A charged particle of mass 0.0040 kg is subjected to a $4.0 - \mathrm { T }$ magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius $0.10 \mathrm {~m}$ at a speed of $2.0 \mathrm {~m} / \mathrm { s } \text {, }$ what is the magnitude of the charge on the particle?
A)0.020 C
B)50 C
C)0.00040 C
D)2500 C

Question

Quizplus · Accepted Answer

The correct answer is A because the magnitude of the charge on the particle is given by the equation q = mv / rB, where m is the mass of the particle, v is its speed, r is the radius of the circle in which it moves, and B is the magnetic field strength. Substituting the given values into this equation, we get q = (0.0040 kg)(2.0 m/s) / (0.10 m)(4.0 T) = 0.020 C.

A Charged Particle of Mass 0 4.0−T4.0 - \mathrm { T }4.0−T

A Charged Particle of Mass 0 $4.0 - \mathrm { T }$