The total force on a closed current loop in a uniform magnetic field is always zero due to the canceling out of forces on opposing segments of the loop, but a non-zero torque acts on the loop, causing it to rotate until its magnetic dipole moment aligns with the magnetic field, with the torque's magnitude being directly proportional to the current flowing through the loop, and its direction described by the cross product $\tau=\mu \times B$, resulting in a maximum torque when the loop's plane is parallel to the magnetic field and zero when perpendicular, and the torque's effect is visually demonstrated by the Current slider.
<aside> 🧄
$\complement\cdots$Counselor
</aside>
The most important takeaway is that the total force ( $F$ ) on any closed current loop in a uniform magnetic field is always zero.
This is because the force on each segment of the loop is given by the Lorentz force, $d F= I(d x \times B)$. For every segment of the loop, there's a corresponding segment on the opposite side where the differential displacement vector $d x$ is in the opposite direction. Since the magnetic field $B$ is uniform and constant, the forces on these opposing segments are equal in magnitude but opposite in direction, causing them to cancel out. The integral of all these canceling forces over the entire closed loop therefore sums to zero.
While the net force is zero, a non-zero torque ( $\tau$ ) will generally act on the loop. This happens because the canceling forces are applied at different points on the loop, creating a turning effect. The torque acts to rotate the loop until its magnetic dipole moment ( $\mu$ ) is aligned with the magnetic field $(B)$. The direction of the torque vector is perpendicular to the plane defined by both the magnetic moment and the magnetic field, a relationship described by the cross product: $\tau=\mu \times B$. The magnitude of this torque is at a maximum when the loop's plane is parallel to the magnetic field and is zero when it is perpendicular.
‣
<aside> 🧄
</aside>