A charged particle of mass 0.0040 kg is subjected to a $4.0 - T$ 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 magnetic force provides the centripetal force: $ qvB = \frac{mv^2}{r} $. Solving for $ q $, $ q = \frac{mv}{rB} = \frac{0.0040 \times 2.0}{0.10 \times 4.0} = 0.020 \, \text{C} $.

A Charged Particle of Mass 0 4.0−T4.0 - T4.0−T Magnetic Field Which Acts at a Right Angle to Its

A Charged Particle of Mass 0 $4.0 - T$ Magnetic Field Which Acts at a Right Angle to Its