Loss of energy via EM waves by a charged particle in a uniform magnetic field A charge moving in a uniform magnetic field travels in a circular path and objects moving in circular paths have a constant acceleration. It is well known that accelerating charges produce electromagnetic waves. So eventually the particle in this setup must move towards the centre of its circular path.
I tried to calculate the time for this to occur and got a rather quite strange answer. Here's the derivation:
Now from the Larmor formula:
$$P=\frac{q^2 a^2}{6\pi \epsilon_0 c^3}$$
Hence the total energy of the system over time can be formulised as:
$$E(t)=E_i-\int_0^t P dt$$
Now differentiating wrt t:
$$\frac{d}{dt}\left(E(t)=E_i-\int_0^t P dt\right)\Rightarrow E'(t)=-P$$
Since the charged particle has only kinetic energy, $E'(t)=\frac{d}{dt}\left(0.5mv^2\right)=mav$
Now inputing and some algebra:
$$mav=-\frac{q^2 a^2}{6\pi \epsilon_0 c^3}\Rightarrow -6\pi \epsilon_0 c^3mv=q^2 a \Rightarrow \int_0^t\frac{-6\pi \epsilon_0 c^3m }{q^2}dt=\int_{v_0}^v\frac{dv}{v} $$
This produces the answer;
$$v=v_0 \exp\left(\frac{-6\pi \epsilon_0 c^3m }{q^2}t\right)$$
Now calculating the half life of the speed,
$$t_{\frac{1}{2}}=\left(\frac{\ln(2)q^2}{6\pi \epsilon_0 c^3m }\right)$$
which for an protonis $2.46 \cdot 10^{-27}$ which is too short. Can someone explain why this is happening? Because in cyclotrons protons can whizz around for longer than that.
 A: After this equation
$mav=-\frac{q^2 a^2}{6\pi \epsilon_0 c^3}$
you seem to have cancelled an $a$, but the $a$s on the left and right are different.
The one on the left is the acceleration tangentially to the motion, i.e the one that reduces the magnitude of $v$, but the one on the right is the acceleration towards the centre of the circle.
Best to start from the equation above and swap the $a$ on the right for $\frac{Bqv}{m}$ , the other steps you did seem ok, it's a similar derivation from there...
for a proton at a speed of about 10% the speed of light, in a 1 Tesla magnetic field, it's giving a deceleration of about $0.001ms^{-2}$ and so would take a long time to halve the speed. The deceleration depends inversely on the mass cubed, so for an electron it's much higher and of the order of seconds to halve the speed, depending on the strength of the magnetic field used.
Your final formula should end up as
$$t_{\frac{1}{2}}=\left(\frac{\ln(2)6\pi \epsilon_0 c^3m^3 }{B^2q^4}\right)$$
The centripetal force comes from the magnetic field
$\frac{mv^2}{r} = Bqv$
so the radius of the circle is proportional to the velocity and the time for the radius to halve is from the same formula.
