A particle of charge 5C moves perpendicularly to a magnetic Held of magnitude C.01 T. If the velocity of the charge is 1.5ms'2, calculate the magnitude of t...
Question 1 Report
A particle of charge 5C moves perpendicularly to a magnetic Held of magnitude C.01 T. If the velocity of the charge is 1.5ms'2, calculate the magnitude of the force exerted on the particle
Answer Details
The force experienced by a charged particle moving in a magnetic field is given by the formula F = Bqv, where B is the magnetic field strength, q is the charge of the particle, and v is the velocity of the particle. In this case, the charge on the particle is 5C, the magnetic field strength is 0.01T, and the velocity of the particle is 1.5ms-1. Thus, substituting the given values into the formula, we get: F = (0.01T)(5C)(1.5ms-1) = 0.075N Therefore, the magnitude of the force exerted on the particle is 0.075N. So, the correct option is (b) 0.075N.
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning