Question 15.6: The iron core shown in Figure 15.13(a) has a cross section o...

The magnetic circuit is depicted in Figure 15.13(b). First, we compute the reluctances of the three paths. For the center path, we have
\mathcal{R}_c = \frac{l_c}{μ_rμ_0A_{core}} =\frac{ 10 × 10^{−2}}{ 1000 × 4π × 10^{−7} × 4 × 10^{−4}}
= 1.989 × 10^5 A·turns/Wb

For the left-hand path, the total reluctance is the sum of the reluctance of the iron core plus the reluctance of gap a. We take fringing into account by adding the gap length to its width and depth in computing area of the gap. Thus, the area of gap a is A_a = 3 cm × 3 cm = 9 × 10^{−4} m². Then, the total reluctance of the left-hand path is
\mathcal{R}_a = \mathcal{R}_{gap} + \mathcal{R}_{core}
= \frac{l_{gap}} {μ_0A_a} + \frac{l_{core}}{ μ_rμ_0A_{core}}
= \frac{1 × 10^{−2}}{4π × 10^{−7} × 9 × 10^{−4}} + \frac{29 × 10^{−2}} {1000 × 4π × 10^{−7} × 4 × 10^{−4}}
= 8.842 × 10^6 + 5.769 × 10^5
= 9.420 × 10^6 A·turns/Wb
Similarly, the reluctance of the right-hand path is

= \frac{l_{gap}} {μ_0A_b} + \frac{l_{core}}{ μ_rμ_0A_{core}}

= \frac{0.5 × 10^{−2}}{4π × 10^{−7} × 6.25 × 10^{−4}} + \frac{29.5 × 10^{−2}} {1000 × 4π × 10^{−7} × 4 × 10^{−4}}

= 6.366 × 10^6 + 5.869 × 10^5
= 6.953 × 10^6 A·turns/Wb

Next, we can combine the reluctances \mathcal{R}_a and \mathcal{R}_b in parallel. Then, the total reluctance is the sum of \mathcal{R}_c and this parallel combination:
\mathcal{R}_{total} = \mathcal{R}_c + \frac{1}{ 1/\mathcal{R}_a + 1/\mathcal{R}_b}
= 1.989 × 10^5 + \frac{1}{1/(9.420 × 10^6) + 1/(6.953 × 10^6)}
= 4.199 × 10^6 A·turns/Wb
Now, the flux in the center leg of the coil can be found by dividing the magnetomotive force by the total reluctance:
\phi_c = \frac{Ni}{ \mathcal{R}_{total}} = \frac{500 × 2} {4.199 × 10^6} = 238.1 μWb
Fluxes are analogous to currents. Thus, we use the current-division principle to determine the flux in the left-hand and right-hand paths, resulting in
\phi_a = \phi_c\frac{\mathcal{R}_b} {\mathcal{R}_a + \mathcal{R}_b}
= 238.1 × 10^{−6} × \frac{6.953 × 10^6}{ 6.953 × 10^6 + 9.420 × 10^6}
= 101.1 μWb

Similarly, for gap b we have
\phi_b = \phi_c\frac{\mathcal{R}_a} {\mathcal{R}_a + v_b}
=238.1 × 10^{−6} × \frac{9.420 × 10^6}{ 6.953 × 10^6 + 9.420 × 10^6}
= 137.0 μWb
As a check on these calculations, we note that \phi_c = \phi_a + \phi_b.
Now, we find the flux densities in the gaps by dividing the fluxes by the areas:
B_a = \frac{\phi_a} {A_a} = \frac{101.1  μWb} {9 × 10^{−4}  m^2} = 0.1123 T
B_b = \frac{\phi_b}{ A_b} = \frac{137.0  μWb} {6.25 × 10^{−4}  m^2} = 0.2192 T