Question

Notify Me Bookmark

A particle of mass 9.35 g and charge 73.3 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is −9.8j^ m/s2 without falling. The velocity of the particle is a constant 16.3i^ km/s, which is perpendicular to the magnetic field. What, then, is the magnetic field?

A particle of mass 9.35 g and charge 73.3 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is −9.8j^ m/s2 without falling. The velocity of the particle is a constant 16.3i^ km/s, which is perpendicular to the magnetic field. What, then, is the magnetic field?

Image text
A particle of mass 9.35 g and charge 73.3 μ C moves through a uniform magnetic field, in a region where the free-fall acceleration is − 9.8 j ^ m / s 2 without falling. The velocity of the particle is a constant 16.3 i ^ km / s , which is perpendicular to the magnetic field. What, then, is the magnetic field?

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

In the figure below, a charged particle moves perpendicular to a uniform B-field (in a vacuum) and the magnetic field is the dominant factor determining the motion. The particle's kinetic energy and speed remain constant, and the direction of motion and speed are unaffected. Because the magnetic force is perpendicular to the direction of travel, the charged particle follows a curved path in the magnetic field. And the particle continues to follow this curved path until it forms a complete circle. The magnetic force is perpendicular to the velocity, where the velocity changes in direction and magnitude. The magnetic force is perpendicular to the velocity and velocity changes in direction not magnitude. In this situation, where a charged particle moves perpendicular to a uniform B-field (in a vacuum), the magnetic force supplies the centripetal force Fc = mv2 r. The magnetic force is always perpendicular to velocity, and it does no work on the charged particle. A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular out of the paper.
Solution
0 0

A positively charged particle of mass 8.45×10−8 kg is traveling due east with a speed of 68.4 m/s and enters a 0.320−T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.12×10−3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.
Solution
0 0

A positively charged particle of mass 6.02×10−8 kg is traveling due east with a speed of 89.3 m/s and enters a 0.424-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.47×10−3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge. (a) Number Units (b) Number Units
Solution
0 0