Chapter 27 Problem 9

Problem

Partially immersed loop

A current-carrying circular loop of wire (radius $r$, current $I$) is partially immersed in a magnetic field of constant magnitude $B_{0}$ directed out of the page. Determine the net force on the loop due to the field in terms of $\theta _{0}$ .

(Note that $\theta _{0}$ points to the dashed line, above which $B=0$ .)

solution

the current in an infinitesimal portion of the loop

$Id{\vec {l}}=I\left(-\cos \theta {\hat {i}}+\sin \theta {\hat {j}}\right)rd\theta$

${\vec {F}}=\int Id{\vec {l}}\times {\vec {B}}=\int _{0}^{2\pi -\theta _{0}}I\left(-\cos \theta {\hat {i}}+\sin \theta {\hat {j}}\right)rd\theta \times B_{0}{\hat {k}}$ $=IB_{0}r\int _{0}^{2\pi -\theta _{0}}\left(\cos \theta {\hat {j}}+\sin \theta {\hat {i}}\right)d\theta$ $=IB_{0}r\left.\left(\sin \theta {\hat {j}}-\cos \theta {\hat {i}}\right)\right|_{0}^{2\pi -\theta _{0}}$ $IB_{0}r\left(\sin(2\pi -\theta _{0}){\hat {j}}-\sin(\theta _{0}){\hat {j}}-\cos(2\pi -\theta _{0}){\hat {i}}+\cos(\theta _{0}){\hat {i}}\right)$

since $\sin(2\pi -\theta )=-\sin \theta$ and $\cos(2\pi -\theta )=\cos \theta$

$=-2IB_{0}r\sin \theta _{0}{\hat {j}}$