Q.

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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a

B10,B20 for conductors carrying equal currents in opposite directions

b

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

c

B10, B2=0 For conductors carrying equal currents in opposite directions

d

B10,B20 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 = 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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