Q54P - Fundamentals Of Physics

Question

Using the approximations given in Problem 61, find (a) the altitude above Earth’s surface where the magnitude of its magnetic field is 50.0% of the surface value at the same latitude; (b) the maximum magnitude of the magnetic field at the core–mantle boundary, 2900 km below Earth’s surface; and the (c) magnitude and (d) inclination of Earth’s magnetic field at the north geographic pole. (e) Suggest why the values you calculated for (c) and (d) differ from measured values

Step-by-Step Solution

Answer

1. The altitude above Earth’s surface where the magnitude of its magnetic field is 50.0% of the surface value at the same latitude is $\mathrm{H}=1.66\times {10}^3\mathrm{km}$ .
2. The maximum magnitude of the magnetic field at the core-mantle boundary 2900km below Earth’s surface is $\mathrm{B}= 3.83\times {10}^{-4}\mathrm{T}$.
3. Magnitude of Earth’s magnetic field at the north geographic pole is $\mathrm{B}=6.10\times {10}^{-5}\mathrm{T}$.
4. Inclination of Earth’s magnetic field at the north geographic pole is $\mathrm{\varphi }={84.2}^0$.
5. Earth’s magnetic field varies with latitude. Therefore, the calculated values for c) and d) are different from the measured values.

Step 1: Listing the given quantities

Magnetic dipole momentum is $\mathrm{\mu }=8\times {10}^{22}{\mathrm{Am}}^2$ .

Core-mantle boundary is  $2900\mathrm{km}$  below the earth surface.

Step 2: Understanding the concepts of magnetic field

We use the concept of magnetic field at latitude. Using the equation, we can find the magnetic fields at the given points. We can find the inclination of the magnetic field at the north geographic pole using the value of magnetic latitude in the equation of tan of angle.

Formulae:

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }$

$\tan \mathrm{\varphi } =2\tan \left({\mathrm{\lambda }}_{\mathrm{m}}\right)$

Step 3: (a) Calculations of the altitude above Earth’s surface where the magnitude of its magnetic field is 50.0% of the surface value at the same latitude

The altitude above Earth’s surface where the magnitude of its magnetic field is 50.0% of the surface value at the same latitude.

We have the equation of magnetic field:

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }$

We can write the ratio of magnetic field magnitudes for two different heights at the same latitude:

${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}_1^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }$

${\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}_2^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }$

Taking ratio of   $\frac{{\mathrm{B}}_2}{{\mathrm{B}}_1}$,

$\begin{array}{c}\frac{{\mathrm{B}}_2}{{\mathrm{B}}_1}=\frac{\left[\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}_2^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }\right]}{\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}_1^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }}\\ =\frac{{\mathrm{r}}_1^3}{{\mathrm{r}}_2^3}\end{array}$

Magnitude of magnetic field at an altitude is 50%. We can write

$\frac{{\mathrm{B}}_2}{{\mathrm{B}}_1}=\frac{1}{2}$

We know  ${\mathrm{r}}_1={\mathrm{R}}_{\mathrm{e}}$ and  ${\mathrm{r}}_2={\mathrm{R}}_{\mathrm{e}}+\mathrm{H}$.

Using these values in equation (1), we get

$\frac{1}{2}=\frac{{\mathrm{R}}_{\mathrm{e}}^3}{{({\mathrm{R}}_{\mathrm{e}}+\mathrm{H})}^3}$

Taking the cube root of both sides, we get

$\frac{1}{2^{\frac{1}{3}}}=\frac{{\mathrm{R}}_{\mathrm{e}}}{{\mathrm{R}}_{\mathrm{e}}+\mathrm{H}}$

Rearranging it, we can write

${\mathrm{R}}_{\mathrm{e}}+\mathrm{H}= 2^{\frac{1}{3}} {\mathrm{R}}_{\mathrm{e}}$

Solving for H, we get

$\begin{array}{c}\mathrm{H}= 2^{\frac{1}{3}} {\mathrm{R}}_{\mathrm{e}}-{\mathrm{R}}_{\mathrm{e}}\\ =(2^{\frac{1}{3}}-1){\mathrm{R}}_{\mathrm{e}}\\ =(2^{\frac{1}{3}}-1)\left(6370\mathrm{km}\right)\\ =1.66\times {10}^3\mathrm{km}\end{array}$

The altitude above Earth’s surface where the magnitude of its magnetic field is 50.0% of the surface value at the same latitude is  $\mathrm{H}=1.66\times {10}^3\mathrm{km}$.

Step 4: (b) Calculations of the maximum magnitude of the magnetic field at the core-mantle boundary 2900km below Earth’s surface:

For maximum magnet field, we can set $\mathrm{sinsin} {\mathrm{\lambda }}_{\mathrm{m}} =1$

Distance from the center is 

$\begin{array}{c}\mathrm{r}=6370-2900\\ =3470\mathrm{km}\end{array}$

Plugging the values in the equation of magnetic field, we get

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }\\ =\frac{4\mathrm{\pi }\times {10}^{-7}(8\times {10}^{22})}{4\mathrm{\pi }{(3.47\times {10}^6)}^3}\sqrt{(1+3{\left(1\right)}^2)}\\ =\frac{(8\times {10}^{15}\sqrt{4})}{{(3.47\times {10}^6)}^3}\\ =\frac{16\times {10}^{15}}{4.1781\times {10}^{19}}\\ = 3.83\times {10}^{-4}\mathrm{T}\end{array}$

The maximum magnitude of the magnetic field at the core-mantle boundary 2900km below Earth’s surface is $\mathrm{B}= 3.83\times {10}^{-4}\mathrm{T}$ .

Step 5: (c) Calculations of the magnitude of Earth’s magnetic field at the north geographic pole

Magnitude of Earth’s magnetic field at the north geographic pole:

We know the angle between the magnetic axis and the rotational axis is ${11.5}^0$ . We can write

$\begin{array}{c}{\mathrm{\lambda }}_{\mathrm{m}}={90}^0-{11.5}^0\\ ={78.5}^0 \end{array}$

Plugging these values in the equation of magnetic field, we get

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}_{\mathrm{e}}^3}\sqrt{(1+3{\mathrm{\lambda }}_{\mathrm{m}}) }\\ =\frac{4\mathrm{\pi }\times {10}^{-7}(8\times {10}^{22})}{4\mathrm{\pi }{(6.37\times {10}^6)}^3}\sqrt{(1+{378.5}^0 )}\\ =\frac{{10}^{-7}(8\times {10}^{22})}{{(6.37\times {10}^6)}^3}\sqrt{(1+{378.5}^0 )}\\ =\frac{8\times {10}^{15}\left(1.9699\right)}{2.5847\times {10}^{20}}\\ =\frac{15.7597\times {10}^{15}}{2.5847\times {10}^{20}}\\ =6.10\times {10}^{-5}\mathrm{T}\end{array}$

Magnitude of Earth’s magnetic field at the north geographic pole is $\mathrm{B}=6.10\times {10}^{-5}\mathrm{T}$

Step 6: (d) Calculations of the inclination of Earth’s magnetic field at the north geographic pole

Inclination of Earth’s magnetic field at the north geographic pole:

We use the equation

$\mathrm{tantan} \mathrm{\varphi } =2\mathrm{tantan} \left({\mathrm{\lambda }}_{\mathrm{m}}\right)$

We can write

$\begin{array}{c}\mathrm{\varphi }=(2\mathrm{tantan} \left({\mathrm{\lambda }}_{\mathrm{m}}\right) ) \\ =(2\mathrm{tantan} \left({78.5}^0\right) ) \\ =\left(9.8303\right)\\ ={84.2}^0 \end{array}$

Inclination of Earth’s magnetic field at the north geographic pole is $\varphi ={84.2}^0$  .

Step 7: (e) Explanation

Suggest why values you calculated for c) and d) differ from the measured values:

We know that earth is not a perfect sphere, so the magnetic field is also not uniform at all places.  Earth’s magnetic field varies with latitude. Therefore, the calculated values for c) and d) are different from the measured values.