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Physics 16 Online
OpenStudy (anonymous):

Fm=Fc question regarding graphs

OpenStudy (anonymous):

if the speed of an electron increases then does the radius of its curved path also increase?

OpenStudy (anonymous):

looking like this?|dw:1414194628570:dw|

OpenStudy (sidsiddhartha):

the electron is moving in a magnetic field? or somewhere else ?

OpenStudy (anonymous):

magnetic!

OpenStudy (sidsiddhartha):

if u consider it moving in a magnetic field then it will feel two kind of forces - \[(1) centripetal ~force\\(2)lorentz~force\] and they will balance each other to establish equilibrium-- centripetal force on the electron will be due to circular motion \[F_{centripetal}=\frac{ mv^2 }{ r }\\where\\v=velocity ~of ~the ~electron\\r=radius ~of ~the~path\] and the lorentz force will be-- \[F_{lorentz}=Q*(BXv)=QBv\\Q=charge\\B=magnetic field ~flux\\v=velocity\] so for equilibrium-- \[F_{centripetal}=F_{lorentz}\\ \frac{ mv^2 }{ r }=QBv\\mv=QBr\\so ~we~can~see~r~(radius ~of ~the~path)~~is proportional ~with~the ~velocity\]so graph will look like |dw:1414219765714:dw|

OpenStudy (sidsiddhartha):

got this?

OpenStudy (anonymous):

ok so what you said "we can see r......... with velocity" could you explain what you did with the formula and why it would be proportional?

OpenStudy (sidsiddhartha):

look the final expression we got is \[mv=QBr\] right?

OpenStudy (sidsiddhartha):

so from here \[v=\frac{ QBr }{ m }=(\frac{ QB }{ m})r=K*r\\where~~K= A~~constant ,\]

OpenStudy (sidsiddhartha):

so from here cant we say that v is proportional to r ?

OpenStudy (anonymous):

oh ok ok thnks

OpenStudy (sidsiddhartha):

np :)

OpenStudy (anonymous):

OH WAIT @sidsiddhartha

OpenStudy (anonymous):

my friend said that |dw:1414220801957:dw|

OpenStudy (anonymous):

it can also be this but that means its not directly proportional though

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