A co-axial cable consists of a thin inner current carrying conductor fixed along the axis of a hollow current carrying conductor. Let B1 and B2 be the magnetic fields in the region between the conductors and outside the conductor, respectively

We apply Ampere’s Circuital Law to the co-axial circular loops C1 and C2 with corresponding magnetic fields B1 and B2 respectively. We can write . B = μ0 inet2πr. For optin (2) , inet ≠0 for B1 and inet =0 for B2. For option (3) , inet ≠0 for both B1 and B2. For option (4) , B1 =μ0i2πx and B2 = μ02i2π2x

A co-axial cable consists of a thin inner current carrying conductor fixed along the axis of a hollow current carrying conductor. Let B1 and B2 be the magnetic fields in the region between the conductors and outside the conductor, respectively

We apply Ampere’s Circuital Law to the co-axial circular loops C1 and C2 with corresponding magnetic fields B1 and B2 respectively. We can write . B = μ0 inet2πr. For optin (2) , inet ≠0 for B1 and inet =0 for B2. For option (3) , inet ≠0 for both B1 and B2. For option (4) , B1 =μ0i2πx and B2 = μ02i2π2x

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A co-axial cable consists of a thin inner current carrying conductor fixed along the axis of a hollow current carrying conductor. Let B₁ and B₂ be the magnetic fields in the region between the conductors and outside the conductor, respectively

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$B_{1} \neq 0, B_{2} \neq 0$ for conductors carrying equal currents in opposite directions

B₁ = B₂ for conductors carrying equal currents in same directions at distances x and 2x from the axis respectively.

$B_{1} \neq 0, B_{2} = 0$ For conductors carrying equal currents in opposite directions

$B_{1} \neq 0, B_{2} \neq 0$ for conductors carrying equal currents in the same direction

answer is B, C, D.

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Detailed Solution

We apply Ampere’s Circuital Law to the co-axial circular loops C₁ and C₂ with corresponding magnetic fields B₁ and B₂ respectively. We can write . B = $\frac{μ_{0} i_{n e t}}{2 πr}$ . For $o p t i n (2), i_{n e t} \neq 0 f o r B_{1} a n d i_{n e t} = 0 f o r B_{2} .$ For option (3) , $i_{n e t} \neq 0 f o r b o t h B_{1} a n d B_{2} .$ For option (4) , $B_{1} = \frac{μ_{0} i}{2 πx} a n d B_{2} = \frac{μ_{0} (2 i)}{2 π (2 x)}$