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 electric field 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)`
Integrate the displacement current density across the cross - section of the cable to find the total displacement current `I^(d)`.

Explanation

`I_(d)=int J_(d)sds d theta`
`= int_(s=0)^(s=a) J_(d) sds int_(0)^(2pi)d theta`
`= int_(s=0)^(s=a) J_(d)sds xx 2pi [because int_(0)^(2pi)d theta = [theta]_(0)^(2pi) = 2pi]`
`int_(s=0)^(a)[(2pi)/(lambda^(2))I_(0)ln ((a)/(s))sds sin(2pi vt)hat(k)]xx 2pi " "` [From equation (1)]
`=((2pi)/(lambda))^(2)(I_(0)int_(s=0)^(a)ln((a)/(s))sds)sin(2pi vt)hat(k)`
`=((2pi)/(lambda))^(2)I_(0)[int_(s=0)^(a)ln((a)/(s))(1)/(2)d(s)^(2)]sin (2pi vt)hat(k)`
`=(a^(2))/(lambda)((2pi)/(lambda))^(2)I_(0)[sin (2pi vt)hat(k)] " " {int_(s=0)^(a)[ln((a)/(s)).d((s)/(a))^(2)]ds}`
`=(a^(2))/(2)((2pi)/(lambda))^(2)I_(0)sin 2pi v t hat(k)xx(-(a)/(4)) " " {because int_(s=0)^(a)[ln((s)/(a))d((s)/(a))^(2)]ds=-(a)/(4)}`
`therefore I_(d)=-(a^(3))/(8)((2pi)/(lambda))^(2)I_(0)sin(2pi vt)hat(k)`
`=-(a)/(2)xx(a^(2))/(4)((2pi)/(lambda))^(2)sin (2pi vt)hat(k)`
Negative sign shows that displacement current is in opposite direction of conduction current.
Hence `(a)/(2)((2pi a)/(2lambda))^(4)I_(0)sin(2pi vt)hat(k)`
`I_(d)` is in negative z - direction and `I_(c )` is in position z - direction.