A proton is moving through a magnetic field of magnitude 4.40 × 10–3 T. The proton experiences only a magnetic force of 3.3 × 10–15 N. At a certain instant, it has a speed of 4.25 × 105 m s–1. Determine the angle (less than 90°) between the proton's velocity and the magnetic field.
Question
A proton is moving through a magnetic field of magnitude 4.40 × 10–3 T. The proton experiences only a magnetic force of 3.3 × 10–15 N. At a certain instant, it has a speed of 4.25 × 105 m s–1. Determine the angle (less than 90°) between the proton's velocity and the magnetic field.
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:
sinθ = 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 in gives:
sinθ = 3.3 × 10^-15 N / (1.6 × 10^-19 C * 4.25 × 10^5 m/s * 4.40 × 10^-3 T)
Solving this gives:
sinθ = 0.578
To find θ, we take the inverse sine (sin^-1) of this value:
θ = sin^-1(0.578)
This gives:
θ = 35.5°
So, the angle between the proton's velocity and the magnetic field is 35.5°.
Similar Questions
A proton moves with a velocity of = (3î − 4ĵ + ) m/s in a region in which the magnetic field is = (î + 2ĵ − ) T. What is the magnitude of the magnetic force this particle experiences?
A proton is at x = 0.970 cm and is moving to the left with speed 2.00 × 106 m/s, as shown above. Point A is located at y = 0.970 cm, point B is at the origin, and point C is at y = -0.970 cm. What is the magnitude of the magnetic field produced by the proton at point B?
A proton travels at a speed of 7.20×107 m/s along the x-axis. What is the magnetic field at the point (0.00 cm, -7.00 cm, 0.00 cm) when the proton passes through the origin?
A charged particle of charge 4μC is moving perpendicular to a magnetic field with velocity 3 m/s. if it experiences a magnetic force of 3×10−3𝑁 3×10 −3 N , determine the magnitude of the magnetic field.
The speed of the proton is 2.0 × 106 m s–1 and the magnetic field strength B is 0.35 T.(i) Show that the radius of the path is about 6 cm.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.