$A$ wire of length $L$ carries a current $I$ along the $X$-axis. The magnetic force acting on the wire is given by $\vec{F} = I B_0 L(\hat{k} - \hat{j})$. The existing magnetic field $\vec{B}$ is

$A$ uniform conducting wire $ABC$ has a mass of $10\,g$. $A$ current of $2\,A$ flows through it. The wire is kept in a uniform magnetic field $B = 2\,T$ directed into the plane of the paper. The acceleration of the wire will be

An arrangement of three parallel wires placed perpendicular to the plane of the paper,each carrying the same current $I$ in the same direction,is shown in the figure. The magnitude of the force per unit length on the middle wire $B$ is given by:

$A$ loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible,it will take the shape of:

$A$ wire carrying current '$I$' along the $x$-axis has length '$L$' and is kept in a magnetic field $\vec{B} = B(\hat{i} + 2\hat{j} - 2\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is:

$A$ 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 $\,A$)