A long horizontal wire P carries current of 50 ~A from left to right. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P . The mass of the wire is ' m ' kg and carries a current of ' I ' A. The direction of current in Q and position of wire Q from P so that the wire Q remains suspended are

Step-by-step solution

For the wire Q to remain suspended, the magnetic force acting on it must balance the gravitational force acting downwards. The magnetic force per unit length between two parallel wires carrying currents I₁ and I₂ separated by a distance r is given by F/L = ₀ I₁ I₂ 2 r . Since the wire Q is suspended above wire P, the magnetic force must be directed upwards (repulsive). For the force to be repulsive, the currents in the two parallel wires must flow in opposite directions. Given that wire P carries current from left