A moving proton with a speed of 0.3 m/s experiences a force of 1.5*10^-20N, when it enters at a right angle to the direction of the magnetic field. What is the magnitude of the magnetic field?
Choose one answer.
a. 0.3 Tesla
b. 0.5 Tesla
c. 0.9 Tesla
d. 1.3 Tesla
Since B and the velocity are peperdicualar,
Force = q V B
That is a formula you should have learned before posting this question
Use e for the proton charge and solve for B
d. 1.3 Tesla
To find the magnitude of the magnetic field, we can use the formula for the magnetic force experienced by a moving charged particle:
F = q * v * B * sin(theta)
Where:
- F is the force experienced by the moving proton
- q is the charge of the proton (which is 1.6 * 10^-19 C)
- v is the speed of the proton (0.3 m/s)
- B is the magnitude of the magnetic field we want to find
- theta is the angle between the velocity vector of the proton and the magnetic field (90 degrees, since the proton enters at a right angle)
Rearranging the formula, we get:
B = F / (q * v * sin(theta))
Substituting the values given:
F = 1.5 * 10^-20 N
q = 1.6 * 10^-19 C
v = 0.3 m/s
theta = 90 degrees
B = (1.5 * 10^-20 N) / ((1.6 * 10^-19 C) * (0.3 m/s) * (sin(90 degrees)))
B = (1.5 * 10^-20 N) / ((1.6 * 10^-19 C) * (0.3 m/s) * (1))
Simplifying further, we get:
B = (1.5 * 10^-20 N) / (4.8 * 10^-20 C * m/s)
B = 0.3125 T
Therefore, the magnitude of the magnetic field is approximately 0.3125 Tesla. None of the given answer options match this value, so it seems there might be an error in the question or the answer choices.