Three long straight wires are connected in parallel to each other across a battery of negligible internal resistance. The ratio of their resistances is $3 : 4 : 5$. What is the ratio of the distances of the middle wire from the others if the net force experienced by it is zero?

$A$ 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:

$A$ 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:

$A$ circular conducting loop of radius $R$ carries a current $I$. Another straight infinite conductor carrying current $I$ passes through the diameter of this loop as shown in the figure. The magnitude of the force exerted by the straight conductor on the loop 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,

$A$ massless square loop of wire with resistance $10\,\Omega$ supports a mass of $1\,g$. It hangs vertically with one of its sides in a uniform magnetic field of $10^3\,G$,directed outwards in the shaded region. $A$ $DC$ voltage $V$ is applied to the loop. For what value of $V$ will the magnetic force exactly balance the weight of the supporting mass of $1\,g$? (Given: side length of the loop $= 10\,cm$,$g = 10\,m/s^2$)