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kirza4 [7]
3 years ago
7

The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e

lectromagnetic waves at the top of the atmosphere.
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

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Answer:

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Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

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51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

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