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

$\frac{dK}{dt}=-\left(\frac{{\mu }_0{q}^4}{6{\mathrm{\pi cm}}^2}\right)B^2{v}^2$

How long does it take $\left(a\right)$an electron and $\left(b\right)$a proton to radiate away half its energy while spiraling in a $2.0T$magnetic field

The particle undergoing cyclotron motion with speed is given as

$\frac{dK}{dt}=-\left(\frac{{\mu }_0{q}^4}{6{\mathrm{\pi cm}}^2}\right)B^2{v}^2$

The objectives are to find the proton and electron to radiate away half its energy while spiraling in a $2.0T$magnetic field

The rate at which kinetic energy is lost is determined by

$\frac{dK}{dt}=-\left(\frac{{\mu }_0{q}^4}{6{\mathrm{\pi cm}}^2}\right)B^2{v}^2$

Because kinetic energy equals half the product of mass and velocity squared, we can get an expression for $v^2$by

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ {v}^2=\frac{2K}{m} \end{gathered}$$

So, in equation $\left(1\right)$, we used $v^2$ expression and separated $dK$ and $dt$ on each side.

$$\begin{gathered} \frac{dK}{dt}=-\left(\frac{{\mu }_0{q}^4}{6{\mathrm{\pi cm}}^2}\right)B^2\left(\frac{2K}{m}\right) \\ \frac{dK}{K}=-\left(\frac{{\mu }_0{q}^4{B}^2}{3{\mathrm{\pi cm}}^3}\right)dt \end{gathered}$$

We now integrate both sides of equation $\left(2\right)$to obtain an expression for the time $t$at which the kinetic energy loses half of its energy.

$$\begin{gathered} {\int }_K^{0.5K}\frac{dK}{K}={\int }_0^t\left(\frac{{\mu }_0{q}^4{B}^2}{3{\mathrm{\pi cm}}^3}\right)dt \\ \ln 0.5K-\ln K=-\left(\frac{{\mu }_0{q}^4{B}^2}{3{\mathrm{\pi cm}}^3}\right)t \\ \ln 0.5=-\left(\frac{{\mu }_0{q}^4{B}^2}{3{\mathrm{\pi cm}}^3}\right)t \\ t=\left(\frac{-3{\mathrm{\pi cm}}^3\ln 0.5}{{\mu }_0{q}^4{B}^2}\right) \end{gathered}$$

The mass of the electron is $m=-9.11\times {10}^{-31}kg$, therefore we insert $c,m,{\mu }_0,q,andB$into equation $\left(3\right)$to get $t$.

$$\begin{gathered} t=\left(\frac{-3{\mathrm{\pi cm}}^3\ln 0.5}{{\mu }_0{q}^4{B}^2}\right) \\ =\left(\frac{-3\mathrm{\pi }(3\times {10}^8\mathrm{m}/\mathrm{s}){(9.11\times {10}^{-31}\mathrm{kg})}^3\ln 0.5}{(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}\times {\displaystyle \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}){(1.6\times {10}^{-19}\mathrm{C})}^4(2\mathrm{T}{)}^2}\right) \\ =0.45s \end{gathered}$$

The mass of a proton is $m=1.67\times {10}^{-27}kg$, thus we insert $c,m,{\mu }_0,q,andB$ into equation $\left(3\right)$ to get $t$.

$$\begin{gathered} t=\left(\frac{-3{\mathrm{\pi cm}}^3\ln 0.5}{\hspace{0.17em}{\mu }_0{q}^4{B}^2}\right) \\ =\left(\frac{-3\mathrm{\pi }(3\times {10}^8\mathrm{m}/\mathrm{s}){(1.67\times {10}^{-27}\mathrm{kg})}^2\ln 0.5}{(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}\times {\displaystyle \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}){(1.6\times {10}^{-19}\mathrm{C})}^4(2\mathrm{T}{)}^2}\right) \\ =2.8\times {10}^{9}s \end{gathered}$$