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A proton is moving through a magnetic field of magnitude 4.40 × 10–3 T. The protonexperiences only a magnetic force of 3.3 × 10–15 N. At a certain instant, it has a speed of4.25 × 105 m s–1. Determine the angle (less than 90°) between the proton's velocity and themagnetic field.

Question

A proton is moving through a magnetic field of magnitude 4.40 × 10–3 T. The protonexperiences only a magnetic force of 3.3 × 10–15 N. At a certain instant, it has a speed of4.25 × 105 m s–1. Determine the angle (less than 90°) between the proton's velocity and themagnetic field.

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Solution

The force experienced by a charged particle moving through a magnetic field is given by the equation:

F = qvBsinθ

where: F is the 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 velocity and the magnetic field.

We can rearrange this equation to solve for θ:

θ = arcsin(F / (qvB))

We know that the charge of a proton (q) is 1.6 × 10^-19 C, the velocity (v) is 4.25 × 10^5 m/s, the magnetic field strength (B) is 4.40 × 10^-3 T, and the force (F) is 3.3 × 10^-15 N. Substituting these values into the equation gives:

θ = arcsin((3.3 × 10^-15 N) / ((1.6 × 10^-19 C) * (4.25 × 10^5 m/s) * (4.40 × 10^-3 T)))

Solving this equation will give you the angle θ in radians. To convert to degrees, multiply by (180/π).

This problem has been solved

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