To determine the magnetic force experienced by a proton moving in a given magnetic field, we can use the formula for the magnetic force on a charged particle:
[tex]\[ F = q v B \sin(\theta) \][/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 and the magnetic field.
Given values:
- The velocity of the proton, [tex]\( v = 2.5 \times 10^7 \, \text{m/s} \)[/tex],
- The magnetic field strength, [tex]\( B = 0.12 \, \text{T} \)[/tex],
- The charge of the proton, [tex]\( q = 1.6 \times 10^{-19} \, \text{C} \)[/tex],
- The angle [tex]\( \theta = 90^\circ \)[/tex] (since the proton is moving perpendicular to the magnetic field).
For [tex]\( \theta = 90^\circ \)[/tex], [tex]\( \sin(90^\circ) = 1 \)[/tex].
Substituting the given values into the formula:
[tex]\[ F = (1.6 \times 10^{-19} \, \text{C}) \times (2.5 \times 10^7 \, \text{m/s}) \times (0.12 \, \text{T}) \times 1 \][/tex]
We perform the multiplication step-by-step:
1. Calculate [tex]\( q \times v \)[/tex]:
[tex]\[ 1.6 \times 10^{-19} \times 2.5 \times 10^7 = 4.0 \times 10^{-12} \, \text{(Câ‹…m/s)} \][/tex]
2. Multiply the result by [tex]\( B \)[/tex]:
[tex]\[ 4.0 \times 10^{-12} \times 0.12 = 4.8 \times 10^{-13} \, \text{N} \][/tex]
Therefore, the magnitude of the magnetic force on the proton is [tex]\( 4.8 \times 10^{-13} \, \text{N} \)[/tex].
To determine the direction, we use the right-hand rule for the cross product of velocity and magnetic field vectors. The thumb points in the direction of the velocity (up), the fingers in the direction of the magnetic field (toward you, out of the page), and the palm faces the direction of the force. Hence, the direction of the force will be to the left.
So, the correct answer is:
[tex]\[ \boxed{4.8 \times 10^{-13} \, \text{N} \text{ left}} \][/tex]
