A proton moving with a velocity of $4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$ enters a magnetic field of $0.20 \mathrm {~T}$. If the angle between the velocity of the proton and the direction of the magnetic field is $60 ^ { \circ }$, what is the magnitude of the magnetic force on the proton? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$
A) $0.60 \times 10 ^ { - 15 } \mathrm {~N}$
B) $2.2 \times 10 ^ { - 15 } \mathrm {~N}$
C) $1.1 \times 10 ^ { - 15 } \mathrm {~N}$
D) $3.3 \times 10 ^ { - 15 } \mathrm {~N}$
E) $1.8 \times 10 ^ { - 15 } \mathrm {~N}$

Question

Quizplus · Accepted Answer

The magnitude of the magnetic force on a moving charge is given by the equation $F = qvB\sin\theta$, where $q$ is the charge of the particle, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field direction. Substituting the given values: $F = (1.60 \times 10^{-19} \, \text{C})(4.0 \times 10^4 \, \text{m/s})(0.20 \, \text{T})\sin(60^\circ)$, we find $F \approx 1.1 \times 10^{-15} \, \text{N}$.

A Proton Moving with a Velocity Of 4.0×104 m/s4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }4.0×104 m/s

A Proton Moving with a Velocity Of $4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$