Chapter 29, Problem 78CP

(a)

To determine

The angle between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field.

Answer to Problem 78CP

The angle between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field is $8.798°$.

Explanation of Solution

Consider the free body diagram of the particle as shown below.

Figure-(1)

Write the expression to calculate the initial velocity of the proton.

    $v_{\text{y}}=\frac{BqS}{m_{\text{p}}}$                                                                                                                   (I)

Here, $v_{\text{y}}$ is the initial velocity, $B$ is the magnetic field, $q$ is the charge and $S$ is the distance.

Write the expression to calculate the final velocity of the proton.

    $\begin{array}{l}K=\frac{1}{2}{m}_{\text{p}}{v}_{\text{s}}^2\\ v_{\text{s}}=\sqrt{\frac{2K}{m_{\text{p}}}}\end{array}$                                                                                                               (II)

Here, $v_{\text{s}}$ is the final velocity of the proton and $K$ is the kinetic energy.

Write the expression to calculate the angle between the velocity.

    $\tan \alpha =\frac{v_{\text{y}}}{v_{\text{s}}}$                                                                                                               (III)

Conclusion:

Substitute $0.0500\text{T}$ for $B$, $1.6\times {10}^{19}\text{C}$ for $q$, $1.00\text{m}$ for $S$ and $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ in equation (I) to calculate $v_{\text{y}}$.

    $\begin{array}{c}{v}_{\text{y}}=\frac{\left(0.0500\text{T}\right)\left(1.60\times {10}^{-19}\text{C}\right)\left(1.00\text{m}\right)}{1.67\times {10}^{-27}\text{kg}}\\ =\frac{8\times {10}^{-21}}{1.67\times {10}^{-27}}\text{m}/\text{s}\\ =4.79\times {10}^6\text{m}/\text{s}\end{array}$

Substitute $5.00\text{MeV}$ for $K$ and $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ in equation (II) to calculate $v_{\text{s}}$.

    $\begin{array}{c}{v}_{\text{s}}=\sqrt{\frac{\left(2\right)5.00\text{MeV}\left(\frac{1\text{eV}}{{10}^{-6}\text{MeV}}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{1.67\times {10}^{-27}\text{kg}}}\\ =\sqrt{\frac{1.6\times {10}^{-24}}{1.67\times {10}^{-27}\text{kg}}}\\ =30.95\times {10}^6\text{m}/\text{s}\end{array}$

Substitute $4.79\times {10}^6\text{m}/\text{s}$ for $v_{\text{y}}$ and $30.95\times {10}^6\text{m}/\text{s}$ for $v_{\text{s}}$ in equation (III) to calculate $\alpha$.

    $\begin{array}{c}\tan \alpha =\frac{4.79\times {10}^6\text{m}/\text{s}}{30.95\times {10}^6\text{m}/\text{s}}\\ \alpha ={\tan }^{-1}\left(\frac{4.79\times {10}^6\text{m}/\text{s}}{30.95\times {10}^6\text{m}/\text{s}}\right)\\ \alpha =8.798°\end{array}$

Therefore, the angle between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field is $8.798°$.

(b)

To determine

The y component of the protons momentum as they leave the magnetic field.

Answer to Problem 78CP

The y component of the protons momentum as they leave the magnetic field is $6.19\times {10}^{21}\text{kg}\cdot \text{m}/\text{s}$

Explanation of Solution

Write the expression to calculate the velocity in x direction.

    $\begin{array}{l}K=\frac{1}{2}{m}_{\text{p}}{v}_{\text{x}}^2\\ v_{\text{x}}=\sqrt{\frac{2K}{m_{\text{p}}}}\end{array}$                                                                                                              (IV)

Here, $v_{\text{x}}$ is the final velocity of the proton and $K$ is the kinetic energy.

Write the expression to calculate time required by the proton to travel.

    $t=\frac{S}{v_{\text{x}}}$                                                                                                                       (V)

Here, $t$ is the time.

Write the expression to calculate the acceleration of the proton.

    $a=\frac{Bq{v}_{\text{x}}}{m_{\text{p}}}$                                                                                                                 (VI)

Here, $a$ is the acceleration.

Write the expression to calculate the velocity in the y direction.

    $v_{\text{y}}=u+at$                                                                                                              (VII)

Here, $v_{\text{y}}$ is the velocity in y direction.

Write the expression to calculate the momentum.

    $P_{\text{y}}=m_{\text{p}}{v}_{\text{y}}$                                                                                                              (VIII)

Here, $P_{\text{y}}$ is the momentum is y direction.

Conclusion:

Substitute $5.00\text{MeV}$ for $K$ and $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ in equation (IV) to calculate $v_{\text{x}}$.

    $\begin{array}{c}{v}_{\text{x}}=\sqrt{\frac{\left(2\right)5.00\text{MeV}\left(\frac{1\text{eV}}{{10}^{-6}\text{MeV}}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{1.67\times {10}^{-27}\text{kg}}}\\ =\sqrt{\frac{1.6\times {10}^{-24}}{1.67\times {10}^{-27}\text{kg}}}\\ =30.95\times {10}^6\text{m}/\text{s}\end{array}$

Substitute $1.00\text{m}$ for $S$ and $30.95\times {10}^6\text{m}/\text{s}$ for $v_{\text{x}}$ in equation (V) to calculate $t$.

    $\begin{array}{c}t=\frac{1.00\text{m}}{30.95\times {10}^6\text{m}/\text{s}}\\ =3.23\times {10}^{-8}\text{s}\end{array}$

Substitute $0.0500\text{T}$ for $B$, $1.6\times {10}^{-19}\text{C}$ for $q$, $30.95\times {10}^6\text{m}/\text{s}$ for $v_{\text{x}}$ and $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ in equation (VI) to calculate $a$.

    $\begin{array}{c}a=\frac{\left(0.0500\text{T}\right)\left(1.6\times {10}^{-19}\text{C}\right)\left(30.95\times {10}^6\text{m}/\text{s}\right)}{1.67\times {10}^{-27}\text{kg}}\\ =1.14826\times {10}^{14}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $0$ for $u$, $1.14826\times {10}^{14}\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $3.23\times {10}^{-8}\text{s}$ for $t$ in equation (VII) to calculate $v_{\text{y}}$.

Substitute $1.67\times {10}^{27}\text{kg}$ for $m_{\text{p}}$ and $3.7\times {10}^6\text{m}/\text{s}$ for $v_{\text{y}}$ in equation (VIII) to calculate $P_{\text{y}}$.

    $\begin{array}{c}{P}_{\text{y}}=\left(1.67\times {10}^{-27}\text{kg}\right)\left(3.7\times {10}^6\text{m}/\text{s}\right)\\ =6.19\times {10}^{21}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Therefore, the y component of the protons momentum as they leave the magnetic field is $6.19\times {10}^{21}\text{kg}\cdot \text{m}/\text{s}$