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

In summary, the conversation discusses the concept of cyclotron radiation and its impact on the kinetic energy of charged particles in circular motion. The equations and steps involved in solving for the time it takes for an electron and proton to lose half of their energy while spiraling in a 2.0 T magnetic field are discussed. The conversation also includes some mistakes made during the process and the correct method of solving the differential equation. The final answer obtained by the speaker is close to the correct answer provided in the textbook.
  • #1
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.
 
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  • #2
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?
 
  • #3
Orodruin said:
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?
 

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

1. How long does it take for an electron or proton to radiate away half of their energy?

The time it takes for an electron or proton to radiate away half of their energy is dependent on several factors such as their initial energy, the medium they are in, and their surroundings. Generally, it can take anywhere from a few nanoseconds to several minutes for this process to occur.

2. What is the significance of an electron or proton radiating away half of their energy?

When an electron or proton radiates away half of their energy, it is an indication of the decay of an unstable particle. This process is known as "half-life" and is used to measure the stability of a particle.

3. Can the time of electron/proton to radiate away half of their energy be manipulated?

The time it takes for an electron or proton to radiate away half of their energy is a natural process and cannot be manipulated. However, certain external factors such as high energy collisions or intense magnetic fields can affect this process.

4. How does the time of electron/proton to radiate away half of their energy differ between different particles?

The time it takes for an electron or proton to radiate away half of their energy can vary between different particles depending on their mass and energy. Generally, heavier particles have a longer half-life compared to lighter particles.

5. Is there a limit to how quickly an electron or proton can radiate away half of their energy?

There is no known limit to how quickly an electron or proton can radiate away half of their energy. The time it takes for this process to occur is dependent on the particle's energy and the environment it is in.

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