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?
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.
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
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).
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
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.
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.
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.
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.
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
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.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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