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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)Integrate the displacement current density across the cross - section of the cable to find the total displacement current I^(d). |
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Answer» Solution :`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. |
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