3. A simple magnetic circuit, shown below, has a core consisting of two equal pa

Question

3. A simple magnetic circuit, shown below, has a core consisting of two equal parts of steel and iron with an air gap of I mm. The mean lengths of the flux paths, Ii and /2, are 20 cm and 30 cm, respectively. The cross-sectional area is the same for all elements of the structure, 1.0x1.0 cm2. A magnetic flux density of 1.0T exists in the air gap. a) What is the reluctance of the air gap? b) What is the magnetic flux in the air gap? c) What is the magnetic flux in the steel core and in the iron core Find the reluctances of the steel and iron parts, given that the relative permeability is -4000 for steel and -1000 for iron. e) If the number of turns is 200, what is the magnitude of the voltage E that would establish a 1.0T flux in the air gap?

Explanation / Answer

a)We know that the magnetic reluctance is given by:

R = L/Au u0

for air, u = 1

R = 1 x 10^-3 /(0.0001 x 1 x 4 x 3.14 x 10^-7) = 7.96 x 10^6

Hence, R = 7.96 x 10^6 Ampere turns/Wb

b)Phi = BA

phi = 1 x 0.0001 = 1 x 10^-4 Wb

Hence, phi = 1 x 10^-4 Wb

c)Since the cross sectional Area is same for all the elements,

phi = 1 x 10^-4 Wb

d)R1 = 0.3/(0.0001 x 4000 x 4 x 3.14 x 10^-7) = 5.97 x 10^5 A turns/wb

R2 = 0.2/(0.0001 x 1000 x 4 x 3.14 x 10^-7) = 1.59 x 10^6 A turns/Wb

e)e = N B A

e = 200 x 1 x 0.0004 = 20 mV

Hence, e = 20 mV

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