Problem with solution: An Ion Orbiting Around The Equator in The Magnetic Field

The earth has a radius of 6370 kilometers. The magnetic field at the equator has a value of $ 30 \, \mu\text{T} $ and is directed to the north. You want to get an ion with mass $ 352 \, u $ and charge $ e $ to orbit the Earth at the equator at a height of 30 kilometers.

An ion orbits along the equator in the Earth's magnetic field.

In what direction along the equator must the ion fly to orbit the Earth?
At what velocity must the ion move so that it remains on the circular path? Consider also the gravitational force. You can calculate classically and assume a homogeneous gravitational field.
Reason why the gravitational force on the ion plays only a minor role.
The earth is also negatively charged electrically and has a average homogeneous magnetic field of $ 130 \, \frac{ \mathrm V }{ \mathrm m } $. Does this E-field have to be taken into account?

Negatively charged earth with a homogeneous E-field.

Solution tips

Consider the forces acting on the ion and write an equation for the equilibrium of forces. To determine the velocity, you have to solve a quadratic equation in any case.

Solution for (a)

To find out the direction of motion of the ion, use the right-hand rule. We are dealing with a positive charge here. Therefore you use your right hand for this.

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Right-hand rule for a positively charged particle.

The earth's magnetic field $ B $ points north at the equator. Point your index finger to the north.
The Lorentz force $ F_{\text m} $ (magnetic force) on the ion must point towards the earth so that it orbits around the earth. Point your middle finger towards the center of the earth.
The extended thumb then shows you the direction in which the ion must move.

Magnetic force forces the ion into a circular path along the equator.

Thus the ion must fly to the west, so that it is deflected around the equator. If the ion would fly to the east, it would be deflected away from the earth.

Solution for (b)

In order for the ion to remain on a stable circular orbit with the radius (6370 kilometers Earth radius + 30 kilometers height), the Lorentz force...

$$ \begin{align} \class{green}{F} ~=~ q \, \class{blue}{v} \, \class{violet}{B} \end{align} $$

and the gravitational force...

$$ \begin{align} F_{\text g} ~=~ m \, g \end{align} $$

must correspond exactly to the centripetal force

$$ \begin{align} F_{ \text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{ r } \end{align} $$

Centripetal force is a force acting to the center of the earth and tells us how to get a stable circular orbit with radius $r$ and velocity $ v $ of a particle of mass $ m $. The role of the centripetal force is taken over here by the Lorentz force and the gravitational force, because they are also radially directed.

So we have the following equation of forces:

$$ \begin{align} F_{\text z} &~=~ F_{\text m} ~+~ F_{\text g} \\\\
\frac{m}{r} \, v^2 &~=~ q \, v \, B ~+~ m \, g \end{align} $$

We have to rearrange this force equation with respect to the velocity $ v $. Here you have a quadratic equation that you cannot rearrange uniquely with respect to $v$. This equation has two solutions as possible velocities. We determine the two solutions $ v_1 $ and $ v_2 $ with the pq formula.

To do this, we convert Eq. 4 into a suitable form:

$$ \begin{align} \frac{m}{r} \, v^2 ~-~ q \, v \, B ~-~ m \, g & ~=~ 0 \\\\
v^2 ~-~ \frac{r}{m} \, q \, v \, B ~-~ \frac{r}{\cancel m} \, \cancel{m} \, g & ~=~ 0 \\\\
v^2 ~-~ \frac{r \,q\, B}{m} \, v ~-~ r \, g & ~=~ 0 \end{align} $$

We put everything on the left side in the second step and in the third step we multiplied the whole equation by $ \frac{r}{m} $ to eliminate the factor before the $v^2$. Now we have the p-value $ - \frac{r \,q\, B}{m} $ and q-value $ - r \, g $ for the pq-formula:

$$ \begin{align} v_{1,2} & ~=~ -\frac{p}{2} ~\pm~ \sqrt{ \left(\frac{p}{2}\right)^2 ~-~ q } \\\\
v_{1,2} & ~=~ \frac{r \,q\, B}{2m} ~\pm~ \sqrt{ \left(\frac{r \,q\, B}{ 2m}\right)^2 ~+~ r \, g } \end{align} $$

Now we just have to insert the concrete values to determine $v_1$ and $v_2$.

The radius $r $ is equal to the earth radius plus the height of the ion above the earth:

Earth radius and height
Formula anchor $$ \begin{align} r ~=~ 6370 \, \text{km} ~+~ 30 \, \text{km} ~=~ 6400 \cdot 10^3 \, \text{m} \end{align} $$
The charge $q$ corresponds to the charge of the ion:

Charge of an ion
Formula anchor $$ \begin{align} q ~=~ e ~=~ 1.60 \cdot 10^{-19} \, \text{C} \end{align} $$
The mass $m$ corresponds to the mass of the ion:

Ion mass
Formula anchor $$ \begin{align} m ~=~ 352 \, u ~=~ 352 ~\cdot~ 1.66 \cdot 10^{-27} \, \text{kg} \end{align} $$
The magnetic field $ B $ corresponds to the constant earth magnetic field at the equator:

Earth magnetic field
Formula anchor $$ \begin{align} B ~=~ 30 \, \mu\text{T} ~=~ 30 \cdot 10^{-6} \, \text{T} \end{align} $$
The gravitational acceleration $ g $ on earth is:

Acceleration due to gravity on earth
Formula anchor $$ \begin{align} g ~=~ 9.8 \, \frac{\text m}{ \text{s}^2 } \end{align} $$

Substituting all these values into Eq. 6 yields:

$$ \begin{align} v_{1,2} ~=~ 26 \,286\,966.05 \, \frac{\text m}{ \text s } ~\pm~ 26\,286 \,967 \frac{\text m}{ \text s } \end{align} $$

We get the first solution of the quadratic equation 5 if we add the two values in 12:

$$ \begin{align} v_1 &~=~ 52\,573\,933 \, \frac{\text m}{ \text s } \\\\
&~\approx~ 5.2 \cdot 10^7 \, \frac{\text m}{ \text s } \end{align} $$

We get the second solution if we subtract the two values in Eq. 12:

$$ \begin{align} v_2 ~=~ -1.19 \, \frac{\text m}{ \text s } \end{align} $$

Eq. 14 is not a valid solution. You can check this by substituting 14 into 4. Eq. 13, on the other hand, is a valid solution. Therefore the required velocity is:

$$ \begin{align} v ~=~ 5.2 \cdot 10^7 \, \frac{\text m}{ \text s } \end{align} $$

Solution for (c)

To justify why the gravitational force hardly plays a role, you can look at the ratio of the gravitational force $ F_{\text g} $ to the Lorentz force $ F_{\text m} $:

$$ \begin{align} \frac{ F_{\text g} }{ F_{\text m} } ~=~ \frac{ m \, g }{ q \, v \, B } \end{align} $$

Inserting the values (including the calculated velocity in exercise b), results in:

$$ \begin{align} \frac{ F_{\text g} }{ F_{\text m} } &~=~ \frac{ 5.73\cdot 10^{-24}\, \text{N} }{ 2.5\cdot 10^{-16} \, \text{N} } \\\\
&~\approx~ 2 \cdot 10^{-6} \, \% \end{align} $$

So the gravitational force is just 0.000002 % of the Lorentz force and can be neglected.

Solution for (d)

To check whether the electric field of the earth has to be taken into account, we can calculate the ratio of the electric force $ F_{\text e} = e \, E $ on the ion to the Lorentz force in analogy to the previous exercise:

$$ \begin{align} \frac{ F_{\text e} }{ F_{\text m} } ~=~ \frac{ e \, E }{ q \, v \, B } \end{align} $$

The electric field $ E ~=~ 130 \, \frac{ \text V }{ \text m } $ is given in the exercise. Inserting the E-field results in:

$$ \begin{align} \frac{ F_{\text e} }{ F_{\text m} } &~=~ \frac{ 2.08 \cdot 10^{-17} \, \text{N} }{ 2.5\cdot 10^{-16} \, \text{N} } \\\\
&~\approx~ 8.3 \, \% \end{align} $$

The electric force accounts for 8.3% of the Lorentz force and can be taken into account for a more accurate result.