Question

Question: In Module 28-4, we showed that a particle of charge and mass will move in a circle of radius$\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{v}\mathbf{/}\mathbf{\left|}\mathit{q}\mathbf{\right|}\mathit{B}$when its velocity is perpendicular to a uniform magnetic field . We also found that the period T of the motion is independent of speed v. These two results are approximately correct if v<<c . For relativistic speeds, we must use the correct equation for the radius:

$\mathit{r}\mathbf{=}\frac{\mathbf{p}}{\left|q\right|\mathbf{B}}\mathbf{=}\frac{\mathbf{\gamma }\mathbf{m}\mathbf{v}}{\left|q\right|\mathbf{B}}$

(a) Using this equation and the definition of period ($\mathit{T}\mathbf{}\mathbf{=}\mathbf{2}\mathit{\Pi }\mathit{r}\mathbf{/}\mathit{v}$), find the correct expression for the period. (b) Is independent of v? If a 10.0 MeV electron moves in a circular path in a uniform magnetic field of magnitude 2.20T, what are (c) the radius according to Chapter 28, (d) the correct radius, (e) the period according to Chapter 28, and (f) the correct period?

Short answer

Answer

(a)The correct expression for time period is$T=\frac{2\pi \gamma m}{\left|q\right|B}$

(b)No, the correct time period is dependent on speed.

(c)The radius according to chapter 28 is 4.85 mm.

(d)The correct radius is $~16mm$.

(e)The time period according to Ch 28 is16ps .

(f)The correct time period is0.334 ns .

Step-by-step solution

Charge particle circulating in uniform magnetic field.

A charged particle moving perpendicular to the uniform magnetic field traces a circular path and the radius of the circular path is given by

$r=\frac{mv}{\left|q\right|B}$ …… (1)

Here, is the magnitude of the charge and is the magnitude of uniform magnetic field and is the velocity of thee charged particle.

The time period of the circular motion can be expressed as

$$\begin{gathered} T=\frac{2\pi r}{v} \\ =\frac{2\pi }{v}\left(\frac{mv}{\left|q\right|B}\right) \\ =\frac{2\pi m}{\left|q\right|B} \end{gathered}$$ …… (2)

It is observed that time period is independent of the velocity, this is the case for velocities much lower than speed of light. For relativistic speeds the expressions would be a little different which we see in further steps.

(a) Time period expression for relativistic speeds.  

The radius of the circular path in relativistic speeds is given as

$r=\frac{\gamma mv}{\left|q\right|B}$ …… (3)

Here,$\gamma$ is the Lorentz factor and is defined as

$\gamma =\frac{1}{\sqrt{1-(v^2/c^2)}}$

The time period of the circular motion can be expressed as

$$\begin{gathered} T=\frac{2\pi r}{v} \\ =\frac{2\pi }{v}\left(\frac{\gamma mv}{\left|q\right|B}\right) \\ =\frac{2\pi \gamma m}{\left|q\right|B} \end{gathered}$$ …… (4)

(b) Determine the speed dependency of Time period.

No, for relativistic speeds, the time period is dependent on speed and thus the obtained expression will result in the correct time period of the circulating charged particle.

(c) Calculate the classical radius.

We need to first determine the velocity of particle from classical kinetic energy expression.

$$\begin{gathered} \frac{1}{2}{m}_e{v}^2=10MeV \\ v=\sqrt{\frac{2\left(10\times {10}^{6}eV\right)\left(1.6\times {10}^{-19}J/eV\right)}{9.1\times {10}^{-31}kg}} \\ =\sqrt{3.52\times {10}^{18}}m/s \\ =1.875\times {10}^{9}m/s \end{gathered}$$

Inserting this in the classical expressions (1),

$$\begin{gathered} r=\frac{mv}{\left|q\right|B} \\ =\frac{m_{e}v}{eB} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} r=\frac{\left(9.1\times {10}^{-31}kg\right)\left(1.875\times {10}^{9}m/s\right)}{\left(1.6\times {10}^{-19}C\right)\left(2.20T\right)} \\ =4.85\times {10}^{-3}m \\ =4.85mm \end{gathered}$$

(d) Calculate relativistic radius for 10.0 MeV electron.

Relativistically, the velocity of the electron is determined as

$$\begin{gathered} \left(\gamma -1\right)m{c}^2=10MeV \\ \gamma =\frac{10MeV}{m_e{c}^2}+1 \end{gathered}$$

For Electron, rest mass is$\text{9.1}\times {10}^{-31}kg$ and its energy equivalent is $\approx \text{0.511}MeV$. Therefore substituting$\text{0.511}MeV$ for $m_e{c}^2$in the above equation, we get

$$\begin{gathered} \gamma =\frac{10MeV}{0.511MeV}+1 \\ =19.57+1 \\ =20.57 \end{gathered}$$

And thus, speed can be calculated as

$$\begin{gathered} \gamma =20.57 \\ \frac{1}{\sqrt{1-}(v^2/c^2)} \\ 90=20.57 \\ \frac{v}{c}=\sqrt{1-\frac{1}{{\left(20.57\right)}^2}} \\ =0.9988 \end{gathered}$$

Inserting this in equation (3)

$$\begin{gathered} r=\frac{\gamma mv}{\left|q\right|B} \\ =\frac{\left(20.57\right)\left(9.1\times {10}^{-31}kg\right)\left(0.9988\times 3\times {10}^{8}m/s\right)}{\left(1.6\times {10}^{-19}J\right)\left(2.20T\right)} \\ =159.4\times {10}^{-4}m \\ =15.9mm \end{gathered}$$

(e) Determine time period for classical as well as relativistic case 

Time period, classically as given Chapter 28 is given by

$$\begin{gathered} T=\frac{2\pi m_e}{\left|q\right|B} \\ =\frac{2\pi \left(9.1\times {10}^{-31}kg\right)}{\left(1.6\times {10}^{-19}C\right)\left(2.20T\right)} \\ =16\times {10}^{-12}s \\ =16ps \end{gathered}$$

Time period for classical limit will get $16pico-second$. ($1ps=1\times {10}^{-12}s$)

(f) Determine time period for relativistic case that is the correct time period.

And the correct time period (equation (4)) is

$$\begin{gathered} T=\frac{2\pi \gamma m}{\left|q\right|B} \\ =\gamma \left(\frac{2\pi m}{\left|q\right|B}\right) \\ =\left(20.57\right)\left(16\times {10}^{-12}s\right) \\ =0.334ns \end{gathered}$$

Thus, the correct time period is $0.334nano-second$.