$A$ current $i$ flows through a wire in the positive $X$-direction. The magnetic field is $\overrightarrow{B} = B_0(\hat{i} + \hat{j} + \hat{k}) \ T$. What is the magnitude of the force acting on a wire segment of length $l$?

$A, B$ and $C$ are three parallel conductors of equal lengths carrying currents $I, I$ and $2I$ respectively. The distance between $A$ and $B$ is $x$ and that between $B$ and $C$ is also $x$. $F_1$ is the force exerted by conductor $B$ on $A$. $F_2$ is the force exerted by conductor $C$ on $A$. The current $I$ in $A$ and $I$ in $B$ are in the same direction,and the current $2I$ in $C$ is in the opposite direction. Then:

Wires $1$ and $2$ carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$ (as shown in figure) due to the magnetic field of wire $1$?

Two long parallel wires are at a distance of $1 \ m$. Both of them carry $1 \ A$ of current. The force of attraction per unit length between the two wires is

Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert

$A$ $4.0 \ cm$ long straight wire carrying a current of $8 \ A$ is placed perpendicular to a uniform magnetic field of strength $0.15 \ T$. The magnetic force on the wire is . . . . . . $mN$.