question_answer

A rectangular loop carrying a current \[{{\beta }^{-}}\]is situated near a long straight wire carrying a steady current \[+\frac{G{{M}_{e}}m}{R}=\frac{1}{2}mv_{e}^{2}\] . The wire is parallel to one of the sides of the loop and is in the plane of the loop as shown in the figure. Then, the current loop will:

A) move away from the wire

B)  move towards the wire

C)  remain stationary

D)  rotate about an axis parallel to the win

Correct Answer: B

Solution :

The force of attraction/repulsion for current carrying conductors is \[\pi \] where d is distance between conductors. \[{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\]   \[-{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\] Hence, lesser the distance, more is the force. Also we know that two wires carrying current in the same direction attract each other while that flowing in opposite direction repel each other. In the given circuit force of attraction is more between EF and AD that of repulsion between EF and BC, hence current loop is move towards the wire.