Question

A cosmic ray proton travels in a circle with a radius of 95 m perpendicular to Earth's magnetic field. If the proton's speed is 106 m/s, then what is the field strength at the proton's location?

Answer 1

The strength of the magnetic field at the proton's location is approximately 106.25 T.

Step-by-step explanation:

To find the field strength at the proton's location, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field. The formula is given by F = qvBsinθ, where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.

In this case, the force experienced by the proton is given as 1.70 × 10-16 N, and the angle θ is 90° (since the proton is moving perpendicular to the Earth's magnetic field). The charge of a proton is 1.6 × 10-19 C. Plugging these values into the formula, we get:

1.70 × 10-16 N = (1.6 × 10-19 C) × (106 m/s) × B × 1

Simplifying the equation, we can solve for B:

B = (1.70 × 10-16 N) / [(1.6 × 10-19 C) × (106 m/s)]

B ≈ 106.25 T

Therefore, the field strength at the proton's location is approximately 106.25 T.