A monochromatic light source with power output 60.0W radiates light of wavelength 700 nm uniformly in all directions. Calculate Emax and Bmax for the 700-nm light at a distance of 5.00 m from the source.

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liannemdh · Accepted Answer

The expression for intensity is,  I=PA  Here, p is the power and A is the area.  The expression for amplitude of magnetic field is,  Bmax=Emaxc  Here, c is the speed of light and Emax is the amplitude of electric field.  The expression for intensity in terms of electric field.  I=12ϵ0cE{max}2  Here, ϵ0 is the permittivity of free space.  The expression for area is,  A=4πr2  Here, r is the radius.  Substitute 5.00 m for r in the above equation.  A=4π(5.00m)2  =314.2m2  The expression for intensity is,  I=PA  Substitute 482.80m2 for A and 60 W for p in above equation of intensity.  I=60 W(314.2 m2}  =0.19W⋅m−2  The expression for intensity in terms of electric field.  I=12ϵ0cEmax2  Substitute 0.12W⋅m−2 for I 8.85×10−12 for ϵ0 and 3×108m⋅s−1 for c and solve for Emax.  (0.19 W⋅m−2)=12(8.85×10−12)(3⋅108)E{max}2  0.191.32×10−3=E{max}2  Emax=143.94 Vm2  Emax=12.0 Vm  The expression for amplitude of magnetic field is,  Bmax=Emaxc  Substitute 9.50 V⋅m−1 for Emax and 3×108 m⋅s−1 for c  Bmax=12.0 V⋅m−13×108 m⋅s−1  =4.0×10−8 T  Therefore, the values of Emax and Bmax are 12.0 Vm and 4.0×10−8 T respectively.

A monochromatic light source with power output 60.0W radiates light of wavelength 700 nm uniformly in all directions. Calculate E_{max} and B_{max} for the 700-nm light at a distance of 5.00 m from the source.

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