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The enegry is transferred form a source of constant voltage V to a consumer by means of long straight coaxial cable with negligible active resistance. The consumed current is I. Final the enegry flux across the cross-section of the cable. The conductive sheath is supposed to be thin. |
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Answer» Solution :The electric FIELD `(………rarr)` and the magnetic field `(H rarr)` are as shown. The electirc field by Gauss's therorem is LIKE `E_(r) = (A)/(r )` Intergrating `varphi = A` in `(r_(2))/(r )` so `A = (V)/(In(r_(2))/(r_(1)) (r_(2) gt r_(1))` Then `E = (V)/(rIn(r_(2))/(r_(1)))` Magnetic field is `H_(theta) = (I)/(2pir)` The Poynting vector `S` is along the `Z` axis and non zero between then `(r_(1) lt r ltr_(2))`. the total power FLUX is `= int_(r_(1))^(r_(2)) (IV)/(2pir^(2)In(r_(2))/(r_(1))).2PI r dr = Iv` 
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