<p>A small helium-neonlaser emits red visible light with a power of 2.60 mW in a

Question

<p>A small helium-neonlaser emits red visible light with a power of 2.60 mW in a beam that has a diameter of 2.90 mm.<br /><br />*I found the amplitude of the electric field of the light, E = 790 V/m<br />*I also found the amplitude of the magnetic field of the light, B = 2.63 &#956;T</p>
<p><br />Part C<br />What is the average energy density associated with the electric field?<br /> u = &#160; &#160; &#160; &#160; J/m^3</p>
<p>&#160;</p>
<p>Part D</p>
<p>What is the average energy density associated with the magnetic field?</p>
<p>u = &#160; &#160; &#160; &#160; &#160; J/m^3</p>
<p>&#160;</p>
<p>A brief explanation and steps would be awesome, thanks!!</p>

Explanation / Answer

Solution: part-c: Average energy density associated with electric field: From part-(a), Eo = 790 V/m E(t) = Eo cos t [E(t)]^2 = Eo^2 cos^2 t Time average of cos^t = 1/2 So, Uele = (1/2) o E^2          = (1/4) o Eo^2          = (1/4) * 8.854 x 10^-12 * (790)^2          = 1.34 x 10^-6 J/m^3 Ans: Energy density associated with electric field: Uele = 1.34 x 10^-6 J/m^3 part-d: Average energy density associated with magnetic field: From part-b, Bo = 2.63 x 10^-6 T Average energy density associated with magnetic field: B = ( 1 / 2o ) B^2     = ( 1 / 4o ) Bo^2     = [ 1 / ( 4 * 4 x 10^-7 ) ] * (2.63 x 10^-6)^2     = 1.34 x 10^-6 J/m^3 Ans: Average energy density associated with magnetic field: Bmag = 1.34 x 10-6 J/m^3 Note: Average energy density of electric and magnetic fields are equal to each other.         

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