$A$ 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:

$A$ long straight wire carrying a current of $25 \,A$ rests on a table. Another wire $PQ$ of length $1 \,m$ and mass $2.5 \,g$ carries the same current but in the opposite direction. The wire $PQ$ is free to slide up and down. To what height $h$ will wire $PQ$ rise (in $\,mm$)? (Take $g = 9.8 \,m/s^2$ and $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$)

$A$ wire of length $L$ carries current $I$ along the $x$-axis. $A$ magnetic field $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k}) \text{ T}$ acts on the wire. The magnitude of the magnetic force acting on the wire is:

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:

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:

The force acting on a current-carrying wire joining two fixed points $A$ and $B$ in a uniform magnetic field is: