This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Magnetic Field Intensity”.
1. The H quantity is analogous to which component in the following?
a) B
b) D
c) E
d) V
View Answer
a) B
b) D
c) E
d) V
View Answer
Answer: c
Explanation: The H quantity refers to magnetic field intensity in the magnetic field. This is analogous to the electric field intensity E in the electric field.
Explanation: The H quantity refers to magnetic field intensity in the magnetic field. This is analogous to the electric field intensity E in the electric field.
2. The magnetic flux density is directly proportional to the magnetic field intensity. State True/False.
a) True
b) False
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a) True
b) False
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Answer: a
Explanation: The magnetic field intensity is directly proportional to the magnetic field intensity for a particular material (Permeability). It is given by B = μH.
Explanation: The magnetic field intensity is directly proportional to the magnetic field intensity for a particular material (Permeability). It is given by B = μH.
3. Ampere law states that,
a) Divergence of H is same as the flux
b) Curl of D is same as the current
c) Divergence of E is zero
d) Curl of H is same as the current density
View Answer
a) Divergence of H is same as the flux
b) Curl of D is same as the current
c) Divergence of E is zero
d) Curl of H is same as the current density
View Answer
Answer: d
Explanation: Ampere circuital law or Ampere law states that the closed integral of the magnetic field intensity is same as the current enclosed by it. It is given by Curl(H) = J.
Explanation: Ampere circuital law or Ampere law states that the closed integral of the magnetic field intensity is same as the current enclosed by it. It is given by Curl(H) = J.
4. Given the magnetic field is 2.4 units. Find the flux density in air(in 10-6 order).
a) 2
b) 3
c) 4
d) 5
View Answer
a) 2
b) 3
c) 4
d) 5
View Answer
Answer: b
Explanation: We know that B = μH. On substituting μ = 4π x 10-7 and H = 2.4, we get B = 4π x 10-7 x 2.4 = 3 x 10-6 units.
Explanation: We know that B = μH. On substituting μ = 4π x 10-7 and H = 2.4, we get B = 4π x 10-7 x 2.4 = 3 x 10-6 units.
5. Find the electric field when the magnetic field is given by 2sin t in air.
a) 8π x 10-7 cos t
b) 4π x 10-7 sin t
c) -8π x 10-7 cos t
d) -4π x 10-7 sin t
View Answer
a) 8π x 10-7 cos t
b) 4π x 10-7 sin t
c) -8π x 10-7 cos t
d) -4π x 10-7 sin t
View Answer
Answer: a
Explanation: Given H = 2sin t. We get B = μH = 4π x 10-7 x 2sin t = 8πx10-7sin t.
To get E, integrate B with respect to time, we get 8πx10-7cos t.
Explanation: Given H = 2sin t. We get B = μH = 4π x 10-7 x 2sin t = 8πx10-7sin t.
To get E, integrate B with respect to time, we get 8πx10-7cos t.
6. Find the height of an infinitely long conductor from point P which is carrying current of 6.28A and field intensity is 0.5 units.
a) 0.5
b) 2
c) 6.28
d) 1
View Answer
a) 0.5
b) 2
c) 6.28
d) 1
View Answer
Answer: b
Explanation: The magnetic field intensity of an infinitely long conductor is given by H = I/2πh. Put I = 6.28 and H = 0.5, we get h = 1/0.5 = 2 units.
Explanation: The magnetic field intensity of an infinitely long conductor is given by H = I/2πh. Put I = 6.28 and H = 0.5, we get h = 1/0.5 = 2 units.
7. Find the magnetic field intensity due to a solenoid of length 12cm having 30 turns and current of 1.5A.
a) 250
b) 325
c) 175
d) 375
View Answer
a) 250
b) 325
c) 175
d) 375
View Answer
Answer: d
Explanation: The magnetic field intensity of a solenoid is given by H = NI/L = 30 X 1.5/0.12 = 375 units.
Explanation: The magnetic field intensity of a solenoid is given by H = NI/L = 30 X 1.5/0.12 = 375 units.
8. Find the magnetic field intensity at the radius of 6cm of a coaxial cable with inner and outer radii are 1.5cm and 4cm respectively. The current flowing is 2A.
a) 2.73
b) 3.5
c) 0
d) 1.25
View Answer
a) 2.73
b) 3.5
c) 0
d) 1.25
View Answer
Answer: c
Explanation: The inner radius is 1.5cm and the outer radius is 4cm. It is clear that the magnetic field intensity needs to be calculated outside of the conductor ie, r>4cm. This will lead to zero, since H outside the conductor will be zero.
Explanation: The inner radius is 1.5cm and the outer radius is 4cm. It is clear that the magnetic field intensity needs to be calculated outside of the conductor ie, r>4cm. This will lead to zero, since H outside the conductor will be zero.
9. Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A.
a) 103.45
b) 102
c) 105.7
d) 171
View Answer
a) 103.45
b) 102
c) 105.7
d) 171
View Answer
Answer: a
Explanation: The magnetic field intensity of a toroid is given by H = NI/2πrm. Put N = 40, I = 3.25 and rm = 0.2, we get H = 40 x 3.25/2π x 0.2 = 103.45 units.
Explanation: The magnetic field intensity of a toroid is given by H = NI/2πrm. Put N = 40, I = 3.25 and rm = 0.2, we get H = 40 x 3.25/2π x 0.2 = 103.45 units.
10. The magnetic field intensity of an infinite sheet of charge with charge density 36.5 units in air will be
a) 18.25
b) 11.25
c) 73
d) 1/36.5
View Answer
a) 18.25
b) 11.25
c) 73
d) 1/36.5
View Answer
Answer: a
Explanation: The magnetic field intensity of an infinite sheet of charge is given by H = 0.5 K, for the point above the sheet and –0.5 K, for the point below the sheet. Here k is the charge density. Thus H = 0.5 x 36.5 = 18.25 units.
Explanation: The magnetic field intensity of an infinite sheet of charge is given by H = 0.5 K, for the point above the sheet and –0.5 K, for the point below the sheet. Here k is the charge density. Thus H = 0.5 x 36.5 = 18.25 units.
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