In Fig. $30-10$, the magnetic field is up out of the page and $B=0.80$
T. The wire shown carries a current of 30 A. Find the magnitude and direction of the force on a $5.0 \mathrm{~cm}$ length of the wire.
Fig. $30-10$
We know that
$\Delta F_{M}=I(\Delta L) B \sin \theta=(30 \mathrm{~A})(0.050 \mathrm{~m})(0.80 \mathrm{~T})(1)=1.2 \mathrm{~N}$
By the right-hand rule, the force is perpendicular to both the wire and the field and is directed toward the bottom of the page.