Chapter 29, Problem 27P

A cyclotron (Fig. 28.16) designed to accelerate protons has an outer radius of 0.350 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T. (a) Find the cyclotron frequency for the protons in this cyclotron. Find (b) the speed at which protons exit the cyclotron and (c) their maximum kinetic energy. (d) How many revolutions does a proton make in the cyclotron? (e) For what time interval does the proton accelerate?

To determine

The frequency of the proton.

Answer to Problem 27P

The frequency of the proton is. $7.66\times {10}^7{\text{s}}^{-1}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the frequency is,

$f=\frac{qB}{2\pi m}$

Here,

$B$ is the magnitude of magnetic field.

$m$ is the mass of proton.

$q$ is the charge of proton.

$V$ is the accelerating voltage.

Substitute $0.800\text{T}$ for $B$, $1.6\times {10}^{-27}\text{kg}$ for $m$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $600\text{V}$ for $V$ in above equation to find $f$.

$\begin{array}{c}f=\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)}{2\pi \left(1.6\times {10}^{-27}\text{kg}\right)}\\ =7.66\times {10}^7{\text{s}}^{-1}\end{array}$

Thus, the frequency of the proton is. $7.66\times {10}^7{\text{s}}^{-1}$.

Conclusion:

Therefore, the frequency of the proton is. $7.66\times {10}^7{\text{s}}^{-1}$.

To determine

The exit speed of the proton.

Answer to Problem 27P

The exit speed of the proton is. $2.68\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the speed is,

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

Here,

$r$ is the radius of the outermost orbit.

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

$\begin{array}{c}v=\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)\left(0.350\text{m}\right)}{\left(1.6\times {10}^{-27}\text{kg}\right)}\\ =2.68\times {10}^7\text{m}/\text{s}\end{array}$

Thus, the exit speed of the proton is. $2.68\times {10}^7\text{m}/\text{s}$.

Conclusion:

Therefore, the exit speed of the proton is. $2.68\times {10}^7\text{m}/\text{s}$.

To determine

The maximum kinetic energy.

Answer to Problem 27P

The maximum kinetic energy is $3.75\text{MeV}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the kinetic energy is,

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

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m$ and $2.68\times {10}^7\text{m}/\text{s}$ for $v$ in above equation to find $K{E}_{\max }$.

$\begin{array}{c}K{E}_{\max }=\frac{1}{2}\left(1.67\times {10}^{-27}\text{kg}\right){\left(2.68\times {10}^7\text{m}/\text{s}\right)}^2\\ =5.99\times {10}^{-13}\text{J}\times \frac{\left(6.2\times {10}^{12}\text{MeV}\right)}{1\text{J}}\\ =3.75\text{MeV}\end{array}$

Thus, the maximum kinetic energy is $3.75\text{MeV}$.

Conclusion:

Therefore, the maximum kinetic energy is $3.75\text{MeV}$.

To determine

The number of revolutions.

Answer to Problem 27P

The number of revolutions are $3.13\times {10}^3\text{revolutions}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the number of revolutions is,

$N=\frac{E_{\text{max}}}{2qV}$

Substitute $600\text{V}$ for $V$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $4.4\times {10}^{-13}\text{J}$ for $K{E}_{\max }$ in above equation to find $N$ .

$\begin{array}{c}N=\frac{\left(4.4\times {10}^{-13}\text{J}\right)}{2\left(1.6\times {10}^{-19}\text{C}\right)\left(600\text{V}\right)}\\ =3.13\times {10}^3\text{revolutions}\end{array}$

Thus, the number of revolutions are $3.13\times {10}^3\text{revolutions}$.

Conclusion:

Therefore, the number of revolutions are $3.13\times {10}^3\text{revolutions}$.

To determine

The time of acceleration.

Answer to Problem 27P

The time of acceleration is $2.57\times {10}^{-4}\text{s}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the number of revolutions is,

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

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

$\begin{array}{c}T=\frac{2\pi \left(1.6\times {10}^{-27}\text{kg}\right)}{\left(1.6\times {10}^{-27}\text{kg}\right)\left(0.800\text{T}\right)}\\ =2.57\times {10}^{-4}\text{s}\end{array}$

Thus, the time of acceleration is $2.57\times {10}^{-4}\text{s}$.

Conclusion:

Therefore, the time of acceleration is $2.57\times {10}^{-4}\text{s}$.