Chapter 19, Problem 114P

(a)

To determine

Frequency of oscillation.

Answer to Problem 114P

Frequency of oscillation is $20\text{MHz}$.

Explanation of Solution

Write an expression for the speed.

  $v=\frac{eBr}{m}$                                                                                                                     (I)

Here, $v$ is the speed, $e$ is the charge, $B$ is the magnetic field, $r$ is the radius and $m$ is the mass.

Write an expression for frequency.

  $f=\frac{v}{C}$                                                                                                              (II)

Here, $f$ is the frequency and $C$ is the circumference.

Write an expression for the circumference.

  $C=2\pi r$                                                                                                        (III)

Here, $r$ is the radius.

Rewrite equation (II) by using equation (I) and (III).

  $\begin{array}{c}f=\frac{\left(\frac{eBr}{m}\right)}{2\pi r}\\ =\frac{eB}{2\pi m}\end{array}$                                                                                                       (IV)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.3\text{T}$  for $B$ and $1.673\times {10}^{-27}\text{kg}$ for $m$ in equation (IV) to find $f$.

    $\begin{array}{c}f=\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(1.3\text{T}\right)}{2\pi \left(1.673\times {10}^{-27}\text{kg}\right)}\\ =\left(20\times {10}^6\text{Hz}\right)\left(\frac{1\text{MHz}}{{10}^6\text{Hz}}\right)\\ =20\text{MHz}\end{array}$

Thus, the frequency of oscillation is $20\text{MHz}$.

(b)

To determine

The kinetic energy of the proton.

Answer to Problem 114P

The kinetic energy of the proton is $3.3\times {10}^{-13}\text{J}$.

Explanation of Solution

Write an expression for the kinetic energy.

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

Here, $K$ is the kinetic energy of the proton.

Rewrite equation (V) using equation (I).

  $\begin{array}{c}K=\frac{1}{2}m{\left(\frac{eBr}{m}\right)}^2\\ =\frac{e^2{B}^2{r}^2}{2m}\end{array}$                                                                                          (VI)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.3\text{T}$  for $B$, $16\text{cm}$ for $r$ and $1.673\times {10}^{-27}\text{kg}$ for $m$ in equation (VI) to find $E$.

    $\begin{array}{c}K=\frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^2{\left(1.3\text{T}\right)}^2{\left(\left(16\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2}{2\left(1.673\times {10}^{-27}\text{kg}\right)}\\ =\frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^2{\left(1.3\text{T}\right)}^2{\left(0.16\right)}^2}{2\left(1.673\times {10}^{-27}\text{kg}\right)}\\ =3.3\times {10}^{-13}\text{J}\end{array}$

Thus, the kinetic energy of the proton is $3.3\times {10}^{-13}\text{J}$.

(c)

To determine

The equivalent voltage necessary to accelerate protons.

Answer to Problem 114P

The equivalent voltage necessary to accelerate protons is $\text{2}\text{.1 MV}$.

Explanation of Solution

Write an expression for the equivalent voltage necessary to accelerate protons.

    $\Delta V=\frac{K}{e}$                                                                                                              (VII)

Here, $\Delta V$  is the equivalent voltage necessary to accelerate protons.

Rewrite equation (VII) using equation (VI).

    $\begin{array}{c}\Delta V=\frac{\left(\frac{e^2{B}^2{r}^2}{2m}\right)}{e}\\ =\frac{e{B}^2{r}^2}{2m}\end{array}$ (VIII)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.3\text{T}$  for $B$, $16\text{cm}$ for $r$ and $1.673\times {10}^{-27}\text{kg}$ for $m$ in equation (VIII) to find $\Delta V$.

      $\begin{array}{c}\Delta V=\frac{\left(1.602\times {10}^{-19}\text{C}\right){\left(1.3\text{T}\right)}^2{\left(\left(16\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2}{2\left(1.673\times {10}^{-27}\text{kg}\right)}\\ =\frac{\left(1.602\times {10}^{-19}\text{C}\right){\left(1.3\text{T}\right)}^2{\left(0.16\right)}^2}{2\left(1.673\times {10}^{-27}\text{kg}\right)}\\ =\left(\text{2}\text{.1}\times {10}^6\text{V}\right)\left(\frac{1\text{MV}}{{10}^6\text{V}}\right)\\ =\text{2}\text{.1 MV}\end{array}$

Thus, the equivalent voltage necessary to accelerate protons is $\text{2}\text{.1 MV}$.

(d)

To determine

The minimum number of revolutions each proton has to make in the cyclotron.

Answer to Problem 114P

The minimum number of revolutions each proton has to make in the cyclotron is $100\text{rev}$.

Explanation of Solution

Write an expression for the minimum number of revolutions each proton has to make in the cyclotron.

    $N=\frac{K}{2e\Delta V}$                                                                                                            (IX)

Here, $N$  is the minimum number of revolutions each proton has to make in the cyclotron.

Conclusion:

Substitute $3.3\times {10}^{-13}\text{J}$ for $K$, $1.602\times {10}^{-19}\text{C}$ for $e$ and $10.0\times {10}^3\text{V/rev}$ for $\Delta V$ in equation (IX) to find $N$.

    $\begin{array}{c}N=\frac{3.3\times {10}^{-13}\text{J}}{2\left(1.602\times {10}^{-19}\text{C}\right)\left(10.0\times {10}^3\text{V/rev}\right)}\\ =\frac{3.3\times {10}^{-13}\text{J}}{2\left(1.602\times {10}^{-15}\text{C}\cdot \text{V/rev}\right)}\\ =100\text{rev}\end{array}$

Thus, the minimum number of revolutions each proton has to make in the cyclotron is $100\text{rev}$.