The Earth's Magnetic Field & Energy

[Bjarne]

At the Earths magnetic North (and South) Pole the Earths magnetic field is about 56 000 nT

I wonder hole much energy (approximately ) - kW or Joule should we use if we should produce a similar magnetic field by electromagnetism.

(How can that be calculated?)

 

[James Leighe]

At the Earths magnetic North (and South) Pole the Earths magnetic field is about 56 000 nT

I wonder hole much energy (approximately ) - kW or Joule should we use if we should produce a similar magnetic field by electromagnetism.

(How can that be calculated?)

You could use Maxwell's equations... I'm sure there is an easier way but anyhow I can mention that we can generate much stronger fields without too much trouble... Hell, MRI machine for animals (for experimental use) can be over 20 Tesla, that's 357142 times stronger than the Earth's field. (regular MRI machines for people are only 53571 times stronger than the Earth's field however)

 

[Andy Resnick]

One way is to simply make a solenoid and pass DC current through it. The magnetic field within the solenoid B = k*I*(N/L), where 'k' is a constant (permeability/2*pi), and N/L the number of turns/length (i.e. turns/cm). You can adjust the magnitude of the B field by adjusting the current, the winding density, or the permeability of the core (e.g. iron vs. air).

 

[elect_eng]

(How can that be calculated?)

You can apply the energy density formula $$U={{B^2}\over{2\mu}}$$

This provides energy per unit volume U (in Joules/m^3) using SI units for magnetic field density B and permeability $$\mu$$.

So energy will depend on the volume over which you establish the field. There will be additional energy required due to losses. If you use the energy to make a permanent magnet then you don't need to worry about power. However, if you make an electromagnet, this energy field must be maintained by using continuous power. This power will depend on the losses, such as wire resistance in a coil, or cooling power if you use superconductors.