$A$ 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:

Three long straight wires $A$,$B$,and $C$ are carrying currents as shown in the figure. Then the resultant force on $B$ is directed .....

$A$ horizontal wire carries $160 \ A$ current. Below it, another wire with a linear mass density of $10 \ g \ m^{-1}$ is kept at a distance of $4 \ cm$. If the lower wire hangs in the air, what is the current in this wire when the direction of current in both wires is the same (in $A$)? $(g=10 \ m \ s^{-2} \text{ and } \mu_0=4 \pi \times 10^{-7} \ T \ m \ A^{-1})$

$A$ wire carrying current $I$ is tied between points $P$ and $Q$ and is in the shape of a circular arc of radius $R$ due to a uniform magnetic field $B$ (perpendicular to the plane of the paper,shown by $\times \times \times $) in the vicinity of the wire. If the wire subtends an angle $2\theta_0$ at the centre of the circle (of which it forms an arc),then the tension in the wire is:

Two parallel conductors $A$ and $B$ of equal lengths carry currents $I$ and $10\, I$,respectively,in the same direction. Then

Two long parallel conductors,separated by a distance $d$,carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2I$ and its direction is reversed. The distance between them is also increased to $3d$. The new value of the force between them is: