1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A long straight cable of length l is placed symmetrically along z-axis and has radius a(<<l). the cable consists of a thin wire and co-axial conducting tube. An alternating current I(t) = I0 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) = μ0I0cos(2πνt) ln(sa)^k. , the ratio of conduction current and displacement current is

A
(aππ)2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(aπλ)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(λaπ)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(λaπ)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A (aππ)2Id=∫JDsdsdθ=2πλ2I02π∫ax=0ln(as).sds sin(2πvt)=(2πλ)2I0∫ax=012ds2ln(as).sin(2πvt)=a24(2πλ)2I0∫ax=0d(sa)2ln(as)2.sin(2πvt)=−a24(2πλ)2I0∫10lnξdξ.sin(2πvt)=+(a2)2(2πλ)2I0sin2πvt(∴The integral has value −1)The displacement currentId=(a2,2πλ)2I0sin2πvt=I40sin2πvtratioId0I0=(axλ)2.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Density
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program