Question

The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the earth's outer core. Suppose we model the core current as a 3000-km-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 3.0 A current in a 2.0-mm-diameter wire.

Answer 1

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

$m = IS$

Where,

I = Current

S = Area

$m = IA \rightarrow m= I( \pi r^2)$

Replacing with our values we have that,

$8*10^{22} = I \pi(\frac{3000*10^3}{2})^2$

Re-arrange to find I,

$I = 1.1317*10^{10} A$

Part B) To find the Current density we need to find the cross sectional area of the Wire:

$A = \pi r^2 \\A = \pi (\frac{1000*10^3}{2})^2\\A = 7.854*10^{11}m^2$

Finally the current density is simply J

$J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2$

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

$A = \pi (10-3)^2 \\A = 49\pi$

Therefore the current density would be

$J = \frac{I}{A}\\J = \frac{30A}{49\pi}\\J = 9.549*10^5 A/m^2$

Comparing the two values we can see that the 2mm wire has a higher current density.

Answer 2

(A) The current will be "1.1317 × 10¹⁰ A".

(B) The current density will be "0.0144 A/m²".

(C) By comparing, we can see that the wire has a higher current density.

Magnetic field

According to the question,

Magnetic dipole moment, m = 8.0 × 10²² A-m²

(A) As we know,

Magnetic dipole moment, m = Current (I) × Area (S)

or,

m = IA

   = I (πr²)

By substituting the values,

8 × 10²² = Iπ ($\frac{3000\times 10^3}{2}$)²

           I = 1.1317 × 10¹⁰ A

(B) We know the cross-sectional area,

→ A = πr²

      = π($\frac{1000\times 10^3}{2}$)²

      = 7.854 × 10¹¹ m²

So,

The current density be:

→ J = $\frac{I}{A}$

     = $\frac{1.1317\times 10^{10}}{7.854\times 10^{11}}$

     = 0.0144 A/m²

(C) Area will be:

→ A = π(10 - 3)²

      = 49π

and,

The current density be:

J = $\frac{I}{A}$

  = $\frac{30}{49 \pi}$

  = 9.549 × 10⁵ A/m²

We can say that a wire has maximum current density,

Thus the responses above are correct.

       

Find out more information about magnetic field here:

https://brainly.com/question/1594186