Chapter 21, Problem 47P

To determine

To estimate the total energy stored in the field in the first 10 km above the Earth’s surface.

Answer to Problem 47P

Solution:

The total energy stored in the magnetic field is $1.018\times {10}^{20}\text{J}\text{.}$

Explanation of Solution

Given:

Average magnetic field of Earth, $B=0\text{.5}\times {\text{10}}^{-4}T$

Height, $h=10\text{ m}$

Formula used:

Volume over which the field energy needs to be calculated

$V=\frac{4}{3}\pi \left[{\left(R+h\right)}^3-R^3\right]$

Calculation:

We know that the radius of the Earth, $R=6.38\times {\text{10}}^{6}m$

The volume over which the field energy needs to be calculated is the volume enclosed between the two spheres of radius $R$ and $\left(R+h\right)$.

Therefore, $\begin{array}{l}V=\frac{4}{3}\pi \left[{\left(R+h\right)}^3-R^3\right]=\frac{4}{3}\pi \left[R^3+3\times h\times R^2+3R\times h^2+h^3-R^3\right]\\ =\frac{4}{3}\pi \left[3\times h\times R^2+3R\times h^2+h^3\right]\end{array}$

Substituting the value of $R=6.38\times {\text{10}}^{6}m$ and $h=10\text{ m}$

$V=\frac{4}{3}\pi \left[3\times 10\times {10}^3\times {\left(6.38\times {10}^6\right)}^2+3\left(6.38\times {10}^6\right)\times {\left(10\times {10}^3\right)}^2+{\left(10\times {10}^3\right)}^3\right]=5.123\times {10}^{18}{\text{ m}}^3$

Total energy stored in the magnetic field

$U=\frac{B^2}{2{\mu }_0}\times V=\frac{0.5\times {10}^{-4}}{2\times 4\pi \times {10}^{-7}}\times 5.123\times {10}^{18}=1.018\times {10}^{20}\text{ J}$