Magnetic: Circuits Problems And Solutions Pdf

A magnetic circuit is a closed path followed by magnetic flux lines, similar to how an electric circuit provides a path for current

Answer: Flux in each outer limb = 2.26 mWb.

If we had neglected nonlinearity and assumed μ_r constant (e.g., 1000), error would be large.

1. Hysteresis loss density: ( p_h = 200 \times 50 \times (1.2)^1.6 )
   ( (1.2)^1.6 = e^1.6 \ln 1.2 = e^1.6 \times 0.1823 = e^0.2917 ≈ 1.339 ).
   ( p_h = 200 \times 50 \times 1.339 = 13390 , \textW/m^3 ).
2. Eddy current loss density: ( p_e = 0.5 \times (50)^2 \times (1.2)^2 = 0.5 \times 2500 \times 1.44 = 1800 , \textW/m^3 ).
3. Total loss density = 13390 + 1800 = 15190 W/m³.
4. Total loss = 15190 × 0.005 = 75.95 W.

Solution:

Step 1: Draw the equivalent magnetic circuit (MMF source, reluctances in series/parallel). Step 2: Calculate each reluctance: ( \mathcalR = \fracl\mu_0 \mu_r A ). Use mean path length for iron. Step 3: Compute total reluctance ( \mathcalRtotal ). Step 4: Apply Ohm’s law: ( \Phi = \fracNI\mathcalRtotal ). Step 5: If material is non-linear, use B-H curve iteratively:

Essential Formulas

1. Magnetomotive Force (MMF): $$F = NI$$ (Where $N$ = number of turns, $I$ = current)
2. Magnetic Field Intensity ($H$): $$H = \fracNIl \quad \textor \quad H = \fracFl$$ (Where $l$ = mean length of the magnetic path)
3. Magnetic Flux Density ($B$): $$B = \frac\phiA$$ (Where $\phi$ = total flux, $A$ = cross-sectional area)
4. Permeability ($\mu$): $$\mu = \mu_0 \mu_r$$ (Where $\mu_0 = 4\pi \times 10^-7 , \textH/m$, $\mu_r$ = relative permeability)
5. Reluctance ($\mathcalR$): $$\mathcalR = \fracl\mu A$$
6. Hopkinson’s Law (Magnetic Ohm's Law): $$F = \phi \mathcalR \quad \textor \quad NI = \phi \mathcalR$$

The magnetic flux is given by:

• Calculate: The current required to establish a magnetic flux of $0.005 , \textWb$ in the core.