$A$ wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\overrightarrow{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is $..........IL$.

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 :

$A$ and $B$ are two conductors carrying a current $i$ in the same direction. $x$ and $y$ are two electron beams moving in the same direction.

$A$ straight wire of length $50 \text{ cm}$ carrying a current of $2.5 \text{ A}$ is suspended in mid-air by a uniform magnetic field of $0.5 \text{ T}$. Calculate the mass of the wire. (Take $g = 10 \text{ m s}^{-2}$) (in $\text{ g}$)

What is the magnitude of magnetic force per unit length (in $N \;m^{-1}$) on a wire carrying a current of $8 \;A$ and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of $0.15 \;T$?

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: