A source of electromagnetic waves radiates power uniformly in all directions at

Question

A source of electromagnetic waves radiates power uniformly in all directions at a single frequency. At a distance of 3.00 km from the source, a detector measures the intensity of the wave to be 28.0 W/m2. What is the peak value of the magnetic field at the detector due to the wave?

En dyler 12 of EN A source of electromagnetic waves radiates power uniformly in all directions at a single frequency. At a distance of 3.00 km from the source, a detector measures the intensity of the wave to be 2s.o uw/m2 What is the peak valué of the magnetic field at the detector due to the wave Ansaar ries 0/12 What is the average power radiated by the source? Answer ries 0/12 s replaced with a perfectly absorbing shee ncident flux with surface area 2.40 m e force on the sheet du Now the detecto t normal to the i What is the wave LSutamil Answer Tries 0/12 Post Discussion Send Feedback

Explanation / Answer

Intensity of the wave at a distance d is I = P/(4*pi*d^2)

P is power transmitted

and d = 3 km = 3000 m

average intensity = I = c*Bmax^2/(2*mu_o)

c is the speed of EM wave

mu_o is the permittivity of free space = 4*pi*10^-7 H/m

then

I = c*Bmax^2/(2*mu_o)

28*10^-6 = 3*10^8*Bmax^2/(2*4*3.142*10^-7)

Bmax = sqrt((28*10^-6*2*4*3.142*10^-7)/(3*10^8))

Bmax = 4.85*10^-10 T

average power transmitted is P = I*(4*pi*d^2) = 28*10^-6*(4*3.142*3000^2) = 3167.136 W

total energy density is eta = F/A = B^2/(mu_o)

Force F = B^2*A/(mu_o) = Bmax^2*A/(2*mu_o) = (4.85*10^-10)^2*2.4/(2*4*3.142*10^-7)

F = 2.24*10^-13 N

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