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R.M.K. COLLEGE OF ENGINEERING AND TECHNOLOGY
RSM Nagar, Puduvoyal - 601 206
DEPARTMENT OF ECE
EC 8451
ELECTRO MAGNETIC FIELDS
UNIT IV
TIME - VARYING FIELDS AND MAXWELL’S EQUATIONS
Dr. KANNAN K
AP/ECE
Problems in Inductance
(i) A Solenoid has inductance of 20 mH. If the length of the Solenoid is increased by 2
times and the radius is decreased to half of its original value. Find new inductance
Solution :
Given L1 = 20 mH , 2l1 = l2 and radius r1/2 = r2
Inductance of the Solenoid L1 =
𝜇0 𝑁1
2𝐴1
l1
The new inductance of the solenoid L2 = =
𝜇0 𝑁2
2𝐴2
l2
Here N1 = N2, A1=𝜋 r1
2 ,A2 = 𝜋 (r1
2 /4) = 4 A1
L2
L1
=
𝑁1
24𝐴1
l1
𝑁1
2𝐴1
2l1
=
4
2
L2 = 2 L1 = 2 x 20 mH = 40 mH
(ii) A toroid of mean radius 0.1m and 10cm2 cross sectional area is uniformly wound with 250
turns of wire. If B is 1 Tesla and µr =500, what is the current required to be passed in the winding?
Determine also the value of self inductance
Solution:
Available Data : r =0.1m,
A = 10cm2 = 10 x 10-4 m2
N = 250 turns , B = 1 tesla , µr =500
We know that, Φ = B A
= 1 x 10 x 10-4 = 10 x 10-4 Wb
B = µ0 µr H =
µ0 µrNI
2 𝜋 𝑅𝑚
1
𝐼
=
4 𝜋 𝑥 10−7 𝑥 500 𝑥 250
1𝑥2 𝑥 𝜋 𝑥 0.1
= 0.25
I = 4A
L = N Φ/I = =
250 x 10 𝑥 10−4
4
L = 0.0625H
(iii) A loop with magnetic dipole moment 8x10 -3 a z A.m 2 lies in a uniform magnetic field
B=0.2a x +0.4a z wb/m 2 . Calculate the torque.
Solution :
Torque T = M X B = 8x10 -3 a z x (0.2a x +0.4a z )
= 1.6 x 10 -3 a y
(iv) What is the maximum torque on a square loop of 100 turns in a field of uniform flux
density 3 Tesla? The loop has 20cm sides and carries a current of 10A.
Solution :
Torque T = BIA Sin Θ
Maximum torque is obtained when Θ = 90o
Tmax = BIA
Area of square loop = 4 x 20 x10-2 = 80 x 10-2 m2
Square loop = 100 turns , 100 x 80 x 10-2 m2 = 80 m2
Tmax = 3 x 10 x 80 m2 = 2400 N-m
(v) An iron ring of relative permeability 100 is wound uniformly with two coils of 100 and 400 turns of
wire. The cross section of the ring is 4cm 2 . The mean circumference is 50cm. Calculate
(i) The self inductance of each of the two coils
(ii) The mutual inductance
(iii) Total inductance when the coils are connected in series with flux in the same sense
(iv) Total inductance when the coils are connected in series with flux in the opposite sense.
Solution:
(i) Self Inductance of Coil with 1000 turns
L =
µ0 N2A
2 𝜋 𝑅𝑚
= =
4 𝜋 𝑥 10−7 x 1002 x 4
50 𝑥 10−2
= 0.1 Hendry
Self Inductance of Coil with 400 turns
L =
µ0 N2A
2 𝜋 𝑅𝑚
= =
4 𝜋 𝑥 10−7 x 4002 x 4
50 𝑥 10−2
= 1.61 Hendry
(ii) Mutual Inductance
L =
µ0µ1N1N2A
2 𝜋 𝑅𝑚
=
4 𝜋 𝑥 10−7 x 100 x 400 x 4
50 𝑥 10−2
= 0.402 Hendry
(iii) Total Inductance L = L1 + L2 + 2M = 0.1 + 1.61 + 2 x 0.402 = 2.51
(iv) Total Inductance when the coils are in series with flux in opposite sense
L = L1 + L2 - 2M = 0.1 + 1.61 - 2 x 0.402 = 0.91
(vi) A solenoid is 50 cm long, 2 cm in diameter and contains 1500 turns. The cylindrical core has a
diameter of 2 cm and a relative permeability of 75. This coil is coaxial with second solenoid, also 50 cm
long, but 3 cm diameter and 1200 turns. Calculate L for the inner solenoid and L for outer solenoid and M
between two solenoids.
Solution :
Inductance of inner solenoid L1 =
𝜇0 𝜇𝑟𝑁1
2𝐴1
l1
=
4 𝜋 𝑥 10−7 𝑥 75 𝑥 1500 2
𝑥 𝜋 𝑥 1 𝑥 10−2 2
50 𝑥 10−2
= 133.2mH
Inductance of outer solenoid L2 =
𝜇0 𝜇𝑟𝑁2
2𝐴2
l2
=
4 𝜋 𝑥 10−7 𝑥 75 𝑥 1200 2
𝑥 𝜋 𝑥 1.5 𝑥 10−2 2
50 𝑥 10−2
= 191.87mH
Mutual Inductance M =
µ0µ1N1N2A
𝑙
=
4 𝜋 𝑥 10−7 𝑥 75 𝑥 1200 𝑥 1500 𝑥 𝜋 𝑥 0.01 2
50
= 106.6mH.
(vii) A current of 2A flowing in an inductor of inductance 100mH.What is the energy stored in the
inductor?
Solution :
Energy Stored =
1
2
LI2
= 1
2
x 100 x 10 -3 x 22 = 200mv

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Unit iv problems - inductance

  • 1. R.M.K. COLLEGE OF ENGINEERING AND TECHNOLOGY RSM Nagar, Puduvoyal - 601 206 DEPARTMENT OF ECE EC 8451 ELECTRO MAGNETIC FIELDS UNIT IV TIME - VARYING FIELDS AND MAXWELL’S EQUATIONS Dr. KANNAN K AP/ECE
  • 2. Problems in Inductance (i) A Solenoid has inductance of 20 mH. If the length of the Solenoid is increased by 2 times and the radius is decreased to half of its original value. Find new inductance Solution : Given L1 = 20 mH , 2l1 = l2 and radius r1/2 = r2 Inductance of the Solenoid L1 = 𝜇0 𝑁1 2𝐴1 l1 The new inductance of the solenoid L2 = = 𝜇0 𝑁2 2𝐴2 l2 Here N1 = N2, A1=𝜋 r1 2 ,A2 = 𝜋 (r1 2 /4) = 4 A1 L2 L1 = 𝑁1 24𝐴1 l1 𝑁1 2𝐴1 2l1 = 4 2 L2 = 2 L1 = 2 x 20 mH = 40 mH
  • 3. (ii) A toroid of mean radius 0.1m and 10cm2 cross sectional area is uniformly wound with 250 turns of wire. If B is 1 Tesla and µr =500, what is the current required to be passed in the winding? Determine also the value of self inductance Solution: Available Data : r =0.1m, A = 10cm2 = 10 x 10-4 m2 N = 250 turns , B = 1 tesla , µr =500 We know that, Φ = B A = 1 x 10 x 10-4 = 10 x 10-4 Wb B = µ0 µr H = µ0 µrNI 2 𝜋 𝑅𝑚 1 𝐼 = 4 𝜋 𝑥 10−7 𝑥 500 𝑥 250 1𝑥2 𝑥 𝜋 𝑥 0.1 = 0.25 I = 4A L = N Φ/I = = 250 x 10 𝑥 10−4 4 L = 0.0625H
  • 4. (iii) A loop with magnetic dipole moment 8x10 -3 a z A.m 2 lies in a uniform magnetic field B=0.2a x +0.4a z wb/m 2 . Calculate the torque. Solution : Torque T = M X B = 8x10 -3 a z x (0.2a x +0.4a z ) = 1.6 x 10 -3 a y (iv) What is the maximum torque on a square loop of 100 turns in a field of uniform flux density 3 Tesla? The loop has 20cm sides and carries a current of 10A. Solution : Torque T = BIA Sin Θ Maximum torque is obtained when Θ = 90o Tmax = BIA Area of square loop = 4 x 20 x10-2 = 80 x 10-2 m2 Square loop = 100 turns , 100 x 80 x 10-2 m2 = 80 m2 Tmax = 3 x 10 x 80 m2 = 2400 N-m
  • 5. (v) An iron ring of relative permeability 100 is wound uniformly with two coils of 100 and 400 turns of wire. The cross section of the ring is 4cm 2 . The mean circumference is 50cm. Calculate (i) The self inductance of each of the two coils (ii) The mutual inductance (iii) Total inductance when the coils are connected in series with flux in the same sense (iv) Total inductance when the coils are connected in series with flux in the opposite sense. Solution: (i) Self Inductance of Coil with 1000 turns L = µ0 N2A 2 𝜋 𝑅𝑚 = = 4 𝜋 𝑥 10−7 x 1002 x 4 50 𝑥 10−2 = 0.1 Hendry Self Inductance of Coil with 400 turns L = µ0 N2A 2 𝜋 𝑅𝑚 = = 4 𝜋 𝑥 10−7 x 4002 x 4 50 𝑥 10−2 = 1.61 Hendry (ii) Mutual Inductance L = µ0µ1N1N2A 2 𝜋 𝑅𝑚 = 4 𝜋 𝑥 10−7 x 100 x 400 x 4 50 𝑥 10−2 = 0.402 Hendry (iii) Total Inductance L = L1 + L2 + 2M = 0.1 + 1.61 + 2 x 0.402 = 2.51 (iv) Total Inductance when the coils are in series with flux in opposite sense L = L1 + L2 - 2M = 0.1 + 1.61 - 2 x 0.402 = 0.91
  • 6. (vi) A solenoid is 50 cm long, 2 cm in diameter and contains 1500 turns. The cylindrical core has a diameter of 2 cm and a relative permeability of 75. This coil is coaxial with second solenoid, also 50 cm long, but 3 cm diameter and 1200 turns. Calculate L for the inner solenoid and L for outer solenoid and M between two solenoids. Solution : Inductance of inner solenoid L1 = 𝜇0 𝜇𝑟𝑁1 2𝐴1 l1 = 4 𝜋 𝑥 10−7 𝑥 75 𝑥 1500 2 𝑥 𝜋 𝑥 1 𝑥 10−2 2 50 𝑥 10−2 = 133.2mH Inductance of outer solenoid L2 = 𝜇0 𝜇𝑟𝑁2 2𝐴2 l2 = 4 𝜋 𝑥 10−7 𝑥 75 𝑥 1200 2 𝑥 𝜋 𝑥 1.5 𝑥 10−2 2 50 𝑥 10−2 = 191.87mH Mutual Inductance M = µ0µ1N1N2A 𝑙 = 4 𝜋 𝑥 10−7 𝑥 75 𝑥 1200 𝑥 1500 𝑥 𝜋 𝑥 0.01 2 50 = 106.6mH. (vii) A current of 2A flowing in an inductor of inductance 100mH.What is the energy stored in the inductor? Solution : Energy Stored = 1 2 LI2 = 1 2 x 100 x 10 -3 x 22 = 200mv