Question: In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is 5× 1014Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joulem3, will be
Options:
| A. | 1.1× 10−11 | |
| B. | 2.2× 10−12 | |
| C. | 3.3× 10−13 | |
| D. | 4.4× 10−14 |
Answer: Option B
: B
Average energy density of electric field is given by
ue=12ϵ0E2=12ϵ0(E0√2)2=14ϵ0E20
=14×8.85×10−12(1)2=2.2×10−12J/m3.
: B
Average energy density of electric field is given by
ue=12ϵ0E2=12ϵ0(E0√2)2=14ϵ0E20
=14×8.85×10−12(1)2=2.2×10−12J/m3.
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Question 1. A TV tower has a height of 100 m. The average population density around the tower is 1000 per km2. The radius of the earth is 6.4× 106m. The population covered by the tower is
- 2× 106
- 3× 106
- 4× 106
- 6× 106
Answer: Option C
: C
Population covered =2πhR×Population density
=2π×100×6.4×106×1000(103)2=4×106
: C
Population covered =2πhR×Population density
=2π×100×6.4×106×1000(103)2=4×106
Question 4. A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity 5× 10−16 W/m2. The maximum instantaneous potential difference across the two ends of the antenna is
- 1.23 mV
- 1.23 mV
- 1.23 V
- 12.3 mV
Answer: Option A
: A
I=12ϵ0CE20
⇒E0=√2Iϵ0C=√2×5×10−168.85=0.61×10−6Vm
Alsop E0=V0d⇒V0=E0d=0.61×10−6×2=1.23μV
: A
I=12ϵ0CE20
⇒E0=√2Iϵ0C=√2×5×10−168.85=0.61×10−6Vm
Alsop E0=V0d⇒V0=E0d=0.61×10−6×2=1.23μV
- An e.m.f.
- Electric current
- Magnetic field
- Pressure radiant
Answer: Option C
: C
According to the Maxwell’s EM theory, the EM waves propagation contains electric and magnetic field vibration in mutually perpendicular direction. Thus the changing of electric field give rise to magnetic field.
: C
According to the Maxwell’s EM theory, the EM waves propagation contains electric and magnetic field vibration in mutually perpendicular direction. Thus the changing of electric field give rise to magnetic field.
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