Question

A square wire frame with side $a$ and a straight conductor carrying a constant current $I$ are located in the same plane (Fig.).
The inductance and the resistance of the frame are equal to $L$ and $R$ respectively. The frame was turned through ${180}^{\circ }$ about the axis $O{O}^{\prime }$ separated from the current-carrying conductor by a distance $b$. Find the electric charge having flown through the frame.

Answer 1

According to Ohm's law and Faraday's law of induction, the current $i_0$ appearing in the frame, during its rotation, is determined by the formula,
$i_0=-\frac{d\Phi }{dt}=-\frac{Ld{i}_0}{dt}$
Hence, the required amount of electricity (charge) is,
$q=\int i_{0}dt=-\frac{1}{R}\int (d\Phi +Ld{i}_0)=-\frac{1}{R}(△\Phi +L△i_0)$
Since the frame has been stopped after rotation, the current in it vanishes, and hence $△i_0=0$.
It remains for us to find the increment of the flux $△\Phi$ through the frame $(△\Phi ={\Phi }_2-{\Phi }_1)$.
Let us choose the normal $\overrightarrow{n}$ to the plane of the frame, for instance, so that in the final position, $\overrightarrow{n}$ is directed behind the plane of the figure (along $\overrightarrow{B})$.
Then it can be easily seen that in the final position, ${\Phi }_2>0$, while in the initial position, ${\Phi }_1<0$ (the normal is opposite to $\overrightarrow{B})$, and $△\Phi$ turns out to be simply equal to the flux through the surface bounded by the final and initial positions of the frame:
$△\Phi ={\Phi }_2+|{\Phi }_1|={\int }_{b-a}^{b+a}Badr$,
where $B$ is a function of $r$, whose form can be easily found with the help of the theorem of circulation. Finally omitting the minus sign, we obtain,
$q=\frac{△\Phi }{R}=\frac{{\mu }_{0}ai}{2\pi R}ln\frac{b+a}{b-a}$.