A Proton Moving At $5.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$

Short Answer

A proton moving at $5.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$ horizontally enters a region where a magnetic field of $0.12 \mathrm {~T}$ is present, directed vertically downward. What magnitude force acts on the proton due to this field?
$\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$