A proton moving eastward with a velocity of $5.0 \mathrm {~km} / \mathrm { s }$ enters a magnetic field of $0.20 \mathrm {~T}$ pointing northward. What are the magnitude and direction of the force that the magnetic field exerts on the proton? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$
A) $1.6 \times 10 ^ { - 16 } \mathrm {~N}$ upwards
B) $4.4 \times 10 ^ { - 16 } \mathrm {~N}$ westwards
C) $0 \mathrm {~N}$
D) $1.6 \times 10 ^ { - 16 } \mathrm {~N}$ downwards
E) $1.1 \times 10 ^ { - 16 } \mathrm {~N}$ eastwards

Question

Quizplus · Accepted Answer

The force on a charged particle moving through a magnetic field is given by $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field direction. Here, $q = 1.6 \times 10^{-19} \mathrm{C}$, $v = 5.0 \times 10^3 \mathrm{m/s}$, $B = 0.20 \mathrm{T}$, and $\theta = 90^\circ$ (since the velocity is eastward and the magnetic field is northward, they are perpendicular). Thus, $F = (1.6 \times 10^{-19} \mathrm{C})(5.0 \times 10^3 \mathrm{m/s})(0.20 \mathrm{T}) = 1.6 \times 10^{-16} \mathrm{N}$. The direction of the force is given by the right-hand rule, which in this case results in an upward force.

A Proton Moving Eastward with a Velocity Of 5.0 km/s5.0 \mathrm {~km} / \mathrm { s }5.0 km/s

A Proton Moving Eastward with a Velocity Of $5.0 \mathrm {~km} / \mathrm { s }$