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Home » A long straight cable of length is placed symmetrically along the z-axis and has radius . The cable consists of a thin wire and a co-axial conducting tube. An alternating current sin flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance from the wire inside the cable is . In , a) calculate the displacement current density inside the cableb) integrate the displacement current density across the cross-section of the cable to find the total displacement current I
A long straight cable of length I is placed symmetrically along the z-axis and has radius a. The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t)=I_{0} sin (2\pi vt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is \mathbf{E}(\mathrm{s}, \mathrm{t})=\mu_{0} \mathrm{l}_{0} \mathrm{~V} cos (2\pi vt). In \left(\frac{s}{a}\right) \hat{k},
a) calculate the displacement current density inside the cable
b) integrate the displacement current density across the cross-section of the cable to find the total displacement current I
CBSE | Class 12 | Electromagnetic Waves | ncert exemplar | Physics
A long straight cable of length I is placed symmetrically along the z-axis and has radius a. The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t)=I_{0} sin (2\pi vt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is \mathbf{E}(\mathrm{s}, \mathrm{t})=\mu_{0} \mathrm{l}_{0} \mathrm{~V} cos (2\pi vt). In \left(\frac{s}{a}\right) \hat{k},
a) calculate the displacement current density inside the cable
b) integrate the displacement current density across the cross-section of the cable to find the total displacement current I

a) The displacement current density is given as

\vec{J}_{d}=\frac{2 \pi I_{0}}{\lambda^{2}} \ln \frac{a}{s} \sin 2 \pi v t \hat{k}

b) Total displacement current will be,

I^{d}=\int J_{d} 2 \pi s d s
I_{d}=(\pi a / \lambda) 2 I_{0} \sin 2 \pi v t

i

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