Question

A cosmic ray proton moving toward the earth at 7.5 × 10⁷ m/s experiences a magnetic force of 1.85 × 10⁻¹⁶ N. What is the magnitude of the magnetic field that produces this force?

1) 2.47 × 10⁻⁹ T
2) 2.47 × 10⁻⁸ T
3) 2.47 × 10⁻⁷ T
4) 2.47 × 10⁻⁶ T

Answer 1

The magnitude of the magnetic field producing a force on a cosmic ray proton moving towards Earth at 7.5 \(\times\) 107 m/s, experiencing a magnetic force of 1.85 \(\times\) 10-16 N, is 1.54 \(\times\) 10-9 T. From the provided options, answer 1) 2.47 \(\times\) 10-9 T is the closest match.

Step-by-step explanation:

To find the magnitude of the magnetic field that produces a force on a moving charged particle, we can use the formula for the magnetic force on a moving charge, which is F = qvBsin(\theta), where F is the force, q is the charge of the particle, v is the velocity, B is the magnetic field strength, and \(\theta\) is the angle between the velocity of the particle and the direction of the magnetic field. In the question, since the charge of a proton is known to be approximately 1.60 \(\times\) 10-19 C, and the angle between the velocity and magnetic field is not provided, we can assume the angle \(\theta\) is 90 degrees or perpendicular, so sin(\(\theta\)) = 1. Rearranging the formula to solve for B, we get B = F / (qv).

Using the given values, we find:

B = (1.85 \(\times\) 10-16 N) / (1.60 \(\times\) 10-19 C \(\times\) 7.5 \(\times\) 107 m/s)

B = 1.54 \(\times\) 10-9 T

Therefore, the magnitude of the magnetic field that produces this force is 1.54 \(\times\) 10-9 T, so the closest answer from the provided options is 2.47 \(\times\) 10-9 T, answer option 1).