Saved Bookmarks
| 1. |
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)Calculate the displacement current density inside the cable. |
|
Answer» Solution :Induced electric field at distance s (which is less then radius of coaxial cable) `vec(E )(s, t)=mu_(0)I_(0)v cos (2pi v t)ln ((s)/(a)) hat(K)` Now displacement current density. `J_(d) =in_(0)(dE)/(DT)` `= in_(0)(d)/(dt)[mu_(0)I_(0)v cos (2pi vt)ln((s)/(a))hat(k)]` `= in_(0)mu_(0)I_(0)v(d)/(dt)[cos (2pi v t) ln ((s)/(a))hat(k)]` `=(1)/(c^(2))I_(0)v^(2)xx 2pi[-sin (2pi vt)ln((s)/(a))hat(k)] "" [because mu_(0)in_(0) =(1)/(c^(2))]` `therefore J_(d)=+(n^(2))/(c^(2))xx 2pi_(0)sin(2 pi v t)ln ((a)/(s))hat(k) "" [because ln.(S)/(a)=-ln.(a)/(S)]` `=(1)/(lambda^(2))xx 2pi_(0)ln((a)/(s))sin(2pi vt)hat(k) "" [because lambda = (c )/(v)]` `therefore J_(d) = (2pi I_(0))/(lambda^(2))ln ((a)/(s))sin(2pi v t)hat(k) ""` .....(1) |
|