$A$ straight wire carrying a current $i_1 \, A$ runs along the axis of a circular loop carrying a current $i_2 \, A$. The force of interaction between the two current-carrying conductors is

$A$ long straight wire is carrying current $I_1$ in the $+z$ direction. The $x-y$ plane contains a closed circular loop carrying current $I_2$ and does not encircle the straight wire. The net force on the loop will be:

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,

Three straight parallel current-carrying conductors are shown in the figure. The force experienced by the middle conductor of length $25\, cm$ is

$A$ 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 $\text{ g}$)

$A$ current-carrying rectangular loop $PQRS$ is made of uniform wire. The lengths are $PR = QS = 5\,cm$ and $PQ = RS = 100\,cm$. If the ammeter current reading changes from $I$ to $2I$,the ratio of magnetic forces per unit length on the wire $PQ$ due to wire $RS$ in the two cases respectively,$f_{PQ}^{I} : f_{PQ}^{2I}$,is: