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Tangential acceleration of a proton in an increasing B

  1. Mar 27, 2015 #1
    1. Consider free protons following a circular path in a uniform magnetic field with a radius of 1meter . At t=0 , the magnitude of the uniform magnetic field begins to increase at 0.001Tesla/second . Enter the tangential acceleration of the protons in meters/second2 : positive if they speed up and negative if they slow down.The problem statement, all variables and given/known data


    2.F=m*a and F=B*q*v
    3. I know the protons will be accelerated up but can't the acceleration?
     
  2. jcsd
  3. Mar 27, 2015 #2

    BiGyElLoWhAt

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    Gold Member

    You need maxwells equations. You have a time variant magnetic field, which produces an E field. Otherwise, since magnetic forces are perpendicular to the velocity, there would be a 0 tangential acceleration. BTW ##\vec{F}=q\vec{v}\times\vec{B}## not B*q*v, that's not even the right magnitude.
     
  4. Mar 27, 2015 #3
    You should use Faraday's law of induction to find the electric field E at distance r from the center and from that the tangential acceleration due to [itex]F=Eq=ma_{tan}[/itex]

    What level of physics is this at? Have you been taught both the integral and differential form of faraday's law?
     
  5. Mar 27, 2015 #4
    I use v=r*q*B/ r then put in m*a= qv x B and a= (B^2 x q^2 x r)/ m^2 but the acceleration seems so big!!!!
     
  6. Mar 27, 2015 #5
    What you find is the centripetal acceleration not the tangential. The force from the magnetic field alone cannot provide tangential acceleration but only centripetal
     
  7. Mar 27, 2015 #6
    How to find the tangential acceleration? By using Faraday's law E= - r/2 x dB/dt so a=E x q/m= -r/2 x dB/dt x q/m so a=- 5.22*10^4m/s^2
     
  8. Mar 27, 2015 #7
    Thanks you Delta
     
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