CodeCogs

Energy stored in a magnetic field, also considering the case of no magnetic saturation

You're viewing an older version of this page (#3617). View the current version.

View versions (1)

If we are to neglect the resistance of the circuit wire, then there would be no energy loss in maintaining a magnetic field. However, energy is required to establish the field, and it can then be recovered when the field is destroyed.

For a toroid, the induced voltage $e$ at any instant is:

$e = N \frac{d\Phi}{dt}$

(4)

where $N$ is the number of turns, and $\Phi$ the magnetic flux.

If the current at any instant is $i$ , then the instantaneous power ( $Watts$ ) is:

$P = e \cdot i$

(5)

The energy ( $Joules$ ) released from the coil in a time $dt$ is:

$E = e \cdot i \cdot dt$

(6)

or, by considering (#1):

$E = N \cdot i \cdot d\Phi$

(7)

The total energy stored in the coil then becomes:

$E_{stored} = \int_0^{\Phi_{Max}} N \cdot i \cdot d\Phi$

(8)

In order to further define the energy stored in a magnetic field, consider a magnetic circuit of length $l$ and cross-sectional area $A$ , as diagramed in Figure 1.

Figure 1

We know that the magnetic flux density $B$ can be defined as:

$B = \frac{\Phi}{A}$

(9)

which leads to:

$\Phi = B \cdot A$

(10)

from which:

$d\Phi = A \cdot dB$

(11)

Taking into account equations (#8) and (#5), we obtain the energy stored in the magnetic circuit:

$E_{stored} = \int N \cdot i \cdot A \cdot dB$

(12)

which can also be written as:

$E_{stored} = \int \frac{N \cdot i}{l} \cdot A \cdot l \cdot dB$

(13)

We know that if the magnetic field strength $H$ is uniform, then:

$H = \frac{N i}{l}$

(14)

Taking into account (#11), equation (#10) becomes:

$E_{stored} = V \int H dB$

(15)

where $V$ ( $=A l$ ) is the volume. Although this equation was proved for a toroid, it can in fact be demonstrated for all magnetic circuits.

For a $BH$ curve as the one diagramed in Figure 2, $\int H dB$ is the blue shaded area:

Figure 2

It can be noted that, if there is no magnetic saturation (i.e. the $BH$ curve is straight), then:

$\int H dB = \frac{1}{2} H B$

(16)

We also know that the magnetic field strength $H$ is related to the magnetic flux density $B$ with the equation:

$B = \mu_0 \mu_r H$

(17)

or:

$H = \frac{B}{\mu_0 \mu_r}$

(18)

\calc{V*(B^2/(8*π*10^(-7)*mu_r))} "Instant calculator eq(16)"

where $\mu_0$ is the magnetic permeability of free space, and $\mu_r$ the relative magnetic permeability.

Taking into account equations (#15), (#13), and (#12), the energy stored in this particular case becomes:

$E_{stored} = V \frac{B^2}{2 \mu_0 \mu_r}$

(19)