Electromagnetic Waves

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} l o g_{e} (\frac{k t}{β t});$ where α and β are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of will be

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$P = \frac{α}{β} l o g_{e} (\frac{k t}{β x})$

$\frac{k t}{β x}$ = Dimensionless

$β = \frac{k t}{x} = \frac{[M L^{2} T^{- 2} k^{- 1}] [k]}{[L]}$

$\frac{α}{β} =$ dimensionless

= dimensionless of

= MLT²

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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)

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | 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

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

When a car is approaching the observer, the frequency of horn is 100Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be $\frac{x}{3}$ Ha where x = _________.

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$f_{1} = 100 = f o (\frac{C}{C - V_{S}})$

C = Speed of sound

Vs = Speed of source

f₂ = 50 = fo = $(\frac{C}{C + V_{s}})$

$\frac{f_{1}}{f_{2}} = 2 = \frac{C + V_{S}}{C - V_{S}}$

$f o = \frac{200}{3} = \frac{x}{3}$

x = 200

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

The oscillating magnetic field in a plane electromagnetic wave is given by $B_{y} = 5 \times 1 0^{- 6} s i n 100 π (5 x - 4 \times 1 0^{8} t) T .$ The amplitude of electric field will be:

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$B_{0} = 5 \times 1 0^{- 6}$

Speed of wave V = $\frac{4 \times 1 0^{8}}{5} = 8 \times 1 0^{7}$

$E ? = V \times B ? = 4 ? \times 10 = 4 \times 1 0^{2} V / m$

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

The magnetic field of a plane electromagnetic wave is given by:

$\vec{B} = 2 \times 1 0^{- 8}$ sin(0.5 × 10³x + 1.5 × 10¹¹t) $\hat{j}$ T.

The amplitude of the electric field would be:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$S p e e d o f l i g h t C = \frac{w}{k} = \frac{1.5 \times 1 0^{11}}{0.5 \times 1 0^{3}} = 3 \times 1 0^{8} m / s$

So, E₀ = B₀ C

= 2 × 10^-8 × 3 × 10⁸

= 6v/m

Direction will be along z – axis

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Sun light falls normally on a surface of area 36cm² and exerts an average force of 7.2 × 10^-⁹N within a time period of 20 minutes. Considering a case of complete adsorption, the energy flux of incident light is

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

A = 36 cm²= 36 × 104 m²

$? F ? = 7.2 \times 1 0^{- 9} N, t = 20 m i n .$

$? F ? = \frac{Δ P}{Δ t} = \frac{E}{C Δ t} \Rightarrow \frac{E}{Δ t} = ? F ? c \Rightarrow \frac{E}{A Δ t} = \frac{? F ? c}{A}$

= 0.06 W/cm²

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Kindly go through the solution

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

A radar sends an electromagnetic single of electric (E₀) = 2.25 V/m and magnetic field (B₀) = 1.5 × 10^-⁸ T which strikes a target on line of sight at a distance of 3km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?

11 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$E_{0} = 2.25 \frac{v}{m}$ Speed of declromagnetic signal, $v = \frac{E_{0}}{B_{0}} = \frac{2.25 v / m}{1.5 \times 1 0^{- 8} T}$

$B_{0} = 1.5 \times 1 0^{- 8} T \Rightarrow v = \frac{225}{150} \times 1 0^{8} = 1.5 \times 1 0^{8} m / s$

Time to reach each to the same

radar, $t = \frac{2 \times 3 \times 1 0^{3} m}{1.5 \times 1 0^{8} m / s} = 4 \times 1 0^{- 5} S$

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New answer posted•a year ago

Electromagnetic Waves Physics NCERT Exemplar Solutions Class 12th Chapter Eight

If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a dept ah(h <<R_e) from the earth surface. The value of a will be_______.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$g_{(a t h)} = g_{(a t d e p t h α h)}$ h << R

$\Rightarrow g (1 - \frac{2 h}{R}) = g (1 - \frac{α h}{R})$

$\begin{array}{l} 1 - \frac{2 h}{R} = 1 - \frac{α h}{R} \Rightarrow \frac{2 h}{R} = α \frac{h}{R} \\ α = 2 \end{array}$

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