Chapter 20, Problem 79GP

To determine

To check: Variation of the beam of protons with the varying speed.

The angle at which beam of protons bent.

Answer to Problem 79GP

If the speed of the beam of protons is greater, then it will exit above the second tube. If it is lesser, then it will move below.

The angle at which the beam of protons bent is $9.1°$

Explanation of Solution

Given:

Magnetic field is extended over a length of $5\text{ cm}$ is $0.33\text{ T}$ .

Speed of the beam of protons is $1\times {10}^7\text{ m/s}$ .

Formula used:

Angle is calculated as

  $\theta ={\sin }^{-1}\left(\frac{\text{Perpendicular}}{\text{Hypotenuse}}\right)$

Calculation:

Beam of protons follows a circular path in the presence of a magnetic field with a radius of curvature $r=\frac{mv}{qB}$ . If the speed of the beam of protons is greater, then the radius of curvature will be large enough to make its path above the second tube. If speed is lesser, then the radius of curvature is so less that its path is bent below the second tube.

Exit velocity as shown in the figure below is perpendicular to the line of radius from the center of curvature, the bending angle will be

  $\begin{array}{c}\sin \theta =\frac{L}{r}\\ \theta ={\sin }^{-1}\frac{LqB}{mv}\\ \theta ={\sin }^{-1}\left[\frac{(5\times {10}^{-2}\text{ m})(1.6\times {10}^{-19}\text{ C})(0.33\text{ T})}{(1.67\times {10}^{-27}\text{ kg})(1\times {10}^7\text{ m/s})}\right]\\ \theta ={\sin }^{-1}0.158\\ =9.1°\end{array}$

Conclusion:

Thus, if the speed of the beam of protons is greater, then it will exit above the second tube. If it is lesser, then it will move below.

The angle at which the beam of protons bent is $9.1°$