$A$ straight horizontal conducting rod of length $L$ and mass $M$ is suspended by two vertical wires at its ends. If $I$ is the current passing through the rod,then in order that the tension in the wires is zero,the magnetic field set up normal to the conductor is (Neglect the mass of the wire,$g$ = acceleration due to gravity).

Two thin long parallel wires separated by a distance $b$ are carrying a current $i$ $A$ each. The magnitude of the force per unit length exerted by one wire on the other is

Two thin long parallel wires,separated by a distance $d$,carry a current of $I$ $A$ in the same direction. They will:

The force on a current-carrying loop (radius $= R$) in a uniform magnetic field $(B)$,which is at an angle $30^{\circ}$ with the normal,will be:

In the figure shown,a current $I_1$ is established in the long straight wire $AB$. Another wire $CD$ carrying current $I_2$ is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire $AB$. The force on the wire $CD$ is:

$A$ thin uniform rod with negligible mass and length $l$ is attached to the floor by a frictionless hinge at point $P$. $A$ horizontal spring with force constant $k$ connects the other end to a wall. The rod is in a uniform magnetic field $B$ directed into the plane of the paper. What is the extension in the spring in equilibrium when a current $i$ is passed through the rod in the direction shown? Assume the spring is initially at its natural length.