This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Poynting Theorem”.

1. Unit of Poynting Vector is _____________

a) Watt

b) Watt/s

c) Watt/m

d) Watt/m^{2}

View Answer

Explanation: Poynting Vector represents the energy moving out per unit area per unit time. Thus, it’s unit is W/m

^{2}. It is a vector quantity.

2. The energy transported by the fields per unit time per unit are is called __________

a) Poynting Energy

b) Electro-magnetic energy

c) Poynting vector

d) Flux density

View Answer

Explanation: The work done on the charges by an electromagnetic field is equal to the decrease in energy stored in the field less the energy that flowed out through the surface. This energy, that is transported by the fields, per unit are per unit time is called Poynting vector.

3. The direction of Poynting vector is perpendicular to the direction of propagation of wave.

a) True

b) False

View Answer

Explanation: The Poynting vector is proportional to the cross product of Electric and magnetic field, E X B. Therefore, its direction is perpendicular to Electric and Magnetic waves, i.e., in the direction of propagation of wave.

4. According to the Poynting theorem, the energy flow per unit time out of any closed surface is ___________

a) Integral of S over the length of the surface

b) Integral of S over the are of the surface

c) Differential of S over the length of the surface

d) Differential of S over the are of the surface

View Answer

Explanation: According to the pointing theorem, the energy flow per unit time out of any closed surface is the integral of S over the surface i.e., P = ∫S.da.

5. The correct expression for the Poynting vector is __________

a) **S = E X B**

b) **S = E X B**/2

c) **S = E X B**/μ_{o}

d) **S = E X B**/2μ_{o}

View Answer

Explanation: Poynting vector can be defined as the rate at which the energy is carried out of the volume across the bounding surface. It is always in the direction of the propagation of wave, as it is perpendicular to both electric and magnetic field.

6. In free space, **E**(z,t) = 10^{3}sin(ωt – βz) \(\hat{y}\). Obtain **H**(z,t).

a) 1.23 sin (ωt – βz) (-\(\hat{y}\))

b) 1.23 sin(ωt – βz) (-\(\hat{x}\))

c) 2.65 sin (ωt – βz) (-\(\hat{y}\))

d) 2.65 sin(ωt – βz) (-\(\hat{x}\))

View Answer

Explanation: In the following case, the wave is propagating in z direction. Therefore, direction of

**S**is in z direction.

We know,

**S = E X B**. As E is in +ve y direction, B should be in negative x direction,

Also we know, \(\mid E\mid /\mid H\mid\) = 376.6

Therefore, \(\mid H\mid\) = 10

^{3}/376.6

\( \mid H\mid\) = 2.65

Hence,

**H**(z,t) = 2.65 sin(wt – az) – \(\hat{x}\).

7. Earth receives 2 cal/min/cm^{2} of solar energy. What is the value of \mid H\mid?

a) 1.98 A/m

b) 2.45 A/m

c) 3.75 A/m

d) 4.13 A/m

View Answer

Explanation: We know, S = E X H. Here E and H would be in perpendicular direction.

Therefore, S = \(\mid E\mid \mid H\mid\)

Now, given S = 2 cal/min/cm

^{2}

= 1400 J/sm

^{2}

We know, \(\mid E\mid /\mid H\mid\) = 376.6

Therefore, \(\mid E\mid = 376.6 \mid H\mid\)

Hence, 1400 = 376.6 \(\mid H\mid^2\)

\(\mid H\mid^2\)= 3.71

\(\mid H\mid\) = 1.98 A/m.

8. In an electromagnetic wave, the electric field of amplitude 4 V/m is oscillating. The Energy density of the wave is ___________

a) 1.41 X 10^{-10} J/m^{3}

b) 2.41 X 10^{-10} J/m^{3}

c) 3.41 X 10^{-10} J/m^{3}

d) 4.41 X 10^{-10} J/m^{3}

View Answer

Explanation: We know, Energy density = ε E

^{2}

Here, E

_{max}= 4 V/m and ε = 8.85 X 10

^{-12}C

^{2}/Nm

^{2}

Therefore, Energy Density = 8.85 X 10

^{-12}X 4 X 4

= 1.41 X 10

^{-10}J/m

^{3}.

9. The amplitude of the electric field in a parallel beam of light of intensity 3 W/m^{2} is __________

a) 23.4 N/C

b) 34.5 N/C

c) 47.53 N/C

d) 51.45 N/C

View Answer

Explanation: We know, Intensity, I = εE

^{2}c/2

Now, E

_{max}= \(\sqrt{\frac{2I}{ε_0c}}\)

Here, I = 3 W/m

^{2}, ε = 8.85 X 10

^{-12}C

^{2}/Nm

^{2}and c = 3 X 10

^{8}m/s

We get, E

_{max}= 47.53 N/c.

10. What will be the direction of Poynting vector?

a) x direction

b) y direction

c) z direction

d) -z direction

View Answer

Explanation: As we can see, the electric field is in +ve Y direction while the magnetic field is in the +ve X direction. Also, S = E X B/μ. Hence, S is in the negative Z direction.

11. Poynting vector gives the energy flow per unit area per unit time through a cross-sectional area along the direction of propagation of the wave.

a) True

b) False

View Answer

Explanation: The Poynting vector is in the direction of the propagation of the wave. However, the cross-sectional area through which the energy is flowing out is perpendicular to the direction of propagation of wave.

12. The magnitude of average value of S at a point is called as __________

a) Amplitude of radiation

b) Frequency of radiation

c) Intensity of radiation

d) Momentum flow

View Answer

Explanation: As the frequencies of electromagnetic waves are very high, the time variation of the Poynting vector is so rapid that we have to deal with the average value. The magnitude of average value of S at a point is called the intensity of the radiation at that point.

13. The radiation pressure is given by __________

a) S

b) Sav

c) S/c

d) Sav/c

View Answer

Explanation: The momentum is transferred per unit surface area per unit time. This momentum transfer is responsible for radiation pressure which is given by Sav/c.

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