Chapter 20, Problem 84GP

To determine

The change in the kinetic energy of the proton.

Answer to Problem 84GP

The change in the kinetic energy of the proton is $\Delta KE=-2.12\times {10}^{-20}\text{J}$ .

Explanation of Solution

Given:

The magnetic field is $B=0.01T$

The charge on the proton is $e=1.6\times {10}^{-19}\text{c}$

The mass of a proton is $m_p=1.67\times {10}^{-27}\text{kg}$ .

The radius of the point P, $r_1=10\text{mm}$ .

The radius of the point Q, $r_2=8.5\text{mm}$ .

Formula used:

The centripetal force on the proton is equal to the magnetic force on it is,

  $\begin{array}{l}\frac{m_p{v}^2}{r}=Bev\\ v=\frac{Ber}{m_p}\end{array}$

Here, $B$ is the magnetic field, e is the charge on the proton, v is the speed of the proton, and $m_p$ is the mass of proton.

The expression of kinetic energy is,

  $K{E}_p=\frac{1}{2}{m}_p{v}^2$

Calculation:

The speed $\left(v\right)$ of the proton is,

  $\begin{array}{c}v=\frac{Ber}{m_p}\\ =\frac{0.01\times \left(1.6\times {10}^{-19}\right)r}{\left(1.67\times {10}^{-27}\right)}\\ =9.58\times {10}^{5}r\end{array}$

The kinetic energy at P is,

  $\begin{array}{c}K{E}_p=\frac{1}{2}{m}_p{v}_1^2\\ =\frac{1}{2}{m}_p{\left(9.58\times {10}^5{r}_1\right)}^2\text{J}\\ =\frac{1}{2}{m}_p{\left(9.58\times {10}^5\times 10\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\text{J}\\ =\frac{1}{2}\left(1.67\times {10}^{-27}\right){\left(9.58\times {10}^5\times 10\times {10}^{-3}\right)}^2\text{J}\\ =7.66\times {10}^{-20}\text{J}\end{array}$

The kinetic energy at Q is,

  $\begin{array}{c}K{E}_Q=\frac{1}{2}{m}_p{v}_2^2\\ =\frac{1}{2}{m}_p{\left(9.58\times {10}^5{r}_2\right)}^2\text{J}\\ =\frac{1}{2}{m}_p{\left(9.58\times {10}^5\times 8.5\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\text{J}\\ =\frac{1}{2}\left(1.67\times {10}^{-27}\right){\left(9.58\times {10}^5\times 8.5\times {10}^{-3}\right)}^2\text{J}\\ =5.54\times {10}^{-20}\text{J}\end{array}$

The change in kinetic energy is,

  $\begin{array}{c}\Delta KE=\left(K{E}_Q-K{E}_p\right)\\ =-\left(7.66-5.54\right)\times {10}^{-20}\text{J}\\ \text{=}-\text{2}\text{.12}\times {\text{10}}^{-20}\text{J}\end{array}$

Conclusion:

Thus, the change in the kinetic energy of the proton is $\Delta KE=-2.12\times {10}^{-20}\text{J}$ .