Field Strength

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Contents
  1. Magnetic Field Strength and Magnetic Flux Density
  2. Reference

Key facts

The magnetic field strength is defined as:

\int{H dl} = i
(1)

For an isolated conductor of radius r, the magnetic field strength is:

H = \frac{i}{2 \pi r}
(2)

The magnetic field strength is related to the magnetic flux density B by the equation:

B = \mu_r \mu_0 H
(3)

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

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Constants

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

A description of the magnetic field strength, also considering its relation with the magnetic flux density

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Physics Magnetism

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The magnetic fields generated by electric currents are characterized by the magnetic flux density B, measured in Weber (Wb) / m^2. But when the generated fields pass through magnetic materials which themselves contribute internal magnetic fields, ambiguities can arise concerning what part of the field comes from the external currents, and what part comes from the material itself. In order to clarify this issue, we introduce another magnetic field quantity, the magnetic fields strength H, which unambiguously designates the magnetic influence from external currents, independent of the material's magnetic contribution.

The magnetic field strength H can be broadly defined by:

\int{H dl} = i
(5)

where H is expressed in amperes/meter, and the current i in amperes.

For the particular case of an isolated conductor of radius r, as the one diagramed in Figure 1:

Figure 1
Figure 1

the magnetic field strength can be defined by writing that:

\int{H dl} = H \cdot 2 \pi r = i
(6)

which leads to:

H = \frac{i}{2 \pi r}
(7)

As was previously noted, magnetic materials, such as iron, can contribute internal magnetic fields. Indeed, iron contains within its atoms a number of tiny current loops. Normally, their random distribution produces no effect. However, when a current flows, these current loops become oriented, the magnetic effect due to the iron becomes more important, and the resulting current is much higher than it would have been without the iron. We can thus write that:

\int{H^i dl} = NI + ni^i
(8)

where I is the current in the coil, and N is the number of turns, while ni represents the equivalent effect of the atomic current loops (see Figure 2).

Figure 2
Figure 2

We redefine H in iron to retain the magnetic circuit law in terms of the known current NI:

\int{H dl} = NI
(9)

Thus, we can write that:

\int{H^i dl} - ni^i = \int{H dl}
(10)

Magnetic Field Strength and Magnetic Flux Density

A commonly used relation between the magnetic flux density B and the magnetic field strength H is:

B = \mu H
(11)

where \mu, the magnetic permeability, is:

\mu = \mu_r \mu_0
(12)

In (#8) \mu_0 is the magnetic permeability of free space, also known as the magnetic constant:

\mu_0 = 4 \pi \cdot 10^{-7} \; \frac{N}{A^2}
(13)

and \mu_r the relative magnetic permeability of the material.

Thus, the magnetic flux density in air can be written as:

B = \mu_0 H
(14)

whereas the magnetic flux density in iron as:

B = \mu_r^i \mu_0 H
(15)

where \mu_r^i is the relative magnetic permeability of iron (for example graphs see Figures 3 and 4).

Figure 3
Figure 3
Figure 4
Figure 4

Reference

Example 1 [metric]
Problem

Given a toroid with the mean diameter of 20 \;cm and the cross section of 2 \; cm^2 (see Figure E1), calculate the number of ampere-turns (At) which would give a magnetic flux of 2.3 \cdot 10^{-4} \; Wb.

Figure E1
Figure E1

The BH curve for iron is given in Figure E2:

Figure E2
Figure E2

Also, calculate the number of ampere-turns required to give the same magnetic flux in the hypothetical case that the toroid has an air gap of length l_g = 0.1 \; cm (see Figure E3).

Figure E3
Figure E3
Workings

Given that the cross section of the toroid is 2 \;cm^2 (= 2 \cdot 10^{-4} \; m^2), the magnetic flux density in the iron should be:

B = \frac{2.3 \cdot 10^{-4}}{2 \cdot 10^{-4}} = 1.15 \; \frac{Wb}{m^2}
(16)

Therefore, from Figure E2, we obtain:

H = 335 \; \frac{At}{m}
(17)

As the mean diameter of the toroid is 20 \; cm (=0.2 \; m), we thus obtain the required ampere-turns:

335 \cdot 0.2 = 67 \; At
(18)

In the case that the toroid has an air gap, we can write the number of ampere-turns for the air gap as H_g l_g, or, furthermore, as \frac{B_g}{\mu_0} l_g. Taking into account that \mu_0 = 4 \pi \cdot 10^{-7} \; \frac{N}{A^2}, B_g = 1.15 \; \frac{Wb}{m^2} (from equation #1), and l_g = 0.1 \cdot 10^{-2} \; m, we get the required ampere-turns for the air gap:

0.8 \cdot 1.15 \cdot 0.1 \cdot 10^{-2} \cdot 10^6 = 920 \; At
(19)

Taking into account (#3) and (#4), we obtain the total required ampere-turns for the toroid with an air gap:

67 + 920 = 987 \; At
(20)

As a side point, it can be noted that the effect of putting an air gap into a toroid is to alter the BI curve to an almost linear relationship, as diagramed in Figure E4.

Figure E4
Figure E4
Solution

For the toroid without an air gap, the required ampere-turns are:

67 \; At

For the toroid with an air gap, the total required ampere-turns are:

987 \; At