Question

In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero
(a) outside the cable
(b) inside the inner conductor
(c) inside the outer conductor
(d) in between the tow conductors.

Answer 1

Answer:

Construct an Ampere’s loop in the concerned regions to see the value of ip.

At a pont outside the co-axial tube at a diatance R from the central axis consider a circular path.

Then from Amperes Law closed integral of B.dl=μ0I(net)

but the net current is zero as the current in the two cables flow in opposite directions.

therefore Magnetic field is zero outside

Ans : (a)

Answer 2

In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero

a) outside the cable

(b) inside the inner conductor

Therefore the correct option are Option (a) and (b)

Explanation:

According to Ampere's law, the outer conductor has a coaxial, straight cable carrying currents I in the inner conductor and –i (The same goes in the opposite direction).

Inside the inner conductor

$\oint \vec{B}.\vec{dl} = \mu_0 i_inside$

$\oint \vec{B}.\vec{dl} = 0$

B.l = 0

B = 0

Between the two conductors

$\oint \vec{B} .\vec{dl} = \mu_0 i$

$B = \frac{\mu_0 i}{2\pi r }$

Outside the main conductor

$\oint \vec{B} .\vec{dl} = \mu_0 (i-i)$

B = 0

Thus the magnetic field outside the cable is zero.