## Magnetic circuit with an air gap examples

**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.

**4) An iron ring of mean circumference of 150 cm and cross sectional area 12 cm^2 is wound with 600 turns of coil. The coil produces flux of 1.25 mWb while carrying a current of 2A. Find the relative permeability of iron.**

Given: l = 150 cm; a = 12 cm^2; N = 600 turns; \phi = 1.25 mWb; I = 2A.

F = \phi S

S = \dfrac{NI}{\phi}

S = \dfrac {600 \times 2}{1.25 \times 10^{-3}}

S = 960000 Wb/AT

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

\mu_r = \dfrac {150 \times 10^{-2}} {4 \times \pi \times 10^{-7} \times 12 \times 10^{-4} \times 960000}

\mu_r = 1036.1649

**5. An iron ring with a mean diameter of 25cm and relative permeability of 1000 is uniformly wound with 500 turns. Find the current required to produce a flux density of 1 T in the ring. If an air gap of 1 mm is cut in the ring Calculate the new value of current to maintain the same flux density in the ring.**

Given:- diameter(d)=25cm; \mu_r=1000; N=500; B=1 T; l_g=1mm

l=\pi \times d

=\pi \times 25 \times 10^{-2}

=0.7853 m

i)F=\dfrac{Bl}{\mu_0 \mu_r}

NI= \dfrac{Bl}{\mu_0 \mu_r}

I =\dfrac{1 \times 0.7853}{4 \times \pi \times 10^{-7} \times 1000 \times 500}

I= 1.2493 A

ii) with an air gap

F_t =\dfrac{Bl}{\mu_0 \mu_r} + \dfrac{B l_g}{\mu_0 \mu_r}

F_t = \dfrac {1 \times 0.7853}{4 \times \pi \times 10^{-7} \times 1000} + \dfrac{1 \times 1 \times 10^{-3}}{4 \times \pi \times 10^{-7} \times 1}

F_t = 1420.6965 AT

F_t = NI

1420.6965=500 \times I

I= 2.8413A

## Self Inductance and Mutual Inductance problems

**1) Two cols A and B have self-inductance of 10 \muH and 40\muH respectively. A current of 2A in coil A produces a flux linkage of 5\muWb- turns in coil B. Calculate i) Mutual inductance between the coils. ii) Coefficient of coupling. iii) Average EMF induced in coil B if the current 1A in coil A is reversed at uniform rate in 0.1 seconds.**

i) M = \dfrac{N_2 \phi_2}{I_1}

M = \dfrac{ 5 \times 10^{-6}}{2} =0.0000025 H

ii) K = \dfrac{M}{\sqrt{L_1 L_2}} = 0.125

iii) e_2 = -M \dfrac{dI_1}{dt}

e_2 = -0.0000025 \times \dfrac {-1-1}{0.1}

e_2 = 50 \times 10^{-6} V

**2. A coil of 500 turns is uniformly wound on an iron ring of mean circumference 25 cm having an area of cross-section is 15 cm^2. When the coil carries a current of 1 A, produces a flux density of 0.8 T. Calculate i) Magnetizing force (H) ii) Flux iii) Inductance iv) Relative permeability of iron.**

Given:- N=500 Turns; l=25 cm; a=15cm^2; I= 1 A; B=0.8 T

i) H= \dfrac{NI}{l}=\dfrac{500 \times 1}{25 \times 10^{-2}}

H= 2000 AT/m

ii) \phi = B \times a

= 0.8 \times 15 \times 10^{-4}

= 0.0012 Wb.

iii) L = \dfrac {N \phi}{I}

= \dfrac {500 \times 0.0012}{1}

= 0.6 H

iv) \mu_r = \dfrac{B}{\mu_0 H}

=\dfrac {0.8}{4 \times \pi \times 10^{-7} \times 2000}

= 318.3098

**3. An iron ring of mean diameter 20 cm has a square area of a cross-section of 2 cm \times 2cm & is uniformly wound with 600 turns. The relative permeability of iron is 1000. Calculate i) Self-inductance of coil ii) If the permeability of iron is doubled, find the new value of inductance.**

Given:- diameter (d) = 20 cm; N=600 Turns; \mu_r =1000; a=2 \times 2 cm= 4 cm^2

l= \pi \times d=0.6283 m

i) L=\dfrac{\mu_0 \mu_r a N^2}{l}

L = \dfrac{4 \times \pi \times 10^{-7} \times 1000 \times 4 \times 10^{-4} \times 600^2}{0.6283}

L = 0.288 H

ii) L \propto \mu_r

\therefore \; \dfrac{L_{new}}{L}=\dfrac {\mu_{r new}}{\mu_r}

L_{new}= L \times \dfrac{\mu_{r new}}{\mu_r}

=0.288 \times \dfrac{2000}{1000}

=0.576 H

**4. Two coils A and B have self-inductance of 120 \mu H and 300 \mu H respectively. A current of 2 A in coil A, produces flux linkage of 200 \mu Wb-turns in coil B. Calculate i) Mutual inductance ii) Coefficient of coupling K, and iii) Average EMF induced in coil B, when the current in coil A is switched off in 0.05 sec.**

Given:- L_A = 120 \mu H; L_B = 300 \mu H; I_A = 2A; N_B \phi_B = 200 \mu Wb;

i) M=\dfrac{N_B\phi_B}{I_A}

M=\dfrac{200 \times 10^{-6}}{2}

M=100\times 10^{-6} H

ii) K=\dfrac{M}{\sqrt{L_A L_B}}

K= \dfrac{100 \times 10^{-6}}{\sqrt{120 \times 10^{-6}\times 300 \times 10^{-6}}}

K=0.5270

iii) e_B = M \dfrac{dI_A}{dt}

e_B = 100 \times 10^{-6} \times \dfrac{2}{0.05}

e_B= 4 \times 10^{-3} V

## Add a Comment