Amperes Circuital Law

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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 Ampereâ€™s Law?

According to Ampereâ€™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 doesnâ€™t change with time.
Ampereâ€™s Law can be stated as:
â€œThe 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.â€
 Electric Circuits Magnetic Circuits Cause E $\mathcal{F}$ Effect I $\Phi$ Opposition R $\mathcal{R}$
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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