Magnetic Circuits Problems And Solutions Pdf Official

Given: Cast steel core (magnetization curve given later). Mean length of steel ( l_s = 0.4 , \textm ), air gap length ( l_g = 1 , \textmm = 0.001 , \textm ), cross-sectional area ( A = 5 \times 10^-4 , \textm^2 ). N = 500 turns. Desired flux density in air gap ( B_g = 0.8 , \textT ). Neglect fringing. Find required current I.

Solution:

Answer: Required current I ≈ 1.59 A.
Note: Air gap dominates reluctance even though it is very short.

Problem Statement: A magnetic core is made of an iron alloy with a constant relative permeability ($\mu_r$) of 1000. The core has a mean length of $50 , \textcm$ and a cross-sectional area of $10 , \textcm^2$. A coil with $500$ turns is wound around the core. magnetic circuits problems and solutions pdf

Solution:

Step 1: Calculate the Reluctance ($\mathcalR$) of the core. First, determine the absolute permeability $\mu$: $$ \mu = \mu_0 \mu_r = (4\pi \times 10^-7) \times 1000 = 4\pi \times 10^-4 , \textH/m $$

Convert dimensions to meters: $$ l = 50 , \textcm = 0.5 , \textm $$ $$ A = 10 , \textcm^2 = 10 \times 10^-4 , \textm^2 = 0.001 , \textm^2 $$ Given: Cast steel core (magnetization curve given later)

Calculate Reluctance: $$ \mathcalR = \fracl\mu A = \frac0.5(4\pi \times 10^-4)(0.001) $$ $$ \mathcalR = \frac0.51.256 \times 10^-6 \approx 398,100 , \textAt/Wb $$

Step 2: Apply Hopkinson’s Law to find MMF ($NI$). $$ NI = \phi \mathcalR $$ $$ NI = (0.005) \times (398,100) $$ $$ NI \approx 1990.5 , \textAmpere-turns $$

Step 3: Calculate Current ($I$). $$ I = \fracNIN = \frac1990.5500 $$ $$ \boxedI \approx 3.98 , \textA $$ Answer: Required current I ≈ 1

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