Magnetic Reluctance

Key facts

The magnetic reluctance is defined as:

$$\mathcal{R} = \frac{\mathcal{F}}{\Phi}$$

(1)

where $\mathcal{F}$ is the magnetomotive force, and $\Phi$ the magnetic flux.

For a magnetic circuit of length $l$, cross-sectional area $A$, and relative magnetic permeability $\mu_r$, the magnetic reluctance can be calculated with:

$$\mathcal{R} = \frac{l}{\mu_0 \mu_r A}$$

(2)

where $\mu_0$ is the magnetic permeability of free space.

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Constants

$$\mu_0 = 4 \pi \cdot 10^{-7} \; \frac{N}{A^2}$$

(3)

A description of the magnetic reluctance, also discussing a way to calculate it

View versions (1)

The magnetic reluctance $\mathcal{R}$ of a magnetic circuit can be regarded as the formal analog of the resistance in an electrical circuit. The magnetic reluctance can be expressed as:

$$\mathcal{R} = \frac{\mathcal{F}}{\Phi}$$

(4)

where $\mathcal{F}$ is the magnetomotive force (mmf), and $\Phi$ is the magnetic flux.

In order to calculate the magnetic reluctance, consider a magnetic circuit of length $l$ and cross-sectional area $A$, as diagramed in Figure 1.

We know that the magnetic field strength $H$ can be written as:

$$\int{H dl} = NI$$

(5)

where $I$ is the current in the coil, and $N$ is the number of turns (for a more detailed discussion on the magnetic field strength see Field Strength ). Furthermore, $H$ can be related to the magnetic flux density $B$ with the equation:

$$B = \mu_0 \mu_r H$$

(6)

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

As the magnetic flux $\Phi$ is defined as:

$$\Phi = B \cdot A$$

(7)

equation (6) can also be written as:

$$\frac{\Phi}{A} = \mu_0 \mu_r H$$

(8)

from which the magnetic field strength becomes:

$$H = \frac{\Phi}{\mu_0 \mu_r A}$$

(9)

Considering that $H$ is uniform, equation (5) becomes:

$$H l = NI$$

(10)

Using the expression form of $H$ from (10) in (9), we get that:

$$NI = \frac{\Phi l}{\mu_0 \mu_r A}$$

(11)

which leads to:

$$\Phi = NI \div \frac{l}{\mu_0 \mu_r A}$$

(12)

As the magnetomotive force $\mathcal{F}$ of a coil is given by:

$$\mathcal{F} = NI$$

(13)

equation (12) becomes:

$$\Phi = \mathcal{F} \div \frac{l}{\mu_0 \mu_r A}$$

(14)

or:

\calc{l/(4*π*10^(-7)*mu_r*A)} "Instant calculator eq(13)"

$$\frac{\mathcal{F}}{\Phi} = \frac{l}{\mu_0 \mu_r A}$$

(15)

Taking into account the definition of the magnetic reluctance from (4), we get that $\mathcal{R}$ can be calculated as:

$$\mathcal{R} = \frac{l}{\mu_0 \mu_r A }$$

(16)

Example 1 [metric]

Problem

Consider a toroid with the mean length of $20 \; cm$, the cross section of $2 \; cm^2$, and the relative magnetic permeability of $6700$. What is the magnetic flux and the magnetic flux density if the coil has 10 turns and the current is 2 amperes ?

Workings

As the magnetic reluctance $\mathcal{R}$ is given by:

$$\mathcal{R} = \frac{l}{\mu_0 \mu_r A}$$

(17)

and, in our case, $l = 20 \; cm$ ($= 0.2 \; m$), and $A = 2 \; cm^2$ ($=2 \cdot 10^{-4} \; m^2$), we get that:

$$\mathcal{R} = \frac{0.2}{6700 \cdot 4 \pi \cdot 10^{-7} \cdot 2 \cdot 10^{-4}}$$

(18)

from which we obtain:

$$\mathcal{R} = 1.19 \cdot 10^5 \; At/Wb$$

(19)

The magnetic flux $\Phi$ can be written as:

$$\Phi = \frac{\mathcal{F}}{\mathcal{R}}$$

(20)

where $\mathcal{F}$, the magnetomotive force, is given by:

$$\mathcal{F} = NI$$

(21)

As, in our case, $N=10$, $I=2$, and also considering (19), we obtain the magnetic flux:

$$\Phi = \frac{20}{1.19 \cdot 10^5} = 1.68 \cdot 10^{-4} \; Wb$$

(22)

Taking into account that the cross-sectional area is $A = 2 \; cm^2$ ($=2\cdot 10^{-4} \; m^2$), the magnetic flux density becomes:

$$B = \frac{\Phi}{A} = \frac{1.68 \cdot 10^{-4}}{2 \cdot 10^{-4}} = 0.84 \; Wb/m^2$$

(23)

As a side note, if the toroid has an air gap of length $l_g$, then its total magnetic reluctance, $\mathcal{R}_t$, would be the magnetic reluctance of the toroid plus the magnetic reluctance of the air gap:

$$\mathcal{R}_t = \frac{l}{\mu_0 \mu_r A} + \frac{l_g}{\mu_0 A}$$

(24)

Thus, in this case, the total magnetic flux, $\Phi_t$, would be given by:

$$\Phi_t = NI \div \frac{l}{\mu_0 \mu_r A} + \frac{l_g}{\mu_0 A}$$

(25)

Solution

$\Phi = 1.68 \cdot 10^{-4} \; Wb$

$B = 0.84 \; Wb/m^2$