$A$ conductor lies along the $z$-axis in the range $-1.5 \le z < 1.5 \ m$ and carries a fixed current of $10.0 \ A$ in the $-\hat{a}_z$ direction (see figure). For a magnetic field $\vec{B} = 3.0 \times 10^{-4} e^{-0.2x} \hat{a}_y \ T$,find the power required to move the conductor at a constant speed to $x = 2.0 \ m, y = 0 \ m$ in $5 \times 10^{-3} \ s$. Assume parallel motion along the $x$-axis.

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^o$. Another straight thin wire with a steady current $I_1$ flowing out of the plane of the paper is kept at the origin. Due to the presence of the current $I_1$ at the origin:

$A$ rectangular loop of sides $25 \text{ cm}$ and $10 \text{ cm}$ carrying a current of $10 \text{ A}$ is placed with its longer side parallel to a long straight conductor $10 \text{ cm}$ apart carrying a current of $25 \text{ A}$. The net force on the loop is

$A$ long wire $AB$ is placed on a table. Another wire $PQ$ of mass $1.0\, g$ and length $50\, cm$ is set to slide on two rails $PS$ and $QR$. $A$ current of $50\, A$ is passed through the wires. At what distance above $AB$ will the wire $PQ$ be in equilibrium? (in $mm$)

Two parallel wires in free space are $10 \,cm$ apart and each carries a current of $10 \,A$ in the same direction. The force exerted by one wire on the other [per unit length] is

Two long 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 $2$ times and its direction is reversed. The distance between them is also increased to $3d$. The new value of force between them is: