Question

What is the total energy transported per hour along a narrow cylindrical laser beam 1.80 mm in diameter, whose B-field has an rms strength of

Answer 1

 

Am assuming the  B-field has an rms strength of 1.10×10−10T which was omitted

Answer:  E=1.686 x10⁻⁸J/hr

Explanation:

Intensity for electromagnetic wave is given as

I= cε₀(Emax)²....... equation 1

Also

B=Emax/c   ....... equation 2

puting the value of B in equation 2 in equation 1 becomes

I= cε₀(Emax)²

I= c³ε₀B²

where

c = speed of light = 3 X 10⁸m/s

ε₀=permittivity of free space = 8.85x 10-12 F m-1

B=1.10×10−10T

I= (3 X 10⁸m/s)³ X (8.85 X 10⁻12F/m) x(1.10×10−10T)²

I=2.891 X 10⁻⁶ W/m²

Energy per unit hour

E= IAt

= 2.891 X 10⁻⁶ W/m² x πr²x 3600

2.891 X 10⁻⁶ W/m² x 3.142 x ( 1.80 x 10⁻³/2) x(3600/1hr)

E=1.686 x10⁻⁸J/hr

Answer 2

The total energy transported per hour will be 6.62 × 10⁻⁹J

Calculation of energy transported:

The Intensity for an electromagnetic wave is given by:

I = cε₀E₀²

where c is the speed of light, and

E₀ is the maximum value of the electric vector

Also

B = E/c  

where B is the magnetic field

Therefore:

I= cε₀E₀²

I= c³ε₀B²

Given that the magnetic field is:

B = 1.10×10⁻¹⁰ T

I= (3 X 10⁸m/s)³ X (8.85 X 10⁻12F/m) x(1.10×10−10T)²

I=2.891 X 10⁻⁶ W/m²

Energy transported per unit hour

E= IAt

where A is the area

and t is the time = 1hr = 3600s

E = 2.891 × 10⁻⁶  × πr² × 3600

E = 2.891 X 10⁻⁶ ×  3.14 × ( 1.80 x 10⁻³/4)² x (3600)

E = 6.62 × 10⁻⁹J

Learn more about electromagnetic waves:

https://brainly.com/question/13803241?referrer=searchResults