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 $35\, \Omega$ and the loop contains an ideal battery with emf $\varepsilon=6.0\, V$. If the magnitude of the field varies with time according to $B=5.0-2.0\, t$, with $B$ in tesla and $t$ in seconds, what is current (in $A$) around the loop?

Answer 1

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
$\varepsilon_{i}=-\frac{d \Phi_{B}}{d t}=-\frac{L^{2}}{2} \frac{d B}{d t}$.
As, $B=5.0-2.0 t$, hence, $\frac{d B}{d t}=-2.0\, T / s$
It means the induced emf
$\varepsilon_{i}=-\frac{(1.0)^{2}}{2}(-2.0)=1.0\, V$.
As $\varepsilon_{i}$ 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
$\varepsilon_{\text {total }}=\varepsilon+\varepsilon_{i}=6.0\, V +1.0\, V =7.0\, V$
The current is in the sense of the total emf (counterclockwise).
The magnitude of the current in the loop
$i=\frac{\varepsilon_{\text {total }}}{R}=\frac{7.0}{35.0}=0.2\, A$