Q.Two long straight wires carry steady currents $I_1$ and $I_2$. The wire carrying $I_1$ lies along the $x$-axis with its current flowing in the $+x$ direction. The wire carrying $I_2$ is parallel to the $y$-axis and passes through the point $x=0,\ z=d$; its point of closest approach to the origin is $O_2=(0,0,d)$, and its current flows in the $+y$ direction. Find the magnetic force exerted on the current element of the second wire located at $O_2$ due to the wire lying along the $x$-axis.

Concept

An infinitely long straight wire carrying current $I_1$ sets up a magnetic field that circles the wire. At perpendicular distance $s$ from the wire the magnitude is $B=\dfrac{\mu_0 I_1}{2\pi s}$, and its direction is tangential (perpendicular to both the wire and the shortest line from the wire to the field point).

Field of wire 1 at $O_2$

Wire 1 lies along the $x$-axis with current in $+\hat{\mathbf{x}}$. The point $O_2=(0,0,d)$ is at perpendicular distance $d$ from the $x$-axis, the shortest line pointing along $+\hat{\mathbf{z}}$. Using $\mathbf{B}\propto \hat{\mathbf{I}}\times\hat{\mathbf{s}}$ with $\hat{\mathbf{I}}=\hat{\mathbf{x}}$ and $\hat{\mathbf{s}}=\hat{\mathbf{z}}$:

So at $O_2$ the field lies entirely along the $y$-axis.

Force on the element of wire 2 at $O_2$

Wire 2 carries $I_2$ along $+\hat{\mathbf{y}}$, so its length element there is $d\mathbf{l}=dl\,\hat{\mathbf{y}}$. The force is …

Method: Finding the Force on One Current Element Due to Another Wire's Field

General two-step technique for any "find the force at point P due to wire X" problem.

Steps

Step 1: Find the field produced by the source wire at the location of interest

For an infinite straight wire, the field magnitude at perpendicular distance $s$ is $B=\dfrac{\mu_0 I}{2\pi s}$, and its direction follows the right-hand rule — always tangential (perpendicular to both the wire and the line joining the wire to the point):

where $\hat I$ is the unit vector along the source current and $\hat s$ points from the wire toward the field point.

Step 2: Apply the force law to the current element sitting at that point

$d\vec F = I\, d\vec l \times \vec B$ …