## College Physics (7th Edition)

a). $3.84\times10^{-18}N$, b). $2.715\times10^{-18}N$, c). 0, d). 0.
$F=qvBsin\theta$ a). $F=1.6\times10^{-19}\times2\times10^{4}\times1.2\times10^{-3}\times sin90^{\circ}=3.84\times10^{-18}N$ b). $F=1.6\times10^{-19}\times2\times10^{4}\times1.2\times10^{-3}\times sin45^{\circ}=2.715\times10^{-18}N$ c). Since, $\theta=0^{\circ}, sin\theta=0$, So, $F=0$ d). Since, $\theta=180^{\circ}, sin\theta=0$, So, $F=0$