$A$ current-carrying wire $LN$ is bent as shown below. If the wire carries a current of $10 \, A$ and is placed in a magnetic field of $5 \, T$ which acts perpendicular to the paper outwards,then it will experience a force of ......... $N$.

$P, Q$ and $R$ are long parallel straight wires in air,carrying currents as shown in the figure. The direction of the resultant force on wire $R$ is

$A$ current $I$ flows along the length of a thin-walled,long metallic hollow cylinder of radius $R$,distributed uniformly on its surface. The pressure on the wall will be:

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 $1$?

Two long conductors separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2$ times and its direction is reversed. The distance between them is also increased to $3d$. The new value of force between them is:

Two straight parallel wires,both carrying $10 \ A$ in the same direction,attract each other with a force of $1 \times 10^{-3} \ N$. If both currents are doubled,the force of attraction will be: