Tangential acceleration of proton due to a changing magnetic field

AI Thread Summary
The discussion focuses on calculating the tangential acceleration of a proton in a changing magnetic field. The initial calculation for acceleration was found to be incorrect due to neglecting the time-dependent changes in both the radius and magnetic field. By considering the relationship between the electric field, electric force, and charge, the correct tangential electric field was derived. Substituting the values for the electric field into the formula for tangential acceleration yielded the correct result of approximately 4.79 x 10^4 m/s². The importance of accounting for all variables in the equations was emphasized throughout the conversation.
Meow12
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Homework Statement
In earlier learning sequences we described how a static magnetic field cannot change the speed (and therefore kinetic energy) of a free charged particle. A changing magnetic field can, and this is one way particle beams are accelerated. Consider free protons following a circular path in a uniform magnetic field with a radius of 1 m. At t=0s, the magnitude of the uniform magnetic field begins to increase at 0.001T/s. Enter the tangential acceleration of the protons in positive if they speed up and negative if they slow down.
Relevant Equations
##\displaystyle R=\frac{mv}{qB}##
##\displaystyle R=\frac{mv}{qB}\implies v=\frac{RqB}{m}## where ##v## is the speed of the proton

##\displaystyle\frac{dv}{dt}=\frac{Rq}{m}\frac{dB}{dt}##

On substituting the values, I get ##\displaystyle\frac{dv}{dt}=9.58\times 10^4\ m/s^2##

This answer, however, is incorrect. Where have I gone wrong?
 
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For a circle of radius R,
E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}
Maybe it is worth considered.
 
anuttarasammyak said:
View attachment 338046
For a circle of radius R,
E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}
Maybe it is worth considered.
But how do we find ##\displaystyle\frac{dv}{dt}##?

Also, why was my solution incorrect?
 
Meow12 said:
Also, why was my solution incorrect?
You should consider not only B, v but also R changes in time.
 
anuttarasammyak said:
You should consider not only B, v but also R changes in time.
So, the radius R also increases as the speed v increases. But how do we find the tangential acceleration dv/dt?
 
You can try to get it from the equation I mentioned.
 
anuttarasammyak said:
E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}
Are you missing an ##R## in the last term?
 
renormalize said:
Are you missing an ##R## in the last term?
I don't think so because
E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{R}{2}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}
 
anuttarasammyak said:
I do not think so because
E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{R}{2}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}
Ah yes, you are correct!
 
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  • #10
Meow12 said:
So, the radius R also increases as the speed v increases. But how do we find the tangential acceleration dv/dt?
Consider the equation that @anuttarasammyak wrote above for the tangential electric field ##E##:$$E=-\frac{R}{2}\frac{dB}{dt}$$What's the relation between an electric field, an electric force and a charge?
 
  • #11
renormalize said:
Consider the equation that @anuttarasammyak wrote above for the tangential electric field ##E##:$$E=-\frac{R}{2}\frac{dB}{dt}$$What's the relation between an electric field, an electric force and a charge?
Plugging in the values, I get ##|E|=5\times 10^{-4}\ N/C##

##\displaystyle a_t=\frac{q|E|}{m}##

Substituting the value of ##|E|##, and the values of ##q## and ##m## for a proton, I get ##a_t=4.79\times 10^4\ m/s^2##, which is the right answer. Thank you so much, both of you. :)
 
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