A square wire frame with side $a$ and a long straight conductor carrying a constant current $I$ are

Question

A square wire frame with side $a$ and a long straight conductor carrying a constant current $I$ are located in the same plane. The inductance and the resistance of the frame are equal to $L$ and $R$ respectively. The frame is turned through $180^\circ$ about the axis $OO'$ separated from the current-carrying conductor by a distance $b$ as shown in the figure. Find the magnitude of the total electric charge having flown through the frame.

Professor PC Thomas Classes · Answer

Let the initial area vector (normal to the loop) be directed into the page, parallel to the magnetic field $\vec{B}$ produced by the long straight wire. The initial magnetic flux through the loop is $\Phi_1 = \int_{b-a}^{b} \frac{\mu_0 I}{2\pi x} a \, dx = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{b}{b-a}ight)$. After a $180^\circ$ rotation about the axis $OO'$, the loop occupies the space from $x=b$ to $x=b+a$. Because of the flip, the normal to the loop now points out of the page (opposite to $\vec{B}$). Therefore, the new flux is $\Phi_2 = -\int_{b}^{b+a} \frac{\mu_0 I}{2\pi x} a \, dx = -\frac{\mu_0 I a}{2\pi} \ln\left(\frac{b+a}{b}ight)$. By Faraday's and Lenz's laws, considering the inductance $L$ of the loop, the circuit equation is $\mathcal{E} - L\frac{di}{dt} = iR \implies -\frac{d\Phi}{dt} - L\frac{di}{dt} = iR$. Integrating this equation with respect to time from $t=0$ to $t 	o \infty$: $-\int_{\Phi_1}^{\Phi_2} d\Phi - L \int_0^0 di = R \int_0^\infty i \, dt$. Since the initial and final currents are zero, the inductance term vanishes: $-(\Phi_2 - \Phi_1) = Rq \implies q = \frac{\Phi_1 - \Phi_2}{R}$. Substituting the flux values, we get $q = \frac{\mu_0 I a}{2\pi R} \left[ \ln\left(\frac{b}{b-a}ight) - \left(-\ln\left(\frac{b+a}{b}ight)ight) ight] = \frac{\mu_0 I a}{2\pi R} \ln\left(\frac{b+a}{b-a}ight)$.