Engineering Physics Multiple Choice Questions on “Poynting Theorem”.
1. Unit of Poynting Vector is _____________
a) Watt
b) Watt/s
c) Watt/m
d) Watt/m2
Answer: d
Clarification: Poynting Vector represents the energy moving out per unit area per unit time. Thus, it’s unit is W/m2. 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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/μo
d) S = E X B/2μo
Answer: c
Clarification: 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) = 103sin(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}))
Answer: d
Clarification: 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 Emid /mid Hmid) = 376.6
Therefore, (mid Hmid) = 103/376.6
( mid Hmid) = 2.65
Hence, H(z,t) = 2.65 sin(wt – az) – (hat{x}).
7. Earth receives 2 cal/min/cm2 of solar energy. What is the value of mid Hmid?
a) 1.98 A/m
b) 2.45 A/m
c) 3.75 A/m
d) 4.13 A/m
Answer: a
Clarification: We know, S = E X H. Here E and H would be in perpendicular direction.
Therefore, S = (mid Emid mid Hmid)
Now, given S = 2 cal/min/cm2
= 1400 J/sm2
We know, (mid Emid /mid Hmid) = 376.6
Therefore, (mid Emid = 376.6 mid Hmid)
Hence, 1400 = 376.6 (mid Hmid^2)
(mid Hmid^2)= 3.71
(mid Hmid) = 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/m3
b) 2.41 X 10-10 J/m3
c) 3.41 X 10-10 J/m3
d) 4.41 X 10-10 J/m3
Answer: a
Clarification: We know, Energy density = ε E2
Here, Emax = 4 V/m and ε = 8.85 X 10-12 C2/Nm2
Therefore, Energy Density = 8.85 X 10-12 X 4 X 4
= 1.41 X 10-10 J/m3.
9. The amplitude of the electric field in a parallel beam of light of intensity 3 W/m2 is __________
a) 23.4 N/C
b) 34.5 N/C
c) 47.53 N/C
d) 51.45 N/C
Answer: c
Clarification: We know, Intensity, I = εE2c/2
Now, Emax = (sqrt{frac{2I}{ε_0c}})
Here, I = 3 W/m2, ε = 8.85 X 10-12 C2/Nm2 and c = 3 X 108 m/s
We get, Emax = 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: d
Clarification: 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.