# Electromagnetic Theory Questions and Answers – Power and Poynting Vector

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Power and Poynting Vector”.

1. The total power of a wave with average power 15 units in a surface density of 0.5 units is
a) 15
b) 30
c) 7.5
d) 0.75

Explanation: The total power is given by the surface integral of the average power. Thus ∫Pavg ds is the total power. On substituting for Pavg = 15 and ∫ds = 0.5, we get total power as 7.5 units.

2. The power of a wave with electric field intensity of 3 units in air is
a) 0.01
b) 0.03
c) 0.05
d) 0.07

Explanation: The Poynting vector gives the power of a wave. It is given as P = E2/2η. On substituting for E = 3 and η = 377 in air, the power is P = 32/(2×377) = 0.01 units.

3. Find the power of an EM wave, given that the cross product of the E and H component is 2 + 3j.
a) 2
b) 1
c) 4
d) 8

Explanation: The Poynting power vector for complex quantity of E x H is P = 0.5 x Re(E x H). In the given data, Re(E x H) = 2, thus we get P = 0.5 x 2 = 1 unit.

4. The power in a electromagnetic wave with electric field and magnetic field intensities 12 and 8 respectively is
a) 96
b) 12
c) 8
d) 48

Explanation: The Poynting vector is given by P = 0.5 EH. Given that E = 12 and H = 8, we get P = 0.5 x 12 x 8 = 48 units.

5. The power in a wave given that H component is 0.82 units in air.
a) 126.74
b) 621.47
c) 216.47
d) 745.62

Explanation: The power of a wave is given by P = ηH2/2. In air medium, η = 377 and given that H = 0.82. We get power P = 377 x 0.822/2 = 126.74 units.

6. Find the power of a wave given that the RMS value of E and H are 6 and 4.5 respectively.
a) 24
b) 27
c) 29
d) 32

Explanation: The power is the product of the RMS electric field and the RMS magnetic field. Thus P = Erms X Hrms. On substituting Erms = 6 and Hrms = 4.5, the power is P = 6 x 4.5 = 27 units.

7. The electric and magnetic fields vary with time in which of the following fields?
a) DC
b) AC
c) Static
d) It does not vary with time

Explanation: The electric and magnetic fields vary with time in oscillating fields. It is certain that such fields are AC fields.

8. The power per unit velocity of a wave with electric field as 8 units and density 10 units is
a) 40
b) 20
c) 80
d) 160

Explanation: The power per unit velocity P/v is given by the product of electric field and the density. Thus P/v = E.d = 8 x 10 = 80 units.

9. The power of a wave in a cylindrical waveguide of radius 2m with electric field 12 units is
a) 2.39
b) 3.92
c) 9.23
d) 9.32

Explanation: The power of a wave is given by ∫P ds, where P = E2/2η and ∫ds = πr2. On substituting for E = 12, η = 377 in air and r = 2, we get P = 2.39 units.

10. The work done in the power transmission with E and H given by 50 and 65 respectively. The velocity of propagation is 20m/s.
a) 162.5
b) 621.5
c) 562.1
d) 261.5

Explanation: The work done is given by W = EH/v, where E = 50, H = 65 and v = 20. On substituting, we get W = 50 x 65/20 = 162.5 units.

11. The Poynting vector is the power component that is calculated by the
a) Product of E and H
b) Ratio of E and H
c) Dot product of E and H
d) Cross product of E and H

Explanation: The Poynting vector P is the cross product of the electric field E and the magnetic field H. It is given by P = E X H. It is used to calculate the power in wave propagation in transmission lines, waveguides and antenna.

12. The maximum power transceived by the antenna is in the region of
a) Aperture
b) Effective aperture
c) Maxima lobe
d) Minima lobe

Explanation: An antenna is a device that transmits and receives power. The transmitting and receiving is in the region called aperture. Maximum or efficient power transmission occurs in the region called effective aperture.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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