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High power lasers in factories are used to cut through cloth and metal. One such laser has a beam diameter of 0.863 mm and generates an electric field at the target having an amplitude 0.955 MV/m. The speed of light is 2.99792 × 10⁸ m/s the permeability of free space is 4π × 10⁻⁷ T· N/A. What is the amplitude of the magnetic field produced? Answer in units of T.

User Wade
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2 Answers

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Answer:

I= 12.09×10^8 W/m^2

Step-by-step explanation:

Em = amplitude of electric field = 0.955 MV/m = 0.955 x 10^6 V/m

B_m = amplitude of magnetic field = ?

c = speed of light = 2.99792 x 10^8 m/s

amplitude of magnetic field is given as

B_m = E_m /c

B_m = (0.955 x 10^6)/(2.99792 x 10^8 )

B_m = 0.00318 T

b)

intensity is given as


I=(0.5)\epsilon* E_m^2 c


I=0.5(8.85*10^(-12))(0.955*10^6)^2(2.99792*10^8)

I= 1209873759.69

I= 12.09×10^8 W/m^2

User Huzzah
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3 votes

Answer:

B_m = 3.186 x 10⁻³ T

Step-by-step explanation:

given,

diameter of the beam = 0.863 mm

Amplitude = 0.955 MV/m

speed of light = 2.99792 × 10⁸ m/s

he permeability of free space= 4π × 10⁻⁷ T· N/A

Amplitude of the magnetic field is given by


B_m = (E_m)/(c)

E_m is amplitude of the electric field.


B_m = (0.955* 10^(6))/(2.99792* 10^8)

B_m = 0.31855 x 10⁻² T

B_m = 3.186 x 10⁻³ T

The amplitude of magnetic field is equal to 3.186 x 10⁻³ T

User Vhurryharry
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