Electromagnetic Theory Questions and Answers – Real Time Applications

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Real Time Applications”.

1. Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.
a) 35.36pF
b) 3.536pF
c) 35.36nF
d) 3.536nF
View Answer

Answer: a
Explanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 = 35.36pF.

2. The resistance of a material with conductivity 2millimho/m2, length 10m and area 50m is
a) 500
b) 200
c) 100
d) 1000
View Answer

Answer: c
Explanation: The resistance is given by, R = ρL/A, where ρ is the resistivity, the inverse of conductivity. R = 10/(0.002 X 50) = 100 ohm.

3. Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.
a) 131.94mH
b) 94.131mH
c) 131.94H
d) 94.131H
View Answer

Answer: a
Explanation: The inductance is given by L = μ N2A/l, where μ= μoμr is the permeability of air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10-7 X 1002 X 6/2 = 131.94mH.
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4. Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.
a) 400
b) 300
c) 200
d) 100
View Answer

Answer: d
Explanation: The current density is given by J = σ E, where σ is the conductivity. Thus resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.

5. Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.
a) 1A
b) 10mA
c) 10A
d) 100mA
View Answer

Answer: a
Explanation: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.

6. In electric fields, D= ε E. The correct expression which is analogous in magnetic fields will be
a) H = μ B
b) B = μ H
c) A = μ B
d) H = μ A
View Answer

Answer: b
Explanation: In electric fields, the flux density is a product of permittivity and field intensity. Similarly, for magnetic fields, the magnetic flux density is the product of permeability and magnetic field intensity, given by B= μ H.

7. Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.
a) 60
b) 120
c) 180
d) 200
View Answer

Answer: b
Explanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L = 12. Force F = 20 X 0.5 x 12 = 120 N.
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8. From the formula F = qE, can prove that work done is a product of force and displacement. State True/False
a) True
b) False
View Answer

Answer: a
Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.

9. Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.
a) 10
b) 20
c) 40
d) 80
View Answer

Answer: b
Explanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.
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10. Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.
a) 100kJ
b) 250kJ
c) 30kJ
d) 15kJ
View Answer

Answer: c
Explanation: Power is given by, P= V X I, where I = J X A is the current.
Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.

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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