Question

Figure CQ19.7 shows a coaxial cable carrying current I in its inner conductor and a return current of the same magnitude in the opposite direction in the outer conductor. The magnetic field strength at r 5 r0 is B0. Find the ratio B /B0 at (a) r 5 2r0 and (b) r 5 4r0.

A little explanation would be greatly appreciated!

A47

Answer 1

According to the Ampere's circuital law

B.d = Ho * Ienclosed

when r = r₀

we can create an amperean loop of radius r_0. now current inside the loop is I.

Now the integration becomes,

φB.di = Βο φal = Βρ.2πο = μο * Ι

hence

Βρ. = μο * Ι /2πιο

Now for (a), when r= 2r₀

then the amperean loop has radius of 2r₀ and current enclosed by the cirle is I

Now,

B. =B di = B.22ro = Ho * I

Β. = μοΙ /4πη.

hence B/B₀= 1/2

for (b), when r = 4r₀, the loop has radius 4r₀ and since two currents are in opposite directions,

then net current inside the loop is zero.

I = 0

Hence,

B.d =B dl = B.24ro = Ho* I = 0

B = 0

hence B/B₀ = 0