Ask & Answer Question

Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight Electromagnetic Waves

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?

Option 1 -

2.0 * 10^-⁵s

Option 2 -

4.0 * 10^-⁵ s

Option 3 -

1.0 * 10^-⁵ s

Option 4 -

8.0 * 10^-⁵ s

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$E_{0} = 2.25 \frac{v}{m}$ 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, $t = \frac{2 \times 3 \times 1 0^{3} m}{1.5 \times 1 0^{8} m / s} = 4 \times 1 0^{- 5} S$

0 Upvotes 0 Downvotes

Similar Questions for you

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.

Vishal Baghel

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

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

Vishal Baghel

According to definition of displacement current, we can write

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

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

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)

Vishal Baghel

$\frac{B_{0}^{2}}{2 μ_{0}} C$

$B_{0}^{2} = \frac{I 2 μ_{0}}{C}$

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

B₀ = 42.9 × 10^-9

42.9 Ans

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

Vishal Baghel

s? = -? + k? ∴ s? = (-? +k? )/√2

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)

Vishal Baghel

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

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question