To answer this question, we need to use the formula for the magnetic force acting on a charged particle moving in a magnetic field. The formula is:
[tex]\[ F = qvB\sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge of the particle,
- [tex]\( v \)[/tex] is the velocity of the particle,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity of the particle and the magnetic field.
In this specific situation, we are dealing with a stationary proton, so the velocity [tex]\( v \)[/tex] is 0. Hence the formula simplifies greatly, because any term multiplied by zero is zero.
Let's analyze step-by-step:
1. Identify the given values:
- The charge of the proton, [tex]\( q = 1.6 \times 10^{-19} \)[/tex] coulombs.
- The magnetic field strength, [tex]\( B = 0.50 \)[/tex] tesla.
- The velocity of the proton, [tex]\( v = 0 \)[/tex] (since the proton is stationary).
2. Substitute the given values into the formula:
[tex]\[
F = (1.6 \times 10^{-19} \text{ C}) \cdot (0 \text{ m/s}) \cdot (0.50 \text{ T}) \cdot \sin(\theta)
\][/tex]
Since the velocity [tex]\( v \)[/tex] is 0,
[tex]\[
F = 1.6 \times 10^{-19} \times 0 \times 0.50 \times \sin(\theta) = 0
\][/tex]
Therefore, the magnetic force [tex]\( F \)[/tex] on a stationary proton in the magnetic field is:
[tex]\[
F = 0 \text{ newtons}
\][/tex]
So, the correct answer is:
A. 0 newtons
