Time of Electron/Proton to Radiate Away Half of Their Energy

[Physistory]

Homework Statement

It is shown in more advanced courses that charged particles in circular orbits radiate electromagnetic waves, called cyclotron radiation. As a result, a particle undergoing cyclotron motion with speed v is actually losing kinetic energy at the rate $$\frac{dK}{dt} = - \frac{μ_0q^4}{6 \pi cm^2}B^2v^2.$$

How long does it take a.) an electron and b.) a proton to radiate away half its energy while spiraling in a 2.0 T magnetic field?

Homework Equations

$K = \frac {mv^2}{2} = Iω$
$v = \frac {l}{t}$
$ω = \frac {v}{r}$
$K(t) = \int \frac{dK}{dt} = \int_0^t K(t)dt$

I have found that μ₀, q, c, m, B and obviously 6$\pi$ are all constants and can thus be pulled out of the integral.

3. The Attempt at a Solution
A hint was provided to replace v² with $\frac {2K}{m}$ and then integrate, so I have attempted to integrate said hint here. (Pun intended.)

I would like to know how to solve part a.) for the electron, as I should be able to apply it to part b.) in general. Please take into consideration that I am moderately wary, and I occasionally make simple mistakes. If I have made any slight or severe errors here, then please point them out and attempt to help me resolve them.

My progress thus far is as follows:

$$\frac{dK}{dt} = - \frac{μ_0q^4}{6 \pi cm^2}B^2\frac{2K}{m}$$
$$= - \frac{μ_0q^4}{6 \pi cm^3}B^22K$$
$$\int_0^t\frac{dK}{dt} = \int_0^t - \frac{μ_0q^4}{6 \pi cm^3}B^22Kdt$$
$$= - \frac{(2)(2 T)(4 \pi *10^-7)(-1.6*10^-19 C)^4}{6 \pi (9.11*10^-31 kg)^3} \int_0^t K(t)dt$$
$$= -2.31*10^-8 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{ml}{2t}dt$$
$$= -2.31*10^-8 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{Iω^2}{2}dt = -2.31*10^-8 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{I \vec v}{2r}dt$$
$$\frac{ml}{2t} = \frac{I \vec v}{2r} = (\frac{1}{2}) (\frac{l}{t})^2$$
I have assumed that l can equal r. Thus I have concluded that m, l, and v cancel each other with these equations.
$$-2.31*10^-8 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{I}{t^2}dt$$
$$= -2.31*10^-8 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{e}{t^2}dt = -2.31*10^-8 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{dt}{t^2}$$
$$(-2.31*10^-8 \frac {T^2mC^4}{Akg^3})(-1.6*10^-19 C) = 3.69*10^-11 \frac {T^2mC^4}{Akg^3}$$
$$3.69*10^-11 \frac {T^2mC^4}{Akg^3} \int_0^t \frac{1}{t^2} \, dt = \left. \frac {1}{t} \right|_0^t$$

I realized at this point that in trying to get K in terms of time, I ended up essentially back to square one, because I want to find a value for the time. I was very confused at first on how to write out the integral, and am still a bit unsure. To my understanding, K must be integrated in terms of t, but a numerical value for t is wanted. I believe I do not understand a key concept. I tried to find some online help, and found (though I will not use this until I find this answer on my own) that the correct time is 0.80 seconds. My current concern, however, is figuring out how to integrate properly.

The correct method of integration is likely very obvious, and I apologize if I am completely missing it. I plan to consult professors in person, but would nevertheless appreciate any and all assistance.

By the way, this is my first time using LaTex. I tried to compose everything perfectly; please forgive me if something is amiss, especially if mistakes are present.

 

[Orodruin]

Integrating really is not the issue. You have a differential equation on the form $y’(t) = -ay(t)$. Do you know how to solve such a differential equation?

 

[Physistory]

Integrating really is not the issue. You have a differential equation on the form $y’(t) = -ay(t)$. Do you know how to solve such a differential equation?

First of all, thank you very much for the hint, and I am sorry for my delayed response. I sought in-person help today, and was led to something similar. (By the way, in the fifth line of my above integration, I mistook the moment of inertia I to be the current, as I had been studying currents previously and was still very focused on that. It does not matter now for this problem, but I just want to point out that I realize my mistake there for the future.

That aside, the work I have now is as follows:

Since $K \rightarrow \frac{1}{2}K_0, v \rightarrow \frac{1}{\sqrt 2}v_0$.

My tutor equated all the constants to the symbol α, so I will be using that here.

$$\frac{dK}{K} = -αB^2v^2 = -αB^2(\frac{2K}{m})$$

Multiplying each side by $dt$ and dividing each side by $K$, I'm left with:

$$\int \frac{dK}{K} = - \frac {2αB^2}{m} \int dt$$ which becomes

$$logK(t) = - \frac{2αB^2}{m}(t-t_0)$$ I feel like this is a form of what you were trying to lead me to. From this,

$$K(t)=e^{- \frac{2αB^2}{m}(t-t_0)}$$

My tutor emphasized the ratio found by dividing $K$ at $t_{\frac{1}{2}}$ by $K$ at $t_0$, which was $$\frac {K(t=t_{\frac{1}{2}})} {K(t=0)} = \frac{1}{2}$$

From this, we found that $K(t_{\frac{1}{2}})=\frac{1}{2}K(t=0)$. I was able to figure out things better from here, and from this gleaned

$$\frac {e^{- \frac{2αB^2}{m}(t_{\frac{1}{2}}-t_0)}}{e^{- \frac{2αB^2}{m}(0-t_0)}} = \frac{1}{2}$$ So,

$$e^{- \frac{2αB^2}{m}(t_{\frac{1}{2}}-t_0)} = \frac{1}{2} e^{- \frac{2αB^2}{m}(-t_0)}$$

From this, several things cancel, and I'm left with

$$e^{- \frac{2αB^2}{m}(t_{\frac{1}{2}})} = \frac{1}{2} \rightarrow log \frac{1}{2} = - \frac{2αB^2}{m}(t_{\frac{1}{2}}) \rightarrow t_{\frac{1}{2}} = - \frac{mlog(\frac{1}{2})}{2αB^2}$$ My tutor said that the final equation I got for $t_{\frac{1}{2}}$ was "exactly right." From here, I assumed that all that was left to do was plug in the values to find $t$.

The back of our textbook said that the exact answers are a.) $0.45$ seconds and b.) $2.8*10^9$ seconds. My answers were a.) $0.40$ seconds and b.) $2.41*10^9$ seconds. Would you say that our progress thus far is correct, and that my answers are close enough to be considered right?