$A$ 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

Obtain the expression for the force between two parallel current-carrying wires.

An arbitrarily shaped closed coil is made of a wire of length $L$ and a current $I$ ampere is flowing in it. If the plane of the coil is perpendicular to a uniform magnetic field $\vec{B}$, the net magnetic force on the coil is

$A$ straight rod of mass $m$ and length $L$ is suspended from two identical springs as shown in the figure. The springs are stretched by a distance of $x_0$ due to the weight of the rod. The circuit has a total resistance $R$. When a magnetic field perpendicular to the plane of the paper is switched on,the springs are observed to extend further by the same distance $x_0$. The magnetic field strength is:

$A$ conducting wire $PQ$ carries a current $10 \ A$ as shown in the figure. It is placed in a uniform magnetic field $5 \ T$ which is acting normally outwards from the paper. The net force experienced by the wire is:

$A$ single turn current loop in the shape of a right-angled triangle with sides $5\,cm$,$12\,cm$,and $13\,cm$ carries a current of $2\,A$. The loop is placed in a uniform magnetic field of magnitude $0.75\,T$,whose direction is parallel to the current in the $13\,cm$ side of the loop. The magnitude of the magnetic force on the $5\,cm$ side is $\frac{x}{130}\,N$. The value of $x$ is $..........$