A long straight cable of length l is placed symmetrically along z-axis and has radius a(<0
The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t) = Io sin (2πνt) 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 E(s,t)= μoIovcos (2πνt) In (s/a)k.
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross section of the cable to find the total displacement current Id.
(iii) Compare the conduction current Io with the displacement current I0d .
A long straight cable of length l is placed symmetrically along z-axis and has radius a(<0
The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t) = Io sin (2πνt) 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 E(s,t)= μoIovcos (2πνt) In (s/a)k.
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross section of the cable to find the total displacement current Id.
(iii) Compare the conduction current Io with the displacement current I0d .
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1 Answer
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This is a long answer type question as classified in NCERT Exemplar
E(s,t)=
Now displacement current Jd=eo
=
=
= ina/s sin2
Id=
=( )2
=
After solving this we get
Id=
Id=
Iod= (
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According to definition of displacement current, we can write
B0 = 42.9 × 10-9
42.9 Ans
s? = -? + k? ∴ s? = (-? +k? )/√2
I = P/ (4πr²) . (i)
and I = (1/2)ε? E? ²c . (ii)
From (i) and (ii)
P/ (4πr²) = (1/2)ε? E? ²c
E? = √ (2P/ (4πε? r²c) ; E? = √ (9×10? ×2×15)/ (15×15×10? ×3×10? ) = 2×10? ³ V/m
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