A charged particle of mass 0.0040 kg is enters a $4.0 - 1$ magnetic field with the direction of the field vector at a right angle to its motion.If the particle moves in a circle of radius $0.10 \mathrm {~m}$ at a speed of $2.0 \mathrm {~m} / \mathrm { s }$ what is the magnitude of the charge on the particle?

Question

Quizplus · Accepted Answer

The magnitude of the charge can be found using the formula for the magnetic force acting on a moving charge, which is also the centripetal force in this case: $F = qvB = \frac{mv^2}{r}$. Solving for $q$, we get $q = \frac{mv^2}{rB}$. Plugging in the given values: $q = \frac{0.0040 \times (2.0)^2}{0.10 \times 4.0} = 0.020$ C.

A Charged Particle of Mass 0 4.0−14.0 - 14.0−1 Magnetic Field with the Direction of the Field Vector at 4.0

A Charged Particle of Mass 0 $4.0 - 1$ Magnetic Field with the Direction of the Field Vector at 4.0