Question 15.5: Consider the magnetic core with an air gap as shown in Figur...
Consider the magnetic core with an air gap as shown in Figure 15.12(a). The core material has a relative permeability of 6000 and a rectangular cross section 2 cm by 3 cm. The coil has 500 turns. Determine the current required to establish a flux density of B_{gap} = 0.25 T in the air gap.
As shown in Figure 15.12(b), this magnetic circuit is analogous to an electrical circuit with one voltage source and two resistances in series. First, we compute the reluctance of the core. Notice that the centerline of the flux path is a square 6 cm by 6 cm. Thus, the mean length of the iron core is
l_{core} = 4 × 6 − 0.5 = 23.5 cm
The cross-sectional area of the core is
A_{core} = 2 cm × 3 cm = 6 × 10^{−4} m²
The permeability of the core is
μ_{core} = μ_rμ_0 = 6000 × 4π × 10^{−7} = 7.540 × 10^{−3}
Finally, the reluctance of the core is
\mathcal{R}_{core} =\frac{l_{core}} {μ_{core} A_{core}} = \frac{23.5 × 10^{−2}}{7.540 × 10^{−3} × 6 × 10^{−4}}
= 5.195 × 10^4 A·turns/Wb
Now, we compute the reluctance of the air gap.
We approximately account for fringing by adding the length of the gap to the depth and width in computing effective gap area.
The flux lines tend to bow out in the air gap as shown in Figure 15.12(a). This is called fringing. Thus, the effective area of the air gap is larger than that of the iron core. Customarily, we take this into account by adding the length of the gap to each of the dimensions of the air-gap cross section. Thus, the effective area of the gap is
A_{gap} = (2 cm + 0.5 cm) × (3 cm + 0.5 cm) = 8.75 × 10^{−4} m²
The permeability of air is approximately the same as that of free space:
μ_{gap} ≅ μ_0 = 4π × 10^{−7}
Thus, the reluctance of the gap is
\mathcal{R}_{gap} = \frac{l_{gap}}{μ_{gap}A_{gap}} = \frac{0.5 × 10^{−2}}{ 4π × 10^{−7} × 8.75 × 10^{−4}}
= 4.547 × 10^6 A·turns/Wb
The total reluctance is the sum of the reluctance of the core and that of the gap:
\mathcal{R}= \mathcal{R}_{gap} + \mathcal{R}_{core} = 4.547 × 10^6 + 5.195 × 10^4 = 4.600 × 10^6
Even though the gap is much shorter than the iron core, the reluctance of the gap is higher than that of the core because of the much higher permeability of the iron.
Most of the magnetomotive force is dropped across the air gap. (This is analogous to the fact that the largest fraction of the applied voltage is dropped across the largest resistance in a series electrical circuit.)
Now, we can compute the flux:
\phi = B_{gap}A_{gap} = 0.25 × 8.75 × 10^{−4} = 2.188 × 10^{−4} Wb
The flux in the core is the same as that in the gap. However, the flux density is higher in the core, because the area is smaller. The magnetomotive force is given by
\mathcal{F} = \phi\mathcal{R} = 4.600 × 10^6 × 2.188 × 10^{−4} = 1006 A·turns
According to Equation 15.20, we have
\mathcal{F} = Ni (15.20)
Solving for the current and substituting values, we get
i =\frac{\mathcal{F}}{N} = \frac{1006}{ 500} = 2.012 A