Two long straight parallel conductors separated by a distance of $0.5\,m$ carry currents of $5\,A$ and $8\,A$ in the same direction. The force per unit length experienced by each other is

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,

$A$ magnetic field is in the $+Y$ direction. $A$ current $i$ flows through the wire $PQRSTU$. Each side has a length $L$. What is the net force on the wire?

$A$ 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:

Write the formula for the magnetic force acting on a current-carrying wire placed in a uniform magnetic field.

Two long straight wires $P$ and $Q$ carrying equal current $10\,A$ each were kept parallel to each other at $5\,cm$ distance. Magnitude of magnetic force experienced by $10\,cm$ length of wire $P$ is $F_1$. If distance between wires is halved and currents on them are doubled,force $F_2$ on $10\,cm$ length of wire $P$ will be :