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A point charge with a mass of 2.5x10-12 kg and a charge of 6.8 C travels...

A point charge with a mass of 2.5x10-12 kg and a charge of 6.8 C travels straight through a velocity selector with crossed electric and magnetic fields. The magnetic field has a strength of .44 T. The charge then enters an area with just the magnetic field, and follows a path with a radius of curvature of 2 mm. What was the electric field of the velocity selector?

Science   Advanced-Physics   
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Answer #1

Solution:

Mass of the point charge, m = 2.5x10-12 kg

Charge of the particle, q = 6.8

Magnetic field strength B = 0.44T

Radius of curvarture of the path followed by the particle under the influence of magnetic field,R= 2 mm

R = 2x10-3 m

Crossed electric field means the electric and magnetic fields are perpendicular to each other and the particle just drifts with a fixed velocity ,

The total force experienced by the particle is given by, F = qE+q(vxB)

The radius, from which we can get the velocity of the particle

Therefore,

Therefore,

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