Relativistic motion in particle accelerator

Question

This question is attempting to simulate the process of a circular particle accelerator. We are given a constant electric field $E$ along the angular direction and a varying magnetic field $B$ along the perpendicular direction so that the radius of the orbit of the particle is always being limited to be $R$. Now assuming the charged particle has mass $m$ and charge $q$, we are asked what is $\frac{v}{c}$ with respect to t. I have difficulty deriving such a relation and I know the following equation $$ \frac{dp}{dt}=m\frac{d\gamma v}{dt}=qvB+qE $$ could help. However, it can't lead me to my expected result. Therefore I wonder if anyone can give me a hint on how to work out the desired result.

Thanks.

Answer 1

Since this problem simulates a particle accelerator, and the magnets in particle accelerators do not change quickly, I'll ignore the extra electric field induced by the varying magnetic field and just treat the electric field as constant. Since the magnetic field is varying slowly, it is basically a static field and does no work on the particle. So, the velocity of the particle is dictated by the electric field. So, start from $dp/dt = qE$ and substitute in the correct expression for $p$ depending on whether you are working relativistically or not. Differentiate and solve for velocity.

Answer 2

That equation is only partially correct. Using vector notation, and assuming $v_r = 0$, we can write the EOM in radial coordinates:

$$ \frac{d \vec p}{d t}= m \frac{d\gamma \vec v}{d t} = q\,v_\theta\,B\,\vec e_r + q\,E\,\vec e_\theta $$

Now, remember the expression for the derivatives of angular versors:

$$ \frac{d}{d t} \vec e_r = \frac{v_\theta}{r} \vec e_\theta \\ \frac{d}{d t} \vec e_\theta = -\frac{v_\theta}{r} e_r $$

We can now split the EOM into the radial and angular parts: $$ \begin{cases} m \gamma \frac{v^ 2}{R} = q v B\\ m \frac{d \gamma v}{dt} = q E \end{cases} $$

Where we set $\vec v$ = $v \vec e_\theta$, as per the assumptions of the problem, i.e. we set all $v_r$ and $\frac{d v_r}{dt}$ to 0

Hence, you get:

$$ \gamma v = \frac{q}{m} E t + (\gamma v)_0 $$

and

$$ B = \frac{m \gamma v}{q R} $$