The energy flow to the earth from sunlight is about 1.4kWm2 . (a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity. (b) The distance from the earth to the sun is about 1.5×1011m. Find the total power radiated by the sun.

Question

soniarus7x · Accepted Answer

Step 1  (a) The given value represents the intensity I=1.4kWm2 of the light. The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude Emax and the amplitude of magnetic field Bmax and it is given by equation (32.29) in the form    I=12ϵocEmax2   (1)    Where ϵo is the electric constant, cc is the speed of light. Solve equation (1) for Emax    Emax=2Iϵoc   (2)    Plug the values for I, c, and ϵo into equation (2) to get Emax    Emax=2Iϵoc    =2(1400Wm2)(8.85×10−12C2N⋅m2)(3×108ms)    =1026Vm    The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation (32.4) in the form    Bmax=Emaxc    Where c is the speed of the light. Now, plug the values for Emax and c to get Bmax    Bmax=Emaxc=1026Vm3×108ms=3.42×10−6T

Lounctirough · Accepted Answer

Step 2&amp;nbsp;(b) The strength I is inversely proportional to&amp;nbsp;Emax2&amp;nbsp;Furthermore, it stands for the incident power P per area A.P=IA&amp;nbsp;(3)&amp;nbsp;The separation between Earth and the Sun corresponds to the sphere-like radius of the Earth&#039;s orbital route around the Sun. Consequently, the area of this path is determined byA=4πr2=4π(1.5×1011m)2=2.82×1023m2&amp;nbsp;Now, plug the values for P and A into equation (2) to get I&amp;nbsp;P=IA=(1400Wm2)(2.82×1023m2)=3.95×1026W

The energy flow to the earth from sunlight is about 1.4kW/m^{2} . (a) Find

Answered question

Answer & Explanation

New Questions in Other