Tangential acceleration of a proton in an increasing B

[Worme]

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.Homework Statement 2.F=m*a and F=B*q*v
3. I know the protons will be accelerated up but can't the acceleration?

 

[BiGyElLoWhAt]

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.

 

[Delta2]

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 $F=Eq=ma_{tan}$

What level of physics is this at? Have you been taught both the integral and differential form of faraday's law?

 

[Worme]

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!

 

[Delta2]

What you find is the centripetal acceleration not the tangential. The force from the magnetic field alone cannot provide tangential acceleration but only centripetal

 

[Worme]

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

 

[Worme]

Thanks you Delta