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(a) A cosmic ray proton moving toward the Earth at 5.00×107m/s experiences a magnetic force of 1.70×10−16N. What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.

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Final answer:

The strength of the magnetic field is approximately 0.0008 T. The value obtained is not consistent with the known strength of the Earth's magnetic field on its surface. Cosmic ray protons are affected by the Earth's magnetic field, but other factors also influence their path.

Step-by-step explanation:

(a) Calculation:

The magnetic force experienced by a charged particle moving through a magnetic field can be calculated using the formula:

F = qvB sinθ, where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the particle's velocity and the magnetic field.

From the given information:

v = 5.00 x 10^7 m/s
F = 1.70 x 10^-16 N
θ = 45°

Substituting these values into the formula, we can solve for B:

1.70 x 10^-16 N = (1.60 x 10^-19 C)(5.00 x 10^7 m/s)B sin 45°

Simplifying the equation:

1.70 x 10^-16 N = (8.00 x 10^-12 C)(5.00 x 10^7 m/s)B / √2

B = (1.70 x 10^-16 N) √2 / (8.00 x 10^-12 C)(5.00 x 10^7 m/s)

B ≈ 0.0008 T

(b) Discussion:

The value obtained in part (a) is not consistent with the known strength of the Earth's magnetic field on its surface, which is approximately 25 - 65 μT (microtesla). Cosmic ray protons are affected by the Earth's magnetic field while traveling through space, but their path is influenced by other factors such as the Sun's magnetic field. Therefore, the value obtained in part (a) cannot be directly compared to the Earth's magnetic field strength on its surface.

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