Electromagnetic Theory Questions and Answers – Magnetic Field Density

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Magnetic Field Density”.

1. Identify which of the following is the unit of magnetic flux density?
a) Weber
b) Weber/m
c) Tesla
d) Weber-1
View Answer

Answer: c
Explanation: The unit of magnetic flux density is weber/m2. It is also called as tesla.
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2. The divergence of H will be
a) 1
b) -1
c) ∞
d) 0
View Answer

Answer: d
Explanation: We know that the divergence of B is zero. Also B = μH. Thus divergence of H is also zero.

3. Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2 units.
a) 23.4
b) 12.3
c) 32.4
d) 21.3
View Answer

Answer: a
Explanation: The total flux is given by φ = ∫ B.ds, where ∫ds is the area. Thus φ = BA. We get φ = 11.7 x 2 = 23.4 units.
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4. Find the current when the magnetic field intensity is given by 2L and L varies as 0->1.
a) 2
b) 1
c) 0.5
d) 0
View Answer

Answer: b
Explanation: From Ampere law, we get ∫ H.dL = I. Put H = 2L and L = 0->1. On integrating H with respect to L, the current will be 1A.

5. Find the magnetic field intensity when the flux density is 8 x 10-6 Tesla in the medium of air.
a) 6.36
b) 3.66
c) 6.63
d) 3.36
View Answer

Answer: a
Explanation: We how that, B = μH. To get H = B/μ, put B = 8 x 10-6 and μ = 4π x 10-7. Thus H = 8 x 10-6/ 4π x 10-7 = 6.36 units.
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6. If ∫ H.dL = 0, then which statement will be true?
a) E = -Grad(V)
b) B = -Grad(D)
c) H = -Grad(Vm)
d) D = -Grad(A)
View Answer

Answer: c
Explanation: The given condition shows that the magnetic field intensity will be the negative gradient of the magnetic vector potential.

7. Find the magnetic flux density of the material with magnetic vector potential A = y i + z j + x k.
a) i + j + k
b) –i – j – k
c) –i-j
d) –i-k
View Answer

Answer: b
Explanation: The magnetic flux density is the curl of the magnetic vector potential. B = Curl(A). Thus Curl(A) = i(-1) – j(1) + k(-1) = -i – j – k. We get B = -i – j – k.
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8. Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm.
a) 178.33
b) 186.67
c) 192.67
d) 124.33
View Answer

Answer: b
Explanation: The total flux is the product of the magnetic flux density and the area. Total flux = B x A. To get B, put flux/area. B = 28/0.15 = 186.67 units.

9. Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.
a) ∫(4sin z i – ex j – 3cos y k)dt
b) -∫(4sin z i – ex j – 3cos y k)dt
c) ∫(4sin y i – ex j + 3cos y k)dt
d) -∫(4sin y i + ex j + 3cos y k)dt
View Answer

Answer: b
Explanation: We know that Curl (E) = -dB/dt. The curl of E is (4sin z i – ex j – 3cos y k). To get B, integrate the -curl(E) with respect to time to get B = -∫(4sin z i – ex j – 3cos y k)dt.
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10. Find current density J when B = 50 x 10-6 units and area dS is 4 units.
a) 9.94
b) 8.97
c) 7.92
d) 10.21
View Answer

Answer: a
Explanation: To get H, H = B/μ = 50 x 10-6/ 4π x 10-7 = 39.78 units. Also H = ∫ J.dS, where H = 39.78 and ∫ dS = 4. Thus J = 39.78/4 = 9.94 units.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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