- #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 μ

_{0}, 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

3. The Attempt at a Solution

A hint was provided to replace v

^{2}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.