# Electromagnetism solved problems

## 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 $\mu$H and 40$\mu$H respectively. A current of 2A in coil A produces a flux linkage of 5$\mu$Wb- 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=15$cm^2$; 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