Knowee
Questions
Features
Study Tools

he 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.

Question

he speed of the proton is 2.0 × 10^6 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.

🧐 Not the exact question you are looking for?Go ask a question

Solution

To calculate the radius of the path of a charged particle moving in a magnetic field, we can use the formula:

r = mv / (qB)

where:

  • r is the radius of the path,
  • m is the mass of the particle,
  • v is the speed of the particle,
  • q is the charge of the particle, and
  • B is the magnetic field strength.

Given in the problem, we have:

  • v = 2.0 × 10^6 m/s,
  • B = 0.35 T,
  • the charge of a proton q = 1.6 × 10^-19 C, and
  • the mass of a proton m = 1.67 × 10^-27 kg.

Substituting these values into the formula, we get:

r = (1.67 × 10^-27 kg * 2.0 × 10^6 m/s) / (1.6 × 10^-19 C * 0.35 T)

Solving this equation gives r ≈ 0.06 m, or about 6 cm. Therefore, the radius of the path is about 6 cm.

This problem has been solved

Similar Questions

The path of a moving charged particle in a magnetic field is ……… * 1 point linear circular spiral downward

Which of the following property of a proton can change while it moves freely in a magnetic field?(a) Mass (b) Speed (c) Charge (d) Momentum

A charged particle is moving along a magnetic field line. The magnetic force on theparticle is

Calcualte the magnetic field strength at point P. The current, I, is 11.0 A, the angle is 21.0° and the radius, R, is 3.60 cm

n a B-H curve, H representsa.Magnetic Field Strengthb.Magnetic Flux densityc.Magnetic Flux

1/1

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.