Answer
$3.5\times10^{3}m/s$
Work Step by Step
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
$v=\frac{10N}{0.050C\times0.080T\times sin45^{\circ}}= 3.5\times10^{3}m/s$
