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A square frame with side length a and a straight conductor carrying constant current $I_{0}$ are located in the same plane. The frame is rotated with constant angular velocity, $ω = 1 r a d / s$ about an axis AD. Find the charge flown through the frame from $t = 0$ to $t = \frac{2 π}{3} s,$ if resistance of frame is R.

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$\frac{μ_{0} I_{0} a}{2 π R} (\ln 2 - \ln 5 / 4)$

$\frac{μ_{0} I_{0} a}{2 π R} (\ln 2 - \ln 3 / 2)$

$\frac{μ_{0} I_{0} a}{2 π R} \ln 2$

$\frac{μ_{0} I_{0} a}{4 π R} \ln 2$

answer is C.

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Detailed Solution

$Q = \frac{Δ ϕ}{R} ϕ_{i} = \int_{a}^{2 a} \frac{μ_{0} I_{0}}{2 π X} a d x = \frac{μ_{0} I_{0} a}{2 π} \ln 2 ϕ_{f} = 0 (b y s y m m e t r y) Q = \frac{μ_{0} I_{0} a}{2 π R} \ln 2$

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