Answer:
4.43 kW
Explanation:
Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.
P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀
= 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A
= 4428.1 W
= 4.4281 kW ≅ 4.43 kW
Answer: (a) and (b) => check attached file.
(c). Picture (a) and (b) will both remain the same.
Explanation:
IMPORTANT: The solution to the question (a) and (b) that is (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.
It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.
Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.
Hence, the solution to option C is given below;
(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.
Do you know what lesson this is ? i can probably help
Answer:
The peak emf in second coil is 1.876 V
Explanation:
Given :
Inductance
H
The current 
We compare above equation with standard equation,

From above equation we have,
and 
Find the inductive resistance,



The peak emf in second coil is,


V
Therefore, the peak emf in second coil is 1.876 V