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DFCCIL Executive S&T 2016 Official Paper

Option 3 : Force between magnetic elements

**Explanation:**

Consider the following Co-axial cable with inner radius 'a' and outer radius b.

By Ampere law -

Enclosed surface-

\(\rm \oint{H\cdot{dl}}={I_{enclosed}}\)

for r < a

\(H(2\pi r) = I_0 \left (\frac{\pi r^2}{\pi a^2}\right)\)

\(H=\dfrac{Ir}{2{π}a^2}\hat{ϕ}\)

Now,

for a ≤ r < b

2πrHϕ = I

\(H=\phi\dfrac{I}{2{\pi}r}\)

Now,

for r > b

I_{enclosed} = I + (-I) = 0

H_{r > b} = 0, B = μH = 0 {outside co-axial cable}

So due to forces between Magnetic elements, current flows in the opposite direction, which leads to zero magnetic field outside the co-axial cable.