Electromagnetism Solved problems

Electromagnetism solved problems

Magnetic circuit with an air gap

1) An iron ring has its mean length of flux path as 60 cm and its cross-sectional area as 15 cm². If relative permeability is 500. Find the current required to be passed through a coil of 300 turns wound uniformly around it to produce a flux density of 1.2 T. What would be the flux density with the same current if the iron ring is replaced by an air core.

Given: l = 60 cm; a = 15 cm²; \mu_r = 500 ; N = 300; B=1.2 T. 

MMF(F) = \dfrac{Bl}{\mu_0 \mu_r}  

F = \dfrac{1.2 \times 60 \times 10^{-2}}{4\pi \times 10^{-7}\times 500}

F = 1145.9155 AT

If iron ring is replaced by air core.

MMF(F) = \dfrac{B_{air}l}{\mu_0 \mu_r}  

B_{air} = 2.4 \times 10^{-3} T

2) A coil of 600 turns and resistance of 10 \Omega is wound uniformly over a steel ring of mean circumference of 30 cm and cross sectional area of 9 cm^2 . It is connected to a supply of 20 V if the relative permeability of the ring is 1600. Find 1) Reluctance 2) Magnetic field intensity 3) MMF 4) Flux.

i) Reluctance:

S = \dfrac{l}{\mu_0 \mu_r a}

S = \dfrac{30 \times 10^{-2}}{4 \times 10^{-7} \times 1600 \times 9 \times 10^{4}}

S = 165786.3991 \;  AT/Wb

ii) Magnetic field intensity:

I = \dfrac{V}{R}

I = \dfrac{20}{10}

I = 2 \; A

H = \dfrac{NI}{l}

H = \dfrac{600 \times 2}{30 \times 10^{-2}}

H = 4000 AT/m

iii) MMF = NI = 600 \times 2 = 1200 AT.

iv) Flux:

\phi = \dfrac{600 \times 2}{165786.3991}

\phi = 7.2382 \times 10^{-3} \; Wb

3) A ring shaped core is made up of two parts of the same material. Part one is a magnetic path of length 25 cm and with cross sectional area 4 cm^2, whereas part two is of length 10 cm and cross section area of 6 cm^2. The flux density in part two is 1.5 T. If the current through the coil, wound over core is 0.5 A. Calculate the number of turns of coil. Assume \mu_r for material is 1000.

Given: l_1 = 25 cm; l_2 = 10 cm; a_1 = 4 cm^2; a_2 = 6 cm^2; B = 1.5 T; \mu_r = 1000

S = S_1 + S_2

S = \dfrac{l_1}{\mu_0 \mu_r A_1} + \dfrac{l_2}{\mu_0 \mu_r A_2}

S = \dfrac{1}{4\pi \times 10^{-7} \times{10^3}} \left[ \dfrac{0.25}{4\times 10^{-4}} + \dfrac{0.1}{6\times 10^{-4}} \right]

S = 629988.3164 AT/Wb.

B_2 = \dfrac{\phi}{A_2}

\phi = 1.5 \times 6 \times 10^{-4} = 9 \times 10^{-4} Wb.

F = \phi \times S

NI = \phi \times S

N = \dfrac{S \phi}{I} = \dfrac{629988.3164 \times 9 \times 10^{-4}}{0.5}

N = 1133.98 = 1134 T.

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