Chapter 20, Problem 17P

(a)

To determine

The radius of curvature if an atom moves in a plane perpendicular to a uniform field.

Answer to Problem 17P

The radius of circular motion is $2.7\times {10}^{-2} \text{m}$

Explanation of Solution

Given:

A doubly charged helium atom has mass of $6.6\times {10}^{-27}\text{ kg}$

Accelerating voltage is 2100 V.

Uniform magnetic field is $\text{0}\text{.340 T}$

Formula used:

According to the law of energy conservation,

  $E_{initial}=E_{final}$

Calculation:

Using energy conservation law, electrical potential energy of ions converted into kinetic energy.

  $\begin{array}{l}qV=\frac{1}{2}m{v}^2\\ v=\sqrt{\frac{2qV}{m}}\end{array}$

Since atom is moving in a perpendicular direction to the magnetic field. Thus, the magnetic force will be maximum at $\theta ={90}^{\circ }$ , which makes the curvature a circular motion.

  $\begin{array}{l}F=q(v\times B)\\ F_{\max }=qvB\end{array}$

Using velocity relation in the force, we get

  $\begin{array}{l}{F}_{\max }=qvB=m\frac{v^2}{r}\\ r=\frac{mv}{qB}=\frac{m\sqrt{\frac{2qV}{m}}}{qB}=\frac{1}{B}\sqrt{\frac{2mV}{q}}\\ r=\frac{\text{1}}{\text{0}\text{.34 T}}\sqrt{\frac{\text{2(6}{\text{.6×10}}^{\text{-27}}\text{kg)(2100V)}}{\text{2(1}{\text{.6×10}}^{\text{-19}}\text{C)}}}\\ r=2.7\times {10}^{-2}\text{m}\end{array}$

Conclusion:

The radius of circular motion adapted by the doubly charged helium atom due to magnetic field is $2.7\times {10}^{-2} \text{m}$

(b)

To determine

The period of revolution.

Answer to Problem 17P

The period of revolution is $3.8\times {10}^{-7}\sec$ .

Explanation of Solution

Given:

A doubly charged helium atom has mass of $6.6\times {10}^{-27}\text{ kg}$

Accelerating voltage is 2100 V

Formula Used:

Using distance relationship $d=2\pi r=vt$

Calculation:

For circular motion, a period is equal to the circumference of a circle. Thus, the time period in revolving a circle is

  $\begin{array}{l}v=\frac{2\pi r}{T}\\ T=\frac{2\pi r}{v}=\frac{2\pi \frac{1}{B}\sqrt{\frac{2mV}{q}}}{\sqrt{\frac{2qV}{m}}}=\frac{2\pi m}{qB}\\ T=\frac{\text{2π(6}{\text{.6×10}}^{\text{-27}}\text{kg)}}{\text{2(1}{\text{.6×10}}^{\text{-19}}\text{C)(0}\text{.34T)}}=3.8\times {10}^{-7}\sec \end{array}$

Conclusion:

The period of revolving in the circular motion is $3.8\times {10}^{-7}\sec$ .