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

The magnetic force on a current-carrying wire in an external uniform magnetic field depends on:

$A$ current of $10 \ A$ is flowing in two straight parallel wires in the same direction. The force of attraction between them is $1 \times 10^{-3} \ N$. If the current is doubled in both the wires,the force will be:

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:

Two long straight wires $A$ and $B$ carrying equal current $I$ were kept parallel to each other at distance $d$ apart. The magnitude of the magnetic force experienced by length $L$ of wire $A$ is $F$. If the distance between the wires is halved and the currents are doubled,the force $F_2$ on length $L$ of wire $A$ will be:

$A$ 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: