A proton, traveling with a velocity of 6.8 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.6 × 10-14 N and direction of due south. What are the magnitude and direction of the magnetic field causing the force? If the field is up, then enter a number greater than zero. If the field is down, then enter a number less than zero.
To find the magnitude and direction of the magnetic field causing the force on the proton, we can use the formula for the magnetic force on a moving charged particle:
F = qvB sin(theta)
Where:
- F is the magnitude of the magnetic force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnitude of the magnetic field
- theta is the angle between the velocity vector and the magnetic field vector
Given:
- F = 8.6 × 10^(-14) N
- q = charge on a proton = 1.6 × 10^(-19) C
- v = 6.8 × 10^6 m/s
- theta (angle between the velocity and the magnetic field) = 90 degrees (due east and due south are perpendicular)
Now we can rearrange the formula to solve for B:
B = F / (qv sin(theta))
Substituting the given values:
B = (8.6 × 10^(-14) N) / ((1.6 × 10^(-19) C) × (6.8 × 10^6 m/s) × sin(90°))
B = (8.6 × 10^(-14) N) / (1.088 × 10^(-12) N/A)
B ≈ 7.93 × 10^(-2) T (Tesla)
To determine the direction of the magnetic field, we need to consider the direction of the force and the velocity of the proton. The force is directed due south, so the magnetic field causing this force should be perpendicular to both the velocity vector and the force vector. Since the force is directed south and the velocity is due east, the magnetic field must be directed either up or down.
Therefore, the direction of the magnetic field causing the force is down (towards the ground), so we enter a number less than zero.