Faculty of Electrical Engineering

Electric Drives ENME 503

Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Tutorial (2)
Magnetic Circuits
Problem 1
Coils A and B have mutual inductance 25 mH. At time t = 0 the current in coil A is zero. Subsequently a
time-varying current is supplied to A. Find an expression for the time-varying current in coil A if the
induced emf in coil B is given by the following equation
𝐸𝐵 = (50 + 0.2𝑡) 𝑉/𝑚𝑠
Answer
𝑑𝑖𝐴
𝐸𝐵 = −𝑀 = 50 + 0.2𝑡
𝑑𝑡
(50 + 0.2𝑡)𝑑𝑡
∴ 𝑑𝑖𝐴 =
−25
(50 + 0.2𝑡)𝑑𝑡
∴ ∫ 𝑑𝑖𝐴 = ∫
−25
50 0.2
∴ 𝑖𝐴 = 𝑡+ 𝑡2 + 𝑐
−25 −25 × 2
at t=0 ms, iA=0 A, therefore c=0
50 0.2
∴ 𝑖𝐴 = 𝑡+ 𝑡2
−25 −25 × 2
∴ 𝑖𝐴 = −2𝑡 − 0.004𝑡 2
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Problem 2
Consider the magnetic core with an air gap as shown in Figure 1. The core material has a relative
permeability of 6000 and a rectangular cross-section of 2cm by 3cm. The coil has 500 turns. Assume that
fringing in the air gap increases the effective cross-sectional area of the air gap by 5 percent. Determine the
current required to establish a flux density of Bgap=0.25 T in the air gap. (Hint: consider the mean length of
the air gap)

Figure 1

Answer

𝐴𝑐 = 2𝑐𝑚 × 3𝑐𝑚 = 2 × 3 × 10−4 = 6 × 10−4 𝑚2

𝐴𝑔𝑎𝑝 = 𝐴𝑐 + 0.05𝐴𝑐 = 𝐴𝑐 (1 + 0.05) = 6 × 10−4 (1 + 0.05) = 6.3 × 10−4 𝑚2
𝜙 = 𝐵𝑔𝑎𝑝 𝐴𝑔𝑎𝑝 = 0.25 × 6.3 × 10−4 = 1.575 × 10−4 𝑊𝑏
𝑙𝑐 𝑙𝑐 23.5 × 10−2
ℛ𝑐 = = = = 51946.41 𝐴. 𝑇/𝑊𝑏
𝜇𝑐 𝐴𝑐 𝜇𝑟 𝜇𝑜 𝐴𝑐 6000 × 4𝜋 × 10−7 × 6 × 10−4
𝑙𝑔𝑎𝑝 0.5 × 10−2
ℛ𝑔𝑎𝑝 = = = 6315672.34 𝐴. 𝑇/𝑊𝑏
𝜇𝑜 𝐴𝑔𝑎𝑝 4𝜋 × 10−7 × 6.3 × 10−4
ℱ = 𝑁𝑖 = 𝜙(ℛ𝑐 + ℛ𝑔𝑎𝑝 )
𝜙(ℛ𝑐 + ℛ𝑔𝑎𝑝 ) 1.575 × 10−4 (51946.41 + 6315672.34)
∴𝑖= = =2𝐴
𝑁 500
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Problem 3
A ferromagnetic core is shown in Figure 2. The depth of the core is 5 cm. The other dimensions of the core are
as shown in Figure 2.
a) Draw the equivalent magnetic circuit of Figure 2.
b) Find the value of the current that will produce a flux of 0.005 Wb?
c) With this current, find the flux density at the top of the core? What is the flux density at the right
side of the core? (Assume that the relative permeability of the core is 1000.)

Figure 2

Answer

𝑙1 = 7.5 𝑐𝑚 + 15 𝑐𝑚 + 7.5 𝑐𝑚 = 0.3 𝑚, 𝐴1 = 10 𝑐𝑚 × 5 𝑐𝑚 = 5 × 10−3 𝑚2

𝑙2 = 5 𝑐𝑚 + 20 𝑐𝑚 + 2.5 𝑐𝑚 = 0.275𝑚, 𝐴2 = 15 𝑐𝑚 × 5 𝑐𝑚 = 7.5 × 10−3 𝑚2
𝑙3 = 7.5 𝑐𝑚 + 15 𝑐𝑚 + 7.5 𝑐𝑚 = 0.3 𝑚, 𝐴3 = 5 𝑐𝑚 × 5 𝑐𝑚 = 2.5 × 10−3 𝑚2
𝑙4 = 5 𝑐𝑚 + 20 𝑐𝑚 + 2.5 𝑐𝑚 = 0.275𝑚, 𝐴4 = 15 𝑐𝑚 × 5 𝑐𝑚 = 7.5 × 10−3 𝑚2
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

𝑙1 0.3
ℛ1 = = = 47748 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴1 1000 × 4𝜋 × 10−7 × 5 × 10−3
𝑙2 0.275
ℛ2 = = = 29179 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴2 1000 × 4𝜋 × 10−7 × 7.5 × 10−3
𝑙3 0.3
ℛ3 = = = 95496 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴3 1000 × 4𝜋 × 10−7 × 2.5 × 10−3
𝑙4 0.275
ℛ4 = = = 29179 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴4 1000 × 4𝜋 × 10−7 × 7.5 × 10−3
𝑇
ℛ𝑡𝑜𝑡𝑎𝑙 = ℛ1 + ℛ2 + ℛ3 + ℛ4 = 201602 𝐴.
𝑚
ℱ = 𝑁𝑖 = 𝜙ℛ𝑡𝑜𝑡𝑎𝑙
𝜙ℛ𝑡𝑜𝑡𝑎𝑙 0.005(201602)
∴𝑖= = =2𝐴
𝑁 500
𝜙 0.005
𝐵𝑡𝑜𝑝 = = = 0.6667 𝑇
𝐴2 7.5 × 10−3
𝜙 0.005
𝐵𝑟𝑖𝑔ℎ𝑡 = = =2𝑇
𝐴3 2.5 × 10−3
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Problem 4
A two-legged core is shown in Figure 3. The winding on the left leg of the core (N1) has 400 turns, and the
winding on the right (N2) has 300 turns. The coils are wound in the directions shown in the Figure 3. If the
dimensions are as shown, then what flux would be produced by currents i1 = 0.5 A and i2 = 0.75 A?
(Assume µr = 1000 and constant.)

Figure 3

Answer

𝑙𝑐 = 4(7.5 𝑐𝑚 + 50 𝑐𝑚 + 7.5 𝑐𝑚) = 2.6 𝑚, 𝐴𝑐 = 15 𝑐𝑚 × 15 𝑐𝑚 = 0.0225 𝑚2

𝑙𝑐 2.6
ℛ𝑐 = = = 91956.2 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴𝑐 1000 × 4𝜋 × 10−7 × 0.0225
ℱ1 = 𝑁1 𝑖1 = 400 × 0.5 = 200 𝐴. 𝑇
ℱ2 = 𝑁2 𝑖2 = 300 × 0.75 = 225 𝐴. 𝑇
ℱ𝑡𝑜𝑡𝑎𝑙 = ℱ1 + ℱ2 = 200 + 225 = 425 𝐴. 𝑇
ℱ𝑡𝑜𝑡𝑎𝑙 425
𝜙= = = 4.62 × 10−3 𝑊𝑏
ℛ𝑐 91956.2
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Problem 5
Figure 4 shows the core of a simple DC motor. The magnetic flux density is 1.2 T while the relative
permeability of the core (stator and rotor) is 3800. Assume that the cross-sectional area of each air gap is 18
cm2 and that the width of each air gap is 0.05 cm. The effective diameter of the rotor core is 4 cm.
a) Draw the magnetic circuit of Figure 4.
b) Find the number of turns if the current inside the coil is 1 A.

Figure 4

Answer

𝐴𝑠𝑡𝑎𝑡𝑜𝑟 = 0.04 × 0.04 = 1.6 × 10−3 𝑚2

𝐴𝑟𝑜𝑡𝑜𝑟 = 0.04 × 0.04 = 1.6 × 10−3 𝑚2
𝐴𝑎𝑖𝑟 𝑔𝑎𝑝 = 1.8 × 10−3 𝑚2
𝑙𝑠𝑡𝑎𝑡𝑜𝑟 0.48
ℛ𝑠𝑡𝑎𝑡𝑜𝑟 = = = 62.824 × 103 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴𝑠𝑡𝑎𝑡𝑜𝑟 3800 × 4𝜋 × 10−7 × 1.6 × 10−3
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

𝑙𝑟𝑜𝑡𝑜𝑟 0.04
ℛ𝑟𝑜𝑡𝑜𝑟 = = = 5.235 × 103 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴𝑟𝑜𝑡𝑜𝑟 3800 × 4𝜋 × 10−7 × 1.6 × 10−3
𝑙𝑎𝑖𝑟 𝑔𝑎𝑝 0.05 × 10−2
ℛ𝑎𝑖𝑟 𝑔𝑎𝑝 = = = 221.0485 × 103 𝐴. 𝑇/𝑊𝑏
𝜇𝑟 𝜇𝑜 𝐴𝑎𝑖𝑟 𝑔𝑎𝑝 4𝜋 × 10−7 × 1.8 × 10−3
ℛ𝑡𝑜𝑡𝑎𝑙 = ℛ𝑠𝑡𝑎𝑡𝑜𝑟 + ℛ𝑟𝑜𝑡𝑜𝑟 + 2 ℛ𝑎𝑖𝑟 𝑔𝑎𝑝 = 62.824 × 103 + 5.235 × 103 + 2(221.0485 × 103 )
= 510.156 × 103 𝐴. 𝑇/𝑊𝑏
𝜙 = 𝐵𝐴 = 1.2 × 1.6 × 10−3 = 1.92 × 10−3 𝑊𝑏
ℱ = 𝑁𝑖 = 𝜙ℛ𝑡𝑜𝑡𝑎𝑙
𝜙ℛ𝑡𝑜𝑡𝑎𝑙 1.92 × 10−3 (510.156 × 103 )
𝑁= = = 979.5 → 1000 𝑇𝑢𝑟𝑛𝑠
𝑖 1
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Problem 6
The magnetic circuit shown in Figure 5 has two windings and two air gaps. The core can be assumed to be
of infinite permeability. Derive an expression for the self-inductances of windings 1 and 2 and the mutual
inductance between the windings.

Figure 5

Answer
➢ Since, the core has infinite permeability, the reluctance of the iron core is neglected.
➢ Only the air gap reluctance will be taken into consideration.

ℱ1 = 𝑁1 𝑖1 , ℱ2 = 𝑁2 𝑖2
𝑔1 𝑔2
ℛ1 = , ℛ2 =
𝜇𝑜 𝐴1 𝜇𝑜 𝐴2
➢ To get the self-inductance of coil (1) (L11):
Φ11: the flux passing through coil (1) generated by coil (1)
➢ Therefore, shut down coil (2) and let only coil (1) to be switched on, and then study the flux passing
through coil (1)
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

ℱ1 ℱ1 𝑁1 𝑖1 𝑁1 𝑖1 𝐴1 𝐴2
𝜙11 = + = 𝑔 + 𝑔 = 𝑁1 𝑖1 𝜇𝑜 ( + )
ℛ1 ℛ 2 ( 1 ) ( 2 ) 𝑔1 𝑔2
𝜇𝑜 𝐴1 𝜇𝑜 𝐴2
2 𝐴1 𝐴2
𝑁1 𝜙11 𝑁1 𝑖1 𝜇𝑜 (𝑔1 + 𝑔2 ) 𝐴1 𝐴2
𝐿11 = = = 𝑁1 2 𝜇𝑜 ( + )
𝑖1 𝑖1 𝑔1 𝑔2
➢ Similarly we can also obtain the self-inductance of coil (2) (L22):
Φ22: the flux passing through coil (2) generated by coil (2)
➢ Therefore, shut down coil (1) and let only coil (2) to be switched on, and then study the flux passing
through coil (2)

ℱ2 𝑁2 𝑖2 𝑁2 𝑖2 𝜇𝑜 𝐴2
𝜙22 = = 𝑔 =
ℛ2 ( 2 ) 𝑔2
𝜇𝑜 𝐴2
𝑁2 𝜙22 𝑁2 2 𝑖2 𝜇𝑜 𝐴2 𝑁2 2 𝜇𝑜 𝐴2
𝐿22 = = =
𝑖2 𝑖2 𝑔2 𝑔2
➢ To get the mutual inductance of coil (1) and coil (2) (L12 or L21):
Φ12: the flux passing through coil (1) generated by coil (2)
➢ Therefore, shut down coil (1) and let only coil (2) to be switched on, and then study the flux passing
through coil (1)

ℱ2 𝑁2 𝑖2 𝑁2 𝑖2 𝜇𝑜 𝐴2
𝜙12 = = 𝑔 =
ℛ2 ( 2 ) 𝑔2
𝜇𝑜 𝐴2
𝑁1 𝜙12 𝑁1 𝑁2 𝑖2 𝜇𝑜 𝐴2 𝑁1 𝑁2 𝜇𝑜 𝐴2
𝐿12 = = =
𝑖2 𝑖2 𝑔2 𝑔2
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

Problem 7
A system of three coils on an ideal core is shown in Figure 6 where the number of turns of each coil and the
airgap lengths are given in the following table.

N1 N2 N3 g1 g2 g3
500 turns 250 turns 500 turns 4 mm 2 mm 2 mm

If the cross-sectional area was constant throughout the core, which is equal to 100 mm2, calculate:
a) The self-inductance of coil N1.
b) The mutual inductance between coils N2 and N3.

Figure 6

Answer
𝑔1 4 × 10−3
ℛ1 = = = 31.831 × 106 𝐴. 𝑇/𝑊𝑏
𝜇𝑜 𝐴 4𝜋 × 10−7 × 100 × 10−6
𝑔2 2 × 10−3
ℛ2 = = = 15.915 × 106 𝐴. 𝑇/𝑊𝑏
𝜇𝑜 𝐴 4𝜋 × 10−7 × 100 × 10−6
𝑔3 2 × 10−3
ℛ3 = = = 15.915 × 106 𝐴. 𝑇/𝑊𝑏
𝜇𝑜 𝐴 4𝜋 × 10−7 × 100 × 10−6
ℱ1 = 𝑁1 𝑖1 = 500 𝑖1 , ℱ2 = 𝑁2 𝑖2 = 250 𝑖2 , ℱ3 = 𝑁3 𝑖3 = 500 𝑖3
➢ To get L11 (coil 1 is on while coil 2 and coil 3 are deactivated, while checking the flux passing through
coil 1)

ℛ2 ℛ3
ℛ𝑒𝑞 = ℛ1 + = 39.789 × 106 𝐴. 𝑇/𝑊𝑏
ℛ2 + ℛ3
 Faculty of Electrical Engineering
Electric Drives ENME 503
Dr. Eng. Noha Shouman
Eng. Ahmed AlKhatib, Eng. Kareem A. Noureldin, M.Sc.

ℱ1 500 𝑖1
𝜙11 = = 𝑊𝑏
ℛ𝑒𝑞 39.789 × 106
500 𝑖1
𝑁1 𝜙11 500 (39.789 × 106 )
𝐿11 = = = 0.006283 𝐻 = 6.283 𝑚𝐻
𝑖1 𝑖1
➢ To get L23 (coil 3 is on while coil 1 and coil 2 are deactivated, while checking the flux passing through
coil 2)

ℛ1 ℛ 2
ℛ𝑒𝑞 = ℛ3 + = 26.525 × 106 𝐴. 𝑇/𝑊𝑏
ℛ1 + ℛ 2
ℱ3 500 𝑖3
𝜙33 = = 𝑊𝑏
ℛ𝑒𝑞 26.525 × 106
ℛ1
𝜙23 = 𝜙33 ( ) = 12.567 × 10−6 𝑖3 𝑊𝑏
ℛ1 + ℛ 2
𝑁2 𝜙23 250(12.567 × 10−6 𝑖3 )
𝐿23 = = = 0.00314 𝐻 = 3.14 𝑚𝐻
𝑖3 𝑖3