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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 about the axis OO separated from the current-carrying conductor by a distance b. Find the electric charge having flown through the frame.
690737_2ffb766802a2418eaa816b5bae49caac.jpg

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Solution

According to Ohm's law and Faraday's law of induction, the current i0 appearing in the frame, during its rotation, is determined by the formula,
i0=dΦdt=L d i0dt
Hence, the required amount of electricity (charge) is,
q=i0dt=1R(dΦ+L di0)=1R(Φ+Li0)
Since the frame has been stopped after rotation, the current in it vanishes, and hence i0=0.
It remains for us to find the increment of the flux Φ through the frame (Φ=Φ2Φ1).
Let us choose the normal n to the plane of the frame, for instance, so that in the final position, n is directed behind the plane of the figure (along B).
Then it can be easily seen that in the final position, Φ2>0, while in the initial position, Φ1<0 (the normal is opposite to B), and Φ turns out to be simply equal to the flux through the surface bounded by the final and initial positions of the frame:
Φ=Φ2+|Φ1|=b+abaB a dr,
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=ΦR=μ0ai2πRlnb+aba.
1797827_690737_ans_e2dbcb246c624432b653f761334d612c.jpg

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