Magnetic Circuit Analysis
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6. The magnetic flux density in the air gap is $B_g = 0.8 \text{ T}$, and the current $I_2 = 8 \text{ A}$. Determine the current $I_1$. The given parameters are: $\ell_1 = 150 \text{ cm}$, $\ell_2 = 20 \text{ cm}$, $\ell_g = 250 \text{ cm}$, $\delta = 3 \text{ mm}$, $N_1 = 4000$, $N_2 = 1000$, and the cross-sectional area of the core is uniform and equal to $S = 25 \text{ cm}^2$. Assume that the magnetic permeability of the core is constant and equal to $\mu_r = 2000$. Note that the length of the middle section is $2\ell_2 + \delta$.
This question includes visual content: The image shows a schematic of a rectangular magnetic core with a center vertical branch containing an air gap. The core has two windings labeled $N_1$ and $N_2$ at the top and bottom of the left rectangular section, carrying currents $I_1$ and $I_2$ respectively. The lengths of various core sections are labeled: $\ell_1$ for the far-left outer vertical section, $\ell_3$ for the far-right outer vertical section, $\ell_g$ for the middle section (including the air gap), and $\delta$ for the width of the air gap itself.
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Magnetic Circuit Analysis
First, we list the given parameters and convert them to SI units.
Parameters:
- $B_g = 0.8 \; \text{T}$
- $I_2 = 8 \; \text{A}$
- $N_1 = 4000, N_2 = 1000$
- $l_1 = 1.5 \; \text{m}, l_3 = 2.5 \; \text{m}, l_2 = 0.2 \; \text{m}$
- $\delta = 0.003 \; \text{m}$
- $S = 25 \times 10^{-4} \; \text{m}^2$
- $\mu_r = 2000, \mu_0 = 4\pi \times 10^{-7} \; \text{H/m}$
The total flux $\Phi$ passing through the air gap is calculated as the product of the flux density and the cross-sectional area.
Next, we find the magnetic field intensity $H$ in both the iron core and the air gap. For the air gap, $H_g = B_g / \mu_0$.
For the core, $H_c = B_g / (\mu_0 \mu_r)$. Using $\mu_r = 2000$, this simplifies to $B_g / (\mu_0 \cdot 2000)$.
We need the total path length of the iron core. Summing the given components, $l_c = l_1 + l_3 + 2l_2$.
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