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A long straight cable of length l is placed symmetrically along z - axis and has radius a(lt lt l). The cable consists of a thin wire and a co-axial conducting tube.An alternating current thin wire and returns along the coaxial conducting tube.The induced electricfield at a distance s from the wire inside the cable is vec(E )(s, t)=mu_(0)I_(0)v cos (2pi vt) ln((s)/(a))hat(k)Compare the conduction current I_(0) with the displacement current I_(0)^(d). |
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Answer» Solution :Displacement current, `I_(d)=(a)/(2)((pi a)/(lambda))^(2)I_(0)SIN(2pi vt)=I_(0)^(d)sin(2pi vt)` so `I_(0)^(d)=(a)/(2)((pi a)/(lambda))^(2)I_(0)` `=((a pi )/(lambda))^(2)I_(0)xx ((a)/(2))` `therefore (I_(0)^(d))/(I_(0))=(((a pi)/(lambda))^(2)I_(0)xx ((a)/(2)))/(I_(0))` `=(a)/(2)((a pi)/(lambda))^(2)` `=(a^(3)pi^(2))/(2 lambda^(2))` Hence required ratio `(I_(0)^(d))/(I_(0))=(a^(3)pi^(2))/(2lambda^(2))` |
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