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3 years ago

If the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave

Physics

1 answer:

tangare [24]3 years ago

4 0

CHECK THE COMPLETE QUESTION BELOW;

the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave? (c=3.0×108m/s, μ0=4π×10−7T⋅m/A, ε0=8.85×10−12C2/N⋅m2)

Answer:

2.4×10^5 N/C

Explanation:

the amplitude can be explained as the maximum field strength of the electric and magnetic fields. Wave energy is proportional to its amplitude squared (E2 or B2).

We were told to calculate the amplitude of the magnetic field, which can be done using expresion below

S=ε²/2cμ

Where S is the intensity intensity of an electromagnetic wave given as 80 MW/m2

ε² is the Amplitude which we are looking for

c= speed of light given as 3×10^8m/s

Substitute the values into above formula we have,

S=ε²/2cμ

Making Amplitude subject of formula

ε²=S×2cμ

ε²=[80×10^6)(2×3×10^8)(4Π×10^-7)

= 245598.44

ε²=2.4×10^5 N/C

Therefore, amplitude of the magnetic field of this wave is S=2.4×10^5 N/C

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Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

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thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

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the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

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$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

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