If the distance between two current-carrying parallel wires is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the magnitude of the force between them (the same current is passing through both wires) is:

Two long straight parallel wires,carrying (adjustable) current $I_1$ and $I_2$,are kept at a distance $d$ apart. If the force $F$ between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other,the graph showing the dependence of $F$ on the product $I_1 I_2$ would be

$A$ rod of mass $M$ and length $l$ is suspended by two wires $P$ and $Q$. $A$ uniform magnetic field $B$ is directed into the page. When current $I$ flows through the rod,the tension in each supporting wire is:

$A$ wire $X$ of length $50 \; cm$ carrying a current of $2 \; A$ is placed parallel to a long wire $Y$ of length $5 \; m$. The wire $Y$ carries a current of $3 \; A$. The distance between the two wires is $5 \; cm$ and the currents flow in the same direction. The force acting on the wire $X$ due to wire $Y$ is:

Two long wires carrying currents $I_1$ and $I_2$ are arranged as shown in the figure. The wire carrying current $I_1$ is along the $x$-axis. The other wire carrying current $I_2$ is parallel to the $y$-axis,passing through the point $(0, 0, d)$. Find the magnetic force exerted on a segment of length $dl$ of the second wire at the point $O_2(0, 0, d)$ due to the wire carrying current $I_1$.

$A$ wire is bent as a parabolic curve and kept in the $x-y$ plane. The curve is described by the equation $x^2 = 6y$. The wire carries a current $i = 2 \ A$. If a uniform magnetic field $\vec{B} = 2 \times 10^{-3} \hat{k} \ T$ is applied,the force experienced by the wire is nearly: (in $\hat{j} \ N$)