question_answer

A long straight wire carrying a current of 10 A is placed parallel to one of the side of square as shown in figure. If the resistance of the square loop is \[2\,\Omega \] and loop is rotated by \[180{}^\circ \] about side AB, then total charge flowing during this interval becomes

A)  \[\frac{{{\mu }_{0}}}{2}\ell n(3)\]                           

B)  \[\frac{{{\mu }_{0}}}{2}\ell n(5)\]

C)  \[{{\mu }_{0}}\ell n(3)\]                              

D)  zero

Correct Answer: A

Solution :

The flux through loop \[Q=\frac{{{\mu }_{0}}i\ell }{2\pi }\ell n\left( \frac{b}{a} \right)\] The charge flow \[Q=\frac{-({{Q}_{f}}-{{Q}_{i}})}{R}=\frac{{{\mu }_{0}}i\ell }{R}\left[ \ell n\left( \frac{20}{10} \right)+\ell n\left( \frac{30}{20} \right) \right]\] \[=\frac{{{\mu }_{0}}i\ell }{R}[\ell n2-\ell n1+\ell n3-\ell n2]=\frac{{{\mu }_{0}}}{2}(\ell n3)\]