Chapter 29, Problem 28P

(a)

To determine

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

Answer to Problem 28P

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

Explanation of Solution

Write the expression to obtain the kinetic energy of the particle.

    $K.E=\frac{1}{2}m{v}^2$

Here, $K.E$ is the kinetic energy, $m$ is the mass of the particle and $v$ is the velocity of the particle.

Write the expression to obtain the rate of change of energy.

    $\frac{d}{dt}\left(K.E\right)=\frac{q^{2}B\Delta V}{\pi m}$

Here, $K.E$ is the kinetic energy, $q$ is the charge, $B$ is the magnetic field, $\Delta V$ is the potential difference and $m$ is the mass of the particle.

Substitute $\frac{1}{2}m{v}^2$ for $K.E$ in the above equation.

    $\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)=\frac{q^{2}B\Delta V}{\pi m}$                                                                                     (I)

Write the expression to obtain the speed of the particle.

    $v=\frac{Bqr}{m}$

Here, $v$ is the velocity of the particle, $B$ is the magnetic field, $q$ is the charge of the particle, $r$ is the radius of the orbit and $m$ is the mass of the particle.

Substitute $\frac{Bqr}{m}$ for $v$ in equation (I).

    $\begin{array}{l}\frac{d}{dt}\left(\frac{1}{2}m{\left(\frac{Bqr}{m}\right)}^2\right)=\frac{q^{2}B\Delta V}{\pi m}\\ \frac{B^2{q}^2}{2m}\left(\frac{d}{dt}{r}^2\right)=\frac{q^{2}B\Delta V}{\pi m}\\ \frac{B}{2}\left(\frac{d}{dt}{r}^2\right)=\frac{\Delta V}{\pi }\\ \frac{1}{2}\left(\frac{d}{dt}{r}^2\right)=\frac{\Delta V}{\pi B}\end{array}$

Further solve the above equation.

    $\begin{array}{l}\frac{1}{2}\left(2r\right)\frac{dr}{dt}=\frac{\Delta V}{\pi B}\\ \frac{dr}{dt}=\frac{1}{r}\frac{\Delta V}{\pi B}\end{array}$                                                                                                (II)

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

(b)

To determine

The comparison of path of the proton in figure 29.16a with the result in part (a).

Answer to Problem 28P

The path of the proton in figure 29.16a is unrealistic according to the result in part (a).

Explanation of Solution

Consider equation (II).

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

From the above expression the rate of change of radius with time is inversely proportional to the radius $r$. Thus the spiral path of the proton is much closer when the proton moves away from the center.

In case of the figure 29.16a, the spiral path of the proton is equally spaced throughout.

Therefore, the path of the proton in figure 29.16a is unrealistic according to the result in part (a).

(c)

To determine

The rate of change of radial position before the proton leaves the cyclone.

Answer to Problem 28P

The rate of change of radial position before the proton leaves the cyclone is $682.44\text{m}/\text{s}$.

Explanation of Solution

Conclusion:

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

    $\begin{array}{c}\frac{dr}{dt}=\frac{1}{\left(0.350\text{m}\right)}\frac{\left(600\text{V}\right)}{\pi \left(0.800\text{T}\right)}\\ =\frac{1}{\left(0.350\text{m}\right)}\frac{\left(600\text{V}\right)}{3.14\left(0.800\text{T}\right)}\\ =682.44\text{m}/\text{s}\end{array}$

Therefore, the rate of change of radial position before the proton leaves the cyclone is $682.44\text{m}/\text{s}$.

(d)

To determine

The increase in radius of the path of the proton during its last full revolution.

Answer to Problem 28P

The increase in radius of the path of the proton during its last full revolution is $5.59\times {10}^{-5}\text{m}$.

Explanation of Solution

Write the expression to obtain the increase in radius of the path of the proton during its last full revolution.

    $dr=\left(\frac{dr}{dt}\right)T$                                                                                                 (III)

Here, $dr$ is the increase in radius of the path of the proton during its last full revolution, $\left(\frac{dr}{dt}\right)$ is the rate of change of radial position before the proton leaves the cyclone and $T$ is the time required to complete one full orbit revolution by the proton.

Write the expression to obtain the time required to complete one full orbit revolution by the proton.

    $T=\frac{2\pi m}{qB}$

Here, $T$ is the time required to complete one full orbit revolution by the proton, $m$ is the mass of the proton, $q$ is the charge on the proton and $B$ is the magnetic field.

Conclusion:

Substitute, $0.800\text{T}$ for $B$, $1.6\times {10}^{-19}\text{C}$ and $1.67\times {10}^{-27}\text{kg}$ in the above equation to calculate $T$.

    $\begin{array}{c}T=\frac{2\pi \left(1.67\times {10}^{-27}\text{kg}\right)}{\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)}\\ =8.19\times {10}^{-8}\text{s}\end{array}$

Substitute $8.19\times {10}^{-8}\text{s}$ for $T$ and $682.44\text{m}/\text{s}$ for $\left(\frac{dr}{dt}\right)$ in equation (III) to calculate $dr$.

    $\begin{array}{c}dr=\left(682.44\text{m}/\text{s}\right)\left(8.19\times {10}^{-8}\text{s}\right)\\ =5.59\times {10}^{-5}\text{m}\end{array}$

Therefore, the increase in radius of the path of the proton during its last full revolution is

$5.59\times {10}^{-5}\text{m}$.