a) we know,
Intensity of radiation,
I = Erms^2/(c*mue)
= 3500^2/(3*10^8*4*pi*10^-7)
= 3.25*10^4 W/m^2
b) we know, Intensity = Power/Area
==> Power = Intensity*Area
= 3.25*10^4*pi*0.033^2
= 111.2 W
c) let t is the time taken.
Q = m*c*dT
P*t = m*c*dT
t = m*c*dT/P
= 0.2*3500*1.5/111.2
= 9.44 s
A heat lamp emits infrared radiation whose electric field is E_rms = 3500 N/C. (a) What...
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CH 24 #15 Assume the mostly infrared radiation from a heat lamp acts like a continuous wave with wavelength 1.50 µm. (a) If the lamp's 205 W output is focused on a person's shoulder, over a circular area 26.0 cm in diameter, what is the intensity in W/m2? (b) What is the peak electric field strength in kV/m? (c) Find the peak magnetic field strength in µT. (d) How long will it take in seconds to increase the temperature of...