Question

The magnetic dipole moment of Earth is 8.00 ✕ 1022 J/T. Assume that this is produced by charges flowing in Earth's molten outer core. If the radius of their circular path is 2350 km, calculate the current they produce.

Answer 1

Answer:

Therefore,

Current produce is

$i=4.61\times 10^{9}\ Ampere$

Explanation:

Given:

Magnetic dipole moment of Earth,

$\mu=8\times 10^{22}\ J/T$

Radius = r = 2350 km = 2.35 × 10⁶ m

To Find:

Current, i =?

Solution:

Magnetic Dipole Moment:

A magnetic moment is a quantity that represents the magnetic strength and orientation of a magnet or any other object that produces a magnetic field.

Magnetic dipole moments have dimensions of current times area.

It is given by,

$\mu=i\times Area$

Where,

$\mu$ =  Magnetic dipole moments

i = Current

A = area = $\pr r^{2}$

Substituting the values we get

$i=\dfrac{\mu}{\pi r^{2}}=\dfrac{8\times 10^{22}}{3.14\times (2.35\times 10^{6})^{2}}$

$i=\dfrac{8\times 10^{22}}{17.34\times 10^{12}}=4.61\times 10^{9}\ Ampere$

Therefore,

Current produce is

$i=4.61\times 10^{9}\ Ampere$