Amperes Circuital Law

Amperes circuital law states that the algebraic sum of the rises and drops of the mmf around a closed loop of a magnetic circuit is equal to zero; that is, the sum of the rises in mmf equals the sum of the drops in mmf around a closed loop.

What Is Ampere’s Law?

According to Ampere’s law, magnetic fields are related to the electric current produced in them. The law specifies the magnetic field that is associated with a given current or vice-versa, provided that the electric field doesn’t change with time.
Ampere’s Law can be stated as:
“The magnetic field created by an electric current is proportional to the size of that electric current with a constant of proportionality equal to the permeability of free space.”
Table. 1
Electric CircuitsMagnetic Circuits
As mentioned in the introduction to this chapter, there is a broad similarity between the analyses of electric and magnetic circuits. This has already been demonstrated to some extent for the quantities in [Table 1]. If we apply the "cause" analogy to Kirchhoff's voltage law ( $ \sum V=0$), we obtain the following:
$$ \sum mmf = 0 \tag{1}$$
Equation (1) is referred to as Ampere's circuital law. When it is applied to magnetic circuits, sources of mmf are expressed by the equation
$$ mmf = NI \tag{2}$$
The equation for the mmf drop across a portion of a magnetic circuit can be found by applying the relationships
$$ mmf = \Phi R \tag{3} $$
where $\Phi$ is the flux passing through a section of the magnetic circuit and $R$ is the reluctance of that section. The reluctance, however, is seldom calculated in the analysis of magnetic circuits. A more practical equation for the mmf drop is
$$ mmf = Hl \tag{4}$$
where $H$ is the magnetizing force on a section of a magnetic circuit and $l$ is the length of the section. As an example of Eq. (1), consider the magnetic circuit appearing in [Fig. 1] constructed of three different ferromagnetic materials.
Applying Ampere's circuital law, we have
$$ \sum mmf = 0$$
$$+NI - H_{ab} l_{ab} - H_{bc} l_{bc} - H_{ca} l_{ca} = 0$$
$$ NI = H_{ab} l_{ab} - H_{bc} l_{bc} - H_{ca} l_{ca}$$
Fig. 1: Series magnetic circuit of three different materials.

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