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svetlana [45]
3 years ago
12

If the radiant energy from the Sun comes in as a plane EM wave of intensity 1340 W/m2, calculate the peak values of E and B

Physics
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

Given that

I = 1340 W/m²

µo = 4πx 10^-7 Tm/A

sc = 3 x 10^8 m/s

So to find the peak values of B we use

I = (c Bm²) / (2 µo)

1340 = (3 x 10^8 x Bm²) / (2 x 4πx 10^-7)

1340 = (3 x 10^8 x Bm²) / (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 1340 x (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 3.366 x 10^-3

Bm² = 3.366 x 10^-3 / 3 x 10^8

Bm² = 1.122 x 10^-11

Bm = 3.4 x 10^-6 T

Also to find the peak values of E we

use Em = c x Bm

Em = (3 x 10^8) (3.4 x 10^-6)

Em = 1.005 N/C

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You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what
pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

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⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

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