Magnetic Circuit Air Gaps

This section of course is related to the effects that an air gap has on a magnetic circuit. Note the presence of air gaps in the magnetic circuits of the motor and meter of Fig. 1. The spreading of the flux lines outside the common area of the core for the air gap in Fig. 2.(a) is known as fringing.
Fig.2: Air gaps: (a) with fringing; (b) ideal.
For our purposes, we shall neglect this effect and assume the flux distribution to be as in Fig. 2(b). The flux density of the air gap in Fig. 2(b) is given by $$\bbox[10px,border:1px solid grey]{B_g = { \Phi_g \over A_g}} \tag{1}$$ where, for our purposes, $$ \Phi_g = \Phi_core$$ $$ A_g = A_core$$ For most practical applications, the permeability of air is taken to be equal to that of free space. The magnetizing force of the air gap is then determined by $$\bbox[10px,border:1px solid grey]{H_g = {B_g \over \mu_o}} \tag{2}$$ and mmf drop across the air gap is equal to $H_gl_g$. An equation for $H_g$ is as follows: $$H_g = {B_g \over \mu_o} = {B_g \over 4 \pi \times 10^{-7}}$$ and $$\bbox[10px,border:1px solid grey]{H_g = (7.9 \times 10^5)B_g \, \text{(At/m)}} \tag{3}$$
Example 1: Find the value of $I$ required to establish a magnetic flux of $\Phi = 0.75 \times 10^{-4} \, \text{Wb}$ in the series magnetic circuit of Fig. 3.
Fig. 3: Relay for Example 1.
Solution: The flux density for each section is $$B = {\Phi \over A}$$ $$B = {0.75 \times 10^{-4} \over 1.5 \times 10^{-4}} = 0.5 \,T$$ From the B-H curves $$H \text{(cast steel)} = 280 \, At/m$$ Applying Eq. (3), $$ \begin {split} H_g &= (7.9 \times 10^5)B_g \\ &= (7.9 \times 10^5) (0.5T) = 3.98 \times 10^5 \,\text{At/m}\\ \end{split}$$ The mmf drops are $$H_{core}l_{core} = (280 At/m)(100 \times 10^{-3} m) = 28 \text{At}$$ $$H_gl_g = (3.98 \times 10^5 At/m)(2 \times 10^{-3} m) = 796 \text{At}$$ Applying Amperes circuital law, $$\begin{split} NI &= H_{core}l_{core} + H_gl_g\\ &= 28 At + 796 At\\ (200 t)I &= 824 At\\ I &=4.12 A\\ \end{split} $$