A particle of mass 3.818x10^-26 kg and charge of 3.2x10^-19 C is accelerated fro

Question

A particle of mass 3.818x10^-26 kg and charge of 3.2x10^-19 C is accelerated from rest in the plane of the page through a potential di?erence of 278 V between two parallel plates as shown. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0.261 T. The particle curves in a semicircular path and strikes a detector. What is the magnitude of the force exerted on the charged particle as it enters the region of the magnetic field B~? Answer in units of N

Explanation / Answer

First, from the conservation of energy, we can find the velocity of the particle

qV = .5mv2

(3.2 X 10-19)(278) = (.5)(3.818 X 10-26)(v2)

v = 68264 m/s

Then, the force is found by

F = qvB

F = (3.2 X 10-19)(68264)(.261)

F = 5.70 X 10-15 N

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