= Solution
Use the source orientation, anticlockwise viewed from $+z$. The position and oriented line element are
$$
\mathbf r=a(\cos\theta,\sin\theta,0),\qquad
d\mathbf l=a(-\sin\theta,\cos\theta,0)\,d\theta.
$$
The <Lorentz force> on the <electric current> element is $d\mathbf F=I\,d\mathbf l\times\mathbf B$. Evaluating the <cross product> gives
$$
\boxed{d\mathbf F=Ia\bigl(B_z\cos\theta,\ B_z\sin\theta,\ -B_y\sin\theta-B_x\cos\theta\bigr)\,d\theta.}
$$
Each component integrates to zero around $0\leq\theta<2\pi$, so $\mathbf F_{\rm net}=0$. Equivalently, uniform <magnetic flux density> permits $I\oint d\mathbf l\times\mathbf B=I(\oint d\mathbf l)\times\mathbf B=0$. Uniformity is essential; a nonuniform <magnetic field> can exert a net <force> on a <magnetic dipole>.
Back to article page
