Two thin,long,parallel wires,separated by a distance $d$,carry a current of $i$ in the same direction. They will

$A$ 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 $..........IL$.

What is the force per unit length between the two wires shown in the figure? [${\mu _0} = 4\pi \times {10^{ - 7}} \text{ T}\,m/A$]

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^o$. Another straight thin wire with a steady current $I_1$ flowing out of the plane of the paper is kept at the origin. Due to the presence of the current $I_1$ at the origin:

$A$ conductor (shown in the figure) carrying constant current $I$ is kept in the $x-y$ plane in a uniform magnetic field $\vec{B}$. If $F$ is the magnitude of the total magnetic force acting on the conductor,then the correct statement$(s)$ is(are):

$A$ metallic rod of mass per unit length $0.5 \,kg \,m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. $A$ magnetic field of strength $0.25 \,T$ is acting on it in the vertical direction. When a current $I$ is flowing through it, the rod is not allowed to slide down. The quantity of current required to keep the rod stationary is (in $\,A$)