A charged particle of mass 0.0020 kg is subjected to a 6.0 T magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius 0.20 m at a speed of 5.0 m/s, what is the magnitude of the charge on the particle?
A) 0.0083 C
B) 120 C
C) 0.00040 C
D) 2500 C

Question

Quizplus · Accepted Answer

Explanation:The magnetic force on the particle is given by:F = qvBwhere q is the charge on the particle, v is its speed, and B is the magnetic field strength.The centripetal force on the particle is given by:F = mv^2/rwhere m is the mass of the particle and r is the radius of the circle.Setting these two forces equal, we get:qvB = mv^2/rSolving for q, we get:q = mv/BrSubstituting the given values, we get:q = (0.0020 kg)(5.0 m/s)/(6.0 T)(0.20 m) = 0.0083 C

A Charged Particle of Mass 0