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 m at a speed of 2.0 m/s, what is
The magnitude of the charge on the particle?
A) 0.00040 C
B) 2500 C
C) 0.020 C
D) 50 C

Question

Quizplus · Accepted Answer

The magnitude of the charge on the particle can be found using the formula for the magnetic force acting on a moving charge, which is given by $F = qvB$, where $F$ is the magnetic force, $q$ is the charge, $v$ is the velocity of the particle, and $B$ is the magnetic field strength. The magnetic force in this scenario also provides the centripetal force required to keep the particle moving in a circle, which is given by $F = \frac{mv^2}{r}$, where $m$ is the mass of the particle, $v$ is its velocity, and $r$ is the radius of the circle.Equating the two expressions for the force gives us $\frac{mv^2}{r} = qvB$. Solving for $q$ gives us $q = \frac{mv^2}{rvB}$.Substituting the given values: $m = 0.0040 \, \text{kg}$, $v = 2.0 \, \text{m/s}$, $r = 0.10 \, \text{m}$, and $B = 4.0 \, \text{T}$, we get:$$q = \frac{(0.0040)(2.0)^2}{(0.10)(4.0)} = \frac{(0.0040)(4.0)}{0.40} = \frac{0.016}{0.40} = 0.040 \, \text{C}$$Thus, the magnitude of the charge on the particle is 0.040 C, which corresponds to choice C.

A Charged Particle of Mass 0