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I was doing it just for fun, I took the magnetic force equation and equated it with $ F=ma$ then I took the mass term on the LHS and differentiated it with time and after using equations of motion I got the change in mass ad equal to $qBsin(theta)*(time elapsed)$ where is this mass coming from is it has do something with Einsteins equation, and yeah I was shocked to see that change in mass is proportional to the time elapsed so if I took an electron and put it in a mag field and moved it then it's mass would increase, so the electrons inside the atom are moving in the magnetic field all the time so have we measured their mass a little more? And yeah I took the initial velocity as 0 so what is math telling me here? Because magnetic force can't act unless you are movin... enter image description here

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  • $\begingroup$ Firstly F=dp/dt if mass is not constant. If you have taken it as a constant then beforehand dm=0. If mass is not constant then F=dp/dt = mdv/dt + vdm/dt . If you have taken m as constant 2nd part vanishes. So F=ma with dm=0 . Also F is not do/dt if you are considering relativistic velocity. $\endgroup$ – Shashaank Apr 4 '17 at 10:00
  • $\begingroup$ Even if you had taken V = u + at , it would imply that v of the particle increases with t Wich is wrong for a circular path. $\endgroup$ – Shashaank Apr 4 '17 at 12:31
  • $\begingroup$ Looks to me that you're just throwing equations and operators around and hoping something comes out, so it's not clear (to me) what it is you're trying to achieve. $\endgroup$ – Kyle Kanos Apr 5 '17 at 10:06
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Firstly F=dp/dt if mass is not constant. If you have taken it as a constant then beforehand dm=0. If mass is not constant then F=dp/dt = mdv/dt + vdm/dt . If you have taken m as constant 2nd part vanishes. So F=ma with dm=0 . Also F is not dp/dt if you are considering relativistic velocity.

Then you write v= u + at . It would imply that v of the particle increases with t , Which is wrong for a circular path . This is the fault here rather than the 0/0 reduction which is off course a mathematical inconsistency ( but then you will probably say L' Hospital)

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  • $\begingroup$ Why would L' Hospital matter a is not tending to zero it is exactly zero $\endgroup$ – Utkarsh futous Apr 4 '17 at 12:46
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    $\begingroup$ @Utkarshfutous it was a pun $\endgroup$ – Shashaank Apr 4 '17 at 12:46
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It is quite funny how you used a*t in place of v with initial velocity being zero since F = qvB and since initially there is no v hence no force hence no acceleration

So in the step you wrote m = qatB/a you are actually dividing zero by zero

Also you got the formula for force wrong as for Time varying mass force is equal to rate of change in momentum or impulse by time

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    $\begingroup$ Even if he had taken V = u + at , it would imply that v of the particle increases with t Wich is wrong for a circular path $\endgroup$ – Shashaank Apr 4 '17 at 12:32
  • $\begingroup$ @shashaank Thanks I saw my mistake so I need to use centripetal acceleration here? Right? And V remains constant right? $\endgroup$ – Sarthak Sharma Apr 4 '17 at 14:01
  • $\begingroup$ @SarthakSharma yup. But forget not , the expression for F $\endgroup$ – Shashaank Apr 4 '17 at 14:02
  • $\begingroup$ @Shashaank I again tried it so I wrote $F=dp/dt$ or $F=mdv/dt + vdm/dt $ now since I enter 90 degrees to the field with a finite velocity which is not zero dv/dt is 0 since only direction changes not the mag. So now when I write $vdm/dt=qvB$ again I can cancel out V assuming that it's not 0 as Utkarsha futous pointed out so again I am getting $dm/dt =qB$ so it seems that if I follow my old method which is extremely wrong and then follow this correct method I get the same result am I doing something wrong again? $\endgroup$ – Sarthak Sharma Apr 4 '17 at 14:17
  • $\begingroup$ @SarthakSharma yes . Wrong Newton's law.....The equation I have written V is a vector not at all a scalar. You cannot take it as a scalar...... $\endgroup$ – Shashaank Apr 4 '17 at 14:20

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