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InterviewSolution
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Calculate the electric anil magnetic fields produced by the radiation coining from a 100 IF bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source. |
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Answer» Solution : The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3 m, the surface area of the surrounding SPHERE is `A = 4pir^(2)=11 3m^(2)` The intensity at this distance is `I=("power")/("area")=(100 w xx2.5 %)/(113 m^(2))=0.022 W//m^(2)` Half of this intensity is provided by the electric field and half by the magnetic field. `E_(rms)=sqrt(0.022)/(8.85xx10^(-12)(3xx10^(8)))V//m=2.9 V//m` The value of E foiund aboveis the root mean squarevalue of the electricfiedl since the electric fieldin a lightbeam s SINUSOIDAL the pealk electric field `2E_(0)` is `E_(0)sqrt(2E_(rms))=sqrt(2)x2.9V//m` `=4.07 V//m` Thus you see that the electricfieldstrengthof the light that you use for reading is fairly large compare ITWITH electricfield strength of TV or Fm waves which isof the order of a fewmicrovolts pre metre now let uscalculate the strength of themagneticfield it is Again, since the field in the light beam is sinusoidal, the peak magnetic field is `B_(0)=sqrt(2B)_(rms)=1.4 xx10^(-8) T`.Note that although the energy in the magnetic field is EQUAL to the energy in the electric field, the magnetic field strength is evidently very weak. |
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