$A$ current of $3\, A$ is flowing in a linear conductor of length $40\, cm$. The conductor is placed in a magnetic field of strength $500\, G$ and makes an angle of $30^\circ$ with the direction of the field. The magnitude of the force experienced by the conductor is:

$A$ 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).

Wires $1$ and $2$ carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$ (as shown in figure) due to the magnetic field of wire $1$?

$A$ thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction,as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper,the wire takes the shape of a circle. The tension in the wire is

$A$ straight horizontal conducting rod of length $0.45\; m$ and mass $60\; g$ is suspended by two vertical wires at its ends. $A$ current of $5.0\; A$ is set up in the rod through the wires.
$(a)$ What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
$(b)$ What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) $g = 9.8\; m s^{-2}.$

Two long wires with no contact are placed perpendicular to each other. $i_1$ and $i_2$ are currents flowing through these wires respectively. The magnetic force on a small length '$d$' of the second wire situated at a distance '$l$' from the first wire is proportional to,