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Fynjy0 [20]
2 years ago
10

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
Physics
1 answer:
emmasim [6.3K]2 years ago
3 0

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

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.

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