## College Physics (7th Edition)

$3.5\times10^{3}m/s$
Given: q= 0.050 C B= 0.080 T $\theta =45^{\circ}$ F= 10 N Recall: $F= qvBsin\theta$ or $v=\frac{F}{qBsin\theta}$ Substituting the values, we get $v=\frac{10N}{0.050C\times0.080T\times sin45^{\circ}}= 3.5\times10^{3}m/s$