Chapter 20, Problem 24P

To determine

The speed of a singly ionized uranium ion.

To explain: Whether gravity can be ignored in this case.

Answer to Problem 24P

The speed of a singly ionized uranium ion is $1\times {10}^8\text{m/s}$ . Yes, gravity can be ignored in this case. Because the magnetic force of the singly ionized uranium ion is much higher as compared to the gravitational force.

Explanation of Solution

Given:

The Earth’s magnetic field is $0.4\times {10}^{-4} \text{T}$ .

The mass of singly ionized helium ion is $238 \text{u}$ and the charge is $e$ .

The height above the equator to circle around the Earth is $5 \text{km}$ .

Formula Used:

According to the energy conservation principle,

  $\begin{array}{l}{F}_{\max }=qvB=\frac{m{v}^2}{r}\\ v=\frac{qrB}{m}\end{array}$

Calculation:

Speed of the singly ionized helium ion is

  $\begin{array}{l}v=\frac{qrB}{m}\\ v=\frac{\text{(1}{\text{.6×10}}^{\text{-19}}\text{ C)(6}{\text{.385×10}}^{\text{6}}\text{ m)(0}{\text{.40×10}}^{\text{-4}}\text{ T)}}{\text{238(1}{\text{.66×10}}^{\text{-27}}\text{ kg)}}\\ v=1\times {10}^8 \text{m/s}\end{array}$

Now, comparing the magnetic force to the gravitational force on the singly ionized helium ion is

  $\begin{array}{l}\frac{F_B}{F_g}=\frac{qvB}{mg}\\ \frac{F_B}{F_g}=\frac{\text{(1}{\text{.6×10}}^{\text{-19}}{\text{ C)(1×10}}^{\text{8}}\text{ m/s)(0}{\text{.40×10}}^{\text{-4}}\text{ T)}}{\text{238(1}{\text{.66×10}}^{\text{-27}}\text{ kg)(9}{\text{.8 m/s}}^{\text{2}}\text{)}}\\ \frac{F_B}{F_g}=1.7\times {10}^8\end{array}$

Hence, the magnetic force is $100$ million times greater than the gravitational force. The gravity can be ignored in this case. Because the magnetic force of the singly ionized uranium ion is much higher as compared to the gravitational force.

Conclusion:

The speed of a singly ionized uranium ion is $1\times {10}^8\text{m/s}$ . Yes, gravity can be ignored.