Question

Notify Me Bookmark

A particle undergoes uniform circular motion of radius 25.8 μm in a uniform magnetic field. The magnetic force on the particle has a magnitude of 2.28×10−17 N. What is the kinetic energy of the particle? Number Units

A particle undergoes uniform circular motion of radius 25.8 μm in a uniform magnetic field. The magnetic force on the particle has a magnitude of 2.28×10−17 N. What is the kinetic energy of the particle? Number Units

Image text
A particle undergoes uniform circular motion of radius 25.8 μ m in a uniform magnetic field. The magnetic force on the particle has a magnitude of 2.28 × 10 − 17 N . What is the kinetic energy of the particle?
Number i Units

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

A particle undergoes uniform circular motion of radius 14.5 μm in a uniform magnetic field. The magnetic force on the particle has a magnitude of 2.40 × 10-17 N. What is the kinetic energy of the particle? Number Units
Solution
0 0

An a-particle has a charge of +2 e and a mass of 6.64×10−27 kg. It is accelerated from rest through a potential difference that has a value of 1.93×106 V and then enters a uniform magnetic field whose magnitude is 2.29 T. The a-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the a-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path? (a) Number Units (b) Number Units (c) Number Units
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