Question

A cosmic ray proton moving toward the Earth at 5.00×10^7 m/s experiences a magnetic force of 1.70×10^(-16) N. What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity?

A) 3.40 × 10^(-9) T
B) 1.20 × 10^(-9) T
C) 2.40 × 10^(-9) T
D) 4.80 × 10^(-9) T

Answer 1

The strength of the magnetic field is approximately 3.40 × 10⁻⁹ T. The value obtained is consistent with the known strength of the Earth's magnetic field on its surface.

Step-by-step explanation:

To find the strength of the magnetic field, we can use the formula:

F = qvBsinθ

Where F is the magnetic force, q is the charge of the proton, v is the velocity, B is the magnetic field strength, and θ is the angle between the magnetic field and the proton's velocity.

Given: F = 1.70 × 10⁻¹⁶ N, v = 5.00 × 10⁷m/s, and θ = 45º

Since the angle is 45º, sinθ = sin(45º) = 1/√2

Plugging in the values into the formula, we have:

1.70 × 10⁻¹⁶) N = (1.60 × 10⁻¹⁹ C) (5.00 × 10⁷ m/s) (B) (1/√2)

Simplifying the equation, we find:

B = (1.70 × 10⁻¹⁶ N) / ((1.60 × 10⁻¹⁹ C) (5.00 × 10⁷ m/s) (1/√2))

B ≈ 3.40 × 10⁻⁹ T

Therefore, the strength of the magnetic field is approximately 3.40 × 10⁻⁹ T.

As for part (b), the value obtained in part (a) is consistent with the known strength of the Earth's magnetic field on its surface. The strength of the Earth's magnetic field on its surface is typically between 25 and 65 microteslas (µT), which is equivalent to 2.5 × 10⁻⁵ to 6.5 × 10⁻⁵ T. The value obtained in part (a) is within the range of the Earth's magnetic field strength, indicating that the magnetic field experienced by the cosmic ray proton could indeed be the Earth's magnetic field.