Question

The earth’s magnetic field, with a magnetic dipole moment of $8.0\times {10}^{22}A{m}^2$, is generated by currents within the molten iron of the earth’s outer core. Suppose we model the core current as a $3000km$-diameter current loop made from a 1000-km-diameter “wire.” The loop diameter is measured from the centers of this very fat wire.

a. What is the current in the current loop?

b. What is the current density $J$ in the current loop?

c. To decide whether this is a large or a small current density, compare it to the current density of a $1.0A$ current in a $1.0mm$-diameter wire.

Short answer

a. The current in the current loop is $1.13\times {10}^{10}A$.

b. The current density $J$in the current loop is $0.014A/m^2$.

c.$1.3\times {10}^{6}A/m^2$The small wire has a very large current density compared to the large one.

Step-by-step solution

Part (a) Step 1 : Given Information

We need to find the current in the current loop.

Part (a) Step 2 : Simplify

The magnetic field dipole moment $M=8\times {10}^{22}A{m}^2$current-carrying loop is just like a magnetic dipole whose magnetic dipole :

$$\begin{gathered} M=i\times A \\ i=\frac{M}{A} \\ i=\frac{8\times {10}^{22}}{\pi {\left(1.5\times {10}^6\right)}^2} \\ i=1.13\times {10}^{10}A \end{gathered}$$

Part (b) Step 1 : Given Information.

We need to find density $J$ in the current loop.

Part (b) Step 2: Simplify

The current density $J=?$

So,

role="math" localid="1650965922291" $$\begin{gathered} i=J\times s \\ J=\frac{i}{s} \\ J=\frac{1.13\times {10}^{10}}{\pi {\left(0.5\times {10}^6\right)}^2} \\ J=0.014A/m^2 \end{gathered}$$

Part (c) Step 1 : Given Information : 

We need to find the current density of a $1.0A$current in a $1.0mm$-diameter wire.

Part (c) Step 2: Simplify

The wire with a diameter $1mm$, the radius is $r_2=0.5mm$and the current flows is $I_2=1A$.

We finding the $J_2$as:

$$\begin{gathered} J_2=\frac{I_2}{\pi r^2} \\ =\frac{\left(1A\right)}{\pi {\left(0.5\times {10}^{-3}m\right)}^2} \\ =1.3\times {10}^{6}A/m^2 \end{gathered}$$

The small wire has a very large current density comparing to the large one.