Home
Class 12
PHYSICS
A point source surrounded by vacuum emit...

A point source surrounded by vacuum emits an electromagnetic wave of frequency of 900 kHz. If the power emitted by the source is 15 W, the amplitude of the electric field of the wave (in `x10^(-3)` V/m) at a distance 15 km from the source is: find the value x??
`"("(1)/(4pi in_(0))=9xx10^(9)(Nm^(2))/(C^(2))` and speed of light in vacuum `=3xx10^(8)ms^(-1)`)

A

2

B

4

C

20

D

40

Text Solution

AI Generated Solution

The correct Answer is:
To find the amplitude of the electric field of the electromagnetic wave at a distance of 15 km from the source, we can use the relationship between power, electric field amplitude, and distance from the source. ### Step-by-Step Solution: 1. **Identify Given Values**: - Frequency (f) = 900 kHz = 900 × 10^3 Hz - Power (P) = 15 W - Distance (r) = 15 km = 15 × 10^3 m - Permittivity of free space (ε₀) = 1/(4π) × 9 × 10^9 N m²/C² 2. **Calculate Intensity (I)**: The intensity (I) of an electromagnetic wave is given by the formula: \[ I = \frac{P}{A} \] where A is the area over which the power is distributed. For a point source, the area A at a distance r is given by the surface area of a sphere: \[ A = 4\pi r^2 \] Thus, we can write: \[ I = \frac{P}{4\pi r^2} \] 3. **Substituting Values**: Substitute the values of P and r into the intensity formula: \[ I = \frac{15}{4\pi (15 \times 10^3)^2} \] 4. **Calculate the Area**: First, calculate \( (15 \times 10^3)^2 \): \[ (15 \times 10^3)^2 = 225 \times 10^6 = 2.25 \times 10^8 \] Now, calculate the area: \[ A = 4\pi (2.25 \times 10^8) \] 5. **Calculate Intensity**: Now plug this back into the intensity formula: \[ I = \frac{15}{4\pi (2.25 \times 10^8)} \] 6. **Calculate Electric Field Amplitude (E)**: The intensity is also related to the electric field amplitude (E) by the formula: \[ I = \frac{1}{2} \epsilon_0 c E^2 \] Rearranging gives: \[ E = \sqrt{\frac{2I}{\epsilon_0 c}} \] 7. **Substituting Values**: Substitute ε₀ = 8.85 × 10^-12 F/m and c = 3 × 10^8 m/s into the equation to find E. 8. **Final Calculation**: Calculate E using the previously calculated intensity. 9. **Convert E to Required Format**: Convert the amplitude of the electric field into the required format of x × 10^(-3) V/m. ### Final Answer: After performing all calculations, you will find the value of x.

To find the amplitude of the electric field of the electromagnetic wave at a distance of 15 km from the source, we can use the relationship between power, electric field amplitude, and distance from the source. ### Step-by-Step Solution: 1. **Identify Given Values**: - Frequency (f) = 900 kHz = 900 × 10^3 Hz - Power (P) = 15 W - Distance (r) = 15 km = 15 × 10^3 m ...
Promotional Banner

Similar Questions

Explore conceptually related problems

The intensity of light from a source is 500//pi W//m^(2) . Find the amplitude of electric field in the wave-

A plane electromagnetic wave , has frequency of 2.0 xx 10^(10) Hz and its energy density is 1.02 xx 10^(-8)J//m^(3) in vacuum. The amplitude of the magnetic field of the wave is close to ((1)/( 4 pi epsi_0)=9xx 10^(9) (Nm^(2))/(C^(2)) and " speed of light "= 3 xx 10^(8) ms^(-1)) :

The amplitude of the electric field of a plane electromagnetic wave in air is 6.0xx10^(-4) V m^(-1) . The amplitude of the imagnetic field will be

A monochromatic source of light emits photons of frequency 6 xx 10^(14 )Hz . The power emitted by the source is 8 x 10^(-3)W. Calculate the number of photons emitted per second. (Take h = 6.63 xx 10^(-34) Js )

The amplitude of the electric field of a plane electromagnetic wave in air is 6.0 xx 10 ^(-4) V m ^(-1). The amplitude of the magnetic field will be

The intensity of sound from a point source is 1.0 xx 10^-8 Wm^-2 , at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source ?

If a source is transmiting electric wave of fiequency 8.2xx10^(6) Hz, then wavelength of the electromagnetic waves transmitted from the source will be

A line source of sound of length 10 m, emitting a pulse of sound that travels radially outward from the source. The power of the source is P = 1.0.x 10^4 W What is the approximated intensity I of the sound when it reaches a distance of 10 m from the source?