Chapter 29, Problem 29.28P

A particle in the cyclotron shown in Figure 28.16a gains energy qΔV from the alternating power supply each time it passes from one dee to the other. The time interval for each full orbit is

$T=\frac{2\pi }{\omega }=\frac{2\pi m}{qB}$

so the particle’s average rate of increase in energy’ is

$\frac{2q\Delta V}{T}=\frac{q^{2}B\Delta V}{\pi m}$

Notice that this power input is constant in time. On the other hand, the rate of increase in the radius r of its path is not constant. (a) Show that the rate of increase in the radius r of the panicle’s path is given by

$\frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}$

(b) Describe how the path of the particles in Figure 28.16a is consistent with the result of part (a). (c) At what rate is the radial position of the protons in a cyclotron increasing immediately before the protons leave the cyclotron? Assume the cyclotron has an outer radius of 0.350 m, an accelerating voltage of ΔV = 600 V, and a magnetic field of magnitude 0.800 T. (d) By how much does the radius of the protons’ path increase during their last full revolution?

Figure 28.16 (a) A cyclotron consists of an ion source at P, two does D₁ and D₂ across which an alternating potential difference is applied, and a uniform magnetic field. (The south pole of the magnet is not shown.) (b) The first cyclotron, invented by E. O. Lawrence and M. S. Livingston in 1934.

To determine

To prove: The rate of increase in the radius $r$ of the particles path is given by $\frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}$.

Answer to Problem 29.28P

The rate of increase in the radius $r$ of the particles path is given by $\frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}$.

Explanation of Solution

Given Info: The time interval of full orbit is $\frac{2\pi m}{qB}$ and the rate of increase in energy is $\frac{2q\Delta V}{T}=\frac{q^{2}B\Delta V}{\pi m}$.

The formula for the energy is,

$E=\frac{1}{2}m{v}^2$ (1)

Here,

$m$ is the mass of proton.

$v$ is the velocity.

Differentiating equation (1) with respect to time

$\frac{dE}{dt}=mv\frac{dv}{dt}$

The above equation can be rewritten as,

$\frac{dv}{dt}=\frac{1}{mv}\frac{dE}{dt}$

Substitute $\frac{q^{2}B\Delta V}{\pi m}$ for $\frac{dE}{dt}$ in the above equation

$\frac{dv}{dt}=\frac{1}{mv}\frac{q^{2}B\Delta V}{\pi m}$ (2)

The formula for the centripetal force is,

$F=\frac{m{v}^2}{r}=qvB$

The above equation can be rewritten as,

$r=\frac{mv}{qB}$ (3)

Differentiating equation (3) with respect to time

$\frac{dr}{dt}=\frac{m}{qB}\frac{dv}{dt}$ (4)

Deducing from equation (2) and equation (4),

$\begin{array}{l}\frac{dr}{dt}=\frac{q}{mV}\frac{\Delta V}{\pi }\\ \frac{dr}{dt}=\frac{1}{rB}\frac{\Delta V}{\pi }\end{array}$

Conclusion:

Therefore, the rate of increase in the radius $r$ of the particles path is given by $\frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}$

To determine

The path of the particle is consistent.

Answer to Problem 29.28P

The path of the particle is consistent.

Explanation of Solution

Given Info: The time interval of full orbit is $\frac{2\pi m}{qB}$ and the rate of increase in energy is $\frac{2q\Delta V}{T}=\frac{q^{2}B\Delta V}{\pi m}$.

The formula of change of radius with time is,

$\frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}$

The value of the path of the particle is consistent with respect to time as according to the above formula the path is dependent on the radius of circle and the magnitude of the magnetic field which remains constant for a path.

Thus, the path of the particle is consistent.

Conclusion:

Therefore, the path of the particle is consistent.

To determine

The rate of increase of the radial direction of proton.

Answer to Problem 29.28P

The rate of increase of the radial direction of proton. is $686.1\text{m}/\text{s}$.

Explanation of Solution

Given Info: The time interval of full orbit is $\frac{2\pi m}{qB}$ and the rate of increase in energy is $\frac{2q\Delta V}{T}=\frac{q^{2}B\Delta V}{\pi m}$. The outer radius is $0.350\text{m}$, the accelerating voltage is $600\text{V}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the change of radius with time is,

$\frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}$

Substitute $0.350\text{m}$ for $r$, $0.800\text{T}$ for $B$ and $600\text{V}$ for $\Delta V$ in above equation to find $\frac{dr}{dt}$.

$\begin{array}{c}\frac{dr}{dt}=\frac{1}{\pi }\frac{\left(600\text{V}\right)}{\left(0.350\text{m}\right)\left(0.800\text{T}\right)}\\ =686.1\text{m}/\text{s}\end{array}$

Thus, the rate of increase of the radial direction of proton is $686.1\text{m}/\text{s}$.

Conclusion:

Therefore, the rate of increase of the radial direction of proton is $686.1\text{m}/\text{s}$.

To determine

The increase in the radius of the path of proton.

Answer to Problem 29.28P

The increase in the radius of the path of proton is $100\text{μm}$.

Explanation of Solution

Given Info: The time interval of full orbit is $\frac{2\pi m}{qB}$ and the rate of increase in energy is $\frac{2q\Delta V}{T}=\frac{q^{2}B\Delta V}{\pi m}$. The outer radius is $0.350\text{m}$, the accelerating voltage is $600\text{V}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the velocity is,

$v=\frac{rqB}{m}$

Substitute $0.350\text{m}$ for $r$, $1.6\times {10}^{-19}\text{C}$ for $q$, $1.6\times {10}^{-27}\text{kg}$ for $m$ and $0.800\text{T}$ for $B$ in above equation to find $v$ .

$\begin{array}{c}v=\frac{\left(0.350\text{m}\right)\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)}{\left(1.6\times {10}^{-27}\text{kg}\right)}\\ =2.8\times {10}^7\text{m}/\text{s}\end{array}$

Thus, the velocity of proton is $2.8\times {10}^7\text{m}/\text{s}$.

The formula for the energy is,

$E=\frac{1}{2}m{v}^2$

Substitute $1.6\times {10}^{-27}\text{kg}$ for $m$ and $2.8\times {10}^7\text{m}/\text{s}$ for $v$ in above equation to find $E$.

$\begin{array}{c}E=\frac{1}{2}\left(1.6\times {10}^{-27}\text{kg}\right){\left(2.8\times {10}^7\text{m}/\text{s}\right)}^2\\ =6.272\times {10}^{-13}\text{J}\end{array}$

The formula for the energy at the end is,

$\begin{array}{c}{E}_{end}=E-\Delta E_{rev}\\ =E-2q\Delta V\end{array}$

Substitute $6.272\times {10}^{-13}\text{J}$ for $E$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $600\text{V}$ for $\Delta V$ in above equation to find $E_{\text{end}}$.

$\begin{array}{c}{E}_{\text{end}}=E-\Delta E_{\text{rev}}\\ =6.272\times {10}^{-13}\text{J}-2\left(1.6\times {10}^{-19}\text{C}\right)\left(600\text{V}\right)\\ =6.272\times {10}^{-13}\text{J}-0.002\times {10}^{-13}\text{J}\\ =\text{6}\text{.270}\times {10}^{-13}\text{J}\end{array}$

The formula for the radius at the end is,

$r_{\text{end}}=\frac{\sqrt{2E_{\text{end}}m}}{qB}$

Substitute $6.27\times {10}^{-13}\text{J}$ for $E_{\text{end}}$, $1.6\times {10}^{-27}\text{kg}$ for $m$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $0.800\text{T}$ for $B$ in above equation to find $r_{\text{end}}$.

$\begin{array}{c}{r}_{\text{end}}=\frac{\sqrt{2\left(6.27\times {10}^{-13}\text{J}\right)\left(1.6\times {10}^{-27}\text{kg}\right)}}{\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)}\\ =0.3499\text{m}\end{array}$

The formula for the increase in the radius is,

$\Delta r=r-r_{\text{end}}$

Substitute $0.350\text{m}$ for $r$ and $0.3499\text{m}$ for $r_{\text{end}}$ in above equation to find $\Delta r$

$\begin{array}{c}\Delta r=0.350\text{m}-0.3499\text{m}\\ =0.0001\text{m}\times \left(\frac{{10}^6\text{μm}}{1\text{m}}\right)\\ =100\text{μm}\end{array}$

Thus the increase in the radius of the path of proton is $100\text{μm}$.

Conclusion:

Therefore, increase in the radius of the path of proton is $100\text{μm}$.