Tangential acceleration of proton due to a changing magnetic field

[Meow12]

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?

 

[anuttarasammyak]

For a circle of radius R,
[tex] E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}[/tex]
Maybe it is worth considered.

 

[Meow12]

View attachment 338046
For a circle of radius R,
[tex] E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}[/tex]
Maybe it is worth considered.

But how do we find ##\displaystyle\frac{dv}{dt}##?

Also, why was my solution incorrect?

 

[anuttarasammyak]

Also, why was my solution incorrect?

You should consider not only B, v but also R changes in time.

 

[Meow12]

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?

 

[anuttarasammyak]

You can try to get it from the equation I mentioned.

 

[renormalize]

[tex] E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}[/tex]

Are you missing an ##R## in the last term?

 

[anuttarasammyak]

Are you missing an ##R## in the last term?

I don't think so because
[tex]E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{R}{2}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}[/tex]

 

[renormalize]

I do not think so because
[tex]E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{R}{2}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}[/tex]

Ah yes, you are correct!

 

[renormalize]

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?

 

[Meow12]

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. :)