Question

A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters the field, with v prependicular to B, but with a higher speed than the first particle. Compared with the radius of the circle for the first particle, the radius of the circular path of the second particle is

Answer 1

Answer:

The second particle will move through the field with a radius greater that the radius of the first particle

Explanation:

For a charged particle, the force on the particle is given as

$F = \frac{mv^{2} }{r}$

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = $2\pi r$

Work on a particle = 2πrF = $2\pi mv^{2}$

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.