question

A square wire loop with side $L=1.0m$ 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 $\epsilon =6.0V$. 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?

Answer 1

Answer: 0.20

Let magnetic field at any time be $B$. Let upward direction of area vector as positive, the magnetic flux associated with the circuit is
${\Phi }_B=\frac{B{L}^2}{2}$ and the induced emf is
${\epsilon }_i=-\frac{d{\Phi }_B}{dt}=-\frac{L^2}{2}\frac{dB}{dt}$.
As, $B=5.0-2.0t$, hence, $\frac{dB}{dt}=-2.0T/s$
It means the induced emf
${\epsilon }_i=-\frac{(1.0{)}^2}{2}(-2.0)=1.0V$.
As ${\epsilon }_i$ is positive, so the induced emf is counterclockwise around the circuit, in the same direction as the emf of the battery. The total emf is
${\epsilon }_{\text{total }}=\epsilon +{\epsilon }_i=6.0V+1.0V=7.0V$
The current is in the sense of the total emf (counterclockwise).
The magnitude of the current in the loop
$i=\frac{{\epsilon }_{\text{total }}}{R}=\frac{7.0}{35.0}=0.2A$