$A$ conducting wire bent in the form of a parabola $y^2=2x$ carries a current of $4 \ A$ as shown in the figure. This wire is placed in a uniform magnetic field of induction $\vec{B}=+6 \hat{k} \ T$. The magnetic force on the wire is

Two long wires with no contact are placed perpendicular to each other. $i_1$ and $i_2$ are currents flowing through these wires respectively. The magnetic force on a small length '$d$' of the second wire situated at a distance '$l$' from the first wire is proportional to,

The magnetic force between the wires as shown in the figure is:

$A$ straight wire is placed in a magnetic field that varies with distance $x$ from the origin as $\vec{B} = B_0 \left( 2 - \frac{x}{a} \right) \hat{k}$. The ends of the wire are at $(a, 0)$ and $(2a, 0)$ and it carries a current $I$ in the positive $x$-direction. If the force on the wire is $\vec{F} = IB_0 \left( \frac{ka}{2} \right) \hat{j}$, then the value of $k$ is:

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 :

$A$ horizontal rod of mass $10 \, g$ and length $10 \, cm$ is placed on a smooth plane inclined at an angle of $60^\circ$ with the horizontal,with the length of the rod parallel to the edge of the inclined plane. $A$ uniform magnetic field of induction $B$ is applied vertically downwards. If the current through the rod is $1.73 \, A$,then the value of $B$ for which the rod remains stationary on the inclined plane is......$T$