A wire is bent into a form $\mathrm{PQRST}$ and carries a current $\mathrm{I}$ as shown in Figure. Straight segment $\mathrm{PQ}$ and $\mathrm{ST}$ are of equal length l and the semi-circular segment $\mathrm{QRS}$ has a radius r. The frame is placed in a region of uniform magnetic field $\mathrm{B}$ directed as shown in the Figure. The magnetic force exerted on the wire frame is:

 

Current segments PQ and ST are perpendicular to B. Therefore, the magnetic forces on PQ and ST are 

 ${\mathrm{F}}_{\mathrm{PQ}}=\mathrm{BI}\mathcal{l} \sin 90°=\mathrm{BI}\mathcal{l}$ 

 ${\mathrm{F}}_{\mathrm{ST}}=\mathrm{BI}\mathcal{l}\sin 90°=\mathrm{BI}\mathcal{l}$

ow, ${\mathrm{F}}_{\mathrm{QRS}}=\left|\mathrm{I}(\overrightarrow{\mathrm{QS}}\times \overrightarrow{\mathrm{B}})\right|=\mathrm{I}\left(2\mathrm{r}\right)\mathrm{B}=2\mathrm{BIr}$. Since $\mathrm{\theta }=90°$

$\therefore$The total force on the wire frame is ( since ${\mathrm{F}}_{\mathrm{PQ}}, {\mathrm{F}}_{\mathrm{ST}}$ and ${\mathrm{F}}_{\mathrm{QRS}}$ are all directed vertically upward) 

$\mathrm{F}={\mathrm{F}}_{\mathrm{PQ}}+ {\mathrm{F}}_{\mathrm{ST}}+{\mathrm{F}}_{\mathrm{QRS}}$

  $=\mathrm{BI}\mathcal{l}+\mathrm{BI}\mathcal{l}+2\mathrm{BIr}$

  $=2\mathrm{BI}(\mathcal{l}+\mathrm{r})$

The direction of F is vertically upward.