Question regarding radius of circular paths (chapter the nucleus)

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The discussion addresses a question about calculating the radius of circular paths for electrons in a magnetic field after they gain kinetic energy from an electric potential. The key formula presented is eV = 1/2 mv², which allows for the calculation of the electrons' velocity. It is explained that when electrons enter a magnetic field perpendicularly, they move in circular paths due to the magnetic force acting as the centripetal force. The relationship between the magnetic force and centripetal force is expressed as F = -e(VxB) = m(v²/R), leading to the calculation of the beam's radius. The participants express appreciation for the clarification provided in the explanation.
Sanosuke Sagara
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I have my doubt,solution and question in the attachment that followed.Thanks for anybody that spend some time on this question.
 

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OK the problem with your solution is that You have made it a bit too complicated.
First of all ,
After accelerating through a potential , the electrons gain certain amount of K.E , which can be calculate din the following way:

eV= \frac{1}{2}mv^2

From The KE , calculate velocity gained by the electron beam.

Now if you have studied how charged particles behave in magnetic fields, you should know that electrons when enter prependicular to a magnetic field, start moving in a circle.

Force due to a Magnetic field on an electron provides the centripedal force necessary for moving in circle. therefore,

<br /> F= -e(VxB) = m \frac{v^2}{R}<br />

From here calculate the radius of the beam...easy ..isnt it?
 
Yes,Dr Brain.Thanks for your help and I really appreciate it.Thanks again for your explanation in detail.
 

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