Electromagnetic Waves

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New answer posted•a year ago

Electromagnetic Waves Class 12th

An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in the this medium is_____× 10⁷ m/s.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$v = \frac{1}{\sqrt[]{ε μ}} = \frac{1}{\sqrt[]{2 ε_{0} . 2 μ_{0}}} = \frac{1}{2 \sqrt[]{ε_{0} μ_{0}}} = \frac{c}{2} = 15 \times 1 0^{7} m / s .$

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New answer posted•a year ago

Electromagnetic Waves Class 12th

The displacement current of 4.425 $μ A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10⁶ Vs^-1. The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is x × 10^-3. The value of x is,

(Permittivity of free space, $E_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2})__________$

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

According to definition of displacement current, we can write

$I_{d} = ε_{0} \frac{d ?}{d t} = ε_{0} \frac{d (E S)}{d t} = ε_{0} \frac{d (\frac{V}{l} S)}{d t} = \frac{ε_{0} S}{l} (\frac{d V}{d t})$

$\Rightarrow l = \frac{8.85 \times 1 0^{- 12} \times 40 \times 1 0^{- 4} \times 1 0^{6}}{4.425 \times 1 0^{- 6}} 8 \times 1 0^{- 3} m$

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New answer posted•a year ago

Electromagnetic Waves Class 12th

The intensity of the light from a bulb incident on a surface is 0.22 W/m². The amplitude of the magnetic field in this light – wave is__________ × 10^-9 T.

(Given : Permittivity of vacuum $\in_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} - m^{- 2},$ speed of light in vacuum c = 3 × 10⁸ ms^-1)

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{B_{0}^{2}}{2 μ_{0}} C$

$B_{0}^{2} = \frac{I 2 μ_{0}}{C}$

$B_{0}^{2} = \frac{0.22 \times 2 \times 4 \times 1 0^{- 7}}{9 \times 108}$

B₀ = 42.9 × 10^-9

42.9 Ans

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New answer posted•a year ago

Electromagnetic Waves Class 12th

An electromagnetic wave is represented by the electric field
E⃗ = E₀n̂[ωt + (y − z)]
The direction of propagation of the wave, ŝ is:

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

s? = -? + k? ∴ s? = (-? +k? )/√2

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New answer posted•a year ago

Electromagnetic Waves Class 12th

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 10^-3V/m) at a distance 15 km from the source is :
( (1 / 4πε₀) = 9×10⁹ Nm²/C² and speed of light in vacuum = 3 × 10⁸ ms^-1)

19 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

I = P/ (4πr²) . (i)
and I = (1/2)ε? E? ²c . (ii)
From (i) and (ii)
P/ (4πr²) = (1/2)ε? E? ²c
E? = √ (2P/ (4πε? r²c) ; E? = √ (9×10? ×2×15)/ (15×15×10? ×3×10? ) = 2×10? ³ V/m

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New answer posted•a year ago

Electromagnetic Waves Class 12th

A TV transmission tower antenna is at a height of 20m. Suppose that the receiving antenna is at
(i) ground level
(ii) a height of 5m
The increase in the antenna range in case (ii) relative to case (i) is n %.
The value of n, to the nearest integer, is.

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

d_lm = Distance covered by transmitting antenna + Distance covered by receiving antenna

⇒ d_lm = √ (2Rh_transmitting) + √ (2Rh_receiving)
⇒ d_lm = √ (2 × 64 × 10? × 20) + √ (2 × 64 × 10? × 5) = 16000 + 8000 = 24000m

When h_receiving = 0 then
d_2m = √ (2 × 64 × 10? × 20) = 16000m

% increment = (8000/16000

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New answer posted•a year ago

Electromagnetic Waves Class 12th

A plane electromagnetic wave propagating along y-direction, can have the following pair of electric field (E) and magnetic field (B) components :

1 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

As we know that direction of propagation of electromagnetic wave is perpendicular to plane containing mutually perpendicular electric field and magnetic field, so option D will be correct answer.

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New answer posted•a year ago

Electromagnetic Waves Class 12th

If 2.5×10^-9 N average force is exerted by a light wave on a non- reflecting surface of 30 cm² area during 40 minutes of time span, the energy flux of light just before it falls on the surface is . W / cm². (Round off to the Nearest Integer) (Assume complete absorption and normal incidence conditions are there)

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The force (F) exerted by radiation is the rate of change of momentum (p).
F = Δp/Δt
For photons, p = E/c. So, F = (1/c) * (ΔE/Δt).
Since Power (P) is ΔE/Δt, F = P/c.
Intensity (I) is Power per unit Area (P/A).
The formula provided in the document is F/A = (nE)/ (Δt c A) which leads to a final calculated

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New answer posted•a year ago

Electromagnetic Waves Class 12th