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,
$$E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}$$
Maybe it is worth considered.

 

[Meow12]

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?

 

[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]

$$E = -\frac{\pi R^2}{2\pi R}\frac{dB}{dt}=-\frac{mv}{2qB}\frac{dB}{dt}$$

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

 

[anuttarasammyak]

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

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

 

[renormalize]

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!

 

[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. :)