Electrical energy in a uniform magnetic field with magnitude 0.620 T.

Question

It is proposed to store 1.00 kWh = 3.60 *10^6 J of electrical energy in a uniform magnetic field with magnitude 0.620 T.

a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy?

b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube 41.0 cm on a side, what magnetic field is required?

Answer 1

The energy density of a magnetic field, B, is:

U/V = B^2 / 2 mu,

where mu0 is the permeability of the material.  Use U instead of E for energy in E&M problems to avoid confusion with the electric field.

Solve for the required volume:

V = 2 EU mu / B^2

They give you the energy and magnetic field strength.  Look up mu0, the permeability of the vacuum.  Plugnchug.

b) Just rearrange terms to solve for the magnetic field:

B = sqrt (2 U mu / V)

= sqrt (2 U mu / L^3)

Now they give you L, so plugnchug again.