Question

The magnetic dipole moment of Earth has magnitude$\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}\mathbf{}}\mathit{J}\mathbf{/}\mathit{T}$ . Assume that this is produced by charges flowing in Earth’s molten outer core. If the radius of their circular path is 3500 km, calculate the current they produce.

Short answer

$i=2.08\times {10}^{9}A$

Step-by-step solution

Given

The radius of the earth=$r=3500km=3500\times {10}^{3}m$

The magnetic dipole moment of the earth = $\mu =8.00\times {10}^{22}J/T$

Understanding the concept

The charges flowing in the molten outer core of the earth produce the magnetic dipole moment. The current produced by the charges can be calculated by substituting the given quantities in the formula for magnetic dipole moment.

$\mu =NiA$

Calculate the current produced

The magnetic dipole moment is calculated as

$\mu =NiA$

Here,$N=1andA=\pi r^2$

$\mu =i{\pi }^2$

Rearranging the equation to find the value of the current.

$$\begin{gathered} i=\frac{\mu }{\pi r^2} \\ i=\frac{8.00\times {10}^{22}}{3.14\times {\left(3500\times {10}^3\right)}^2} \\ i=2.08\times {10}^{9}A \end{gathered}$$

Hence, $i=2.08\times {10}^{9}A$