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Question: A square wire frame of side a is placed a distance b away from a long straight conductor carrying cu...

A square wire frame of side a is placed a distance b away from a long straight conductor carrying current I. The frame has resistance R and self inductance L. The frame is rotated by 1800 about OO¢ as shown in Fig. Find the electric charge flown through the frame–

A

2μ0ia22πRb\frac{2\mu_{0}ia^{2}}{2\pi Rb}

B

μ0i2πR\frac{\mu_{0}i}{2\pi R}logeb+ab−a\frac{b + a}{b - a}

C

μ0ia2πR\frac{\mu_{0}ia}{2\pi R}logeb+ab−a\frac{b + a}{b - a}

D

None of these

Answer

μ0ia2πR\frac{\mu_{0}ia}{2\pi R}logeb+ab−a\frac{b + a}{b - a}

Explanation

Solution

i = 1R\frac { 1 } { \mathrm { R } } [dφdt+Ldidt]\left\lbrack \frac{d\varphi}{dt} + L\frac{di}{dt} \right\rbrack

q =∫idt\int_{}^{}{idt}=1R\frac{1}{R} [Df + 0] =ΔφR\frac{\Delta\varphi}{R}=∫b−ab+aBadx\int_{b - a}^{b + a}{Badx}

=∫b−ab+aμ0ia2πx\int_{b - a}^{b + a}\frac{\mu_{0}ia}{2\pi x}dx =μ0ia2πR\frac{\mu_{0}ia}{2\pi R}logeb+ab−a\frac{b + a}{b - a}