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Figure 29-88 shows a cross section of a long conducting coaxial cable and gives its radii (a,b,c). Equal but opposite currents iare uniformly distributed in the two conductors. Derive expressions for B (r) with radial distance rin the ranges (a) r < c, (b) c< r <b , (c) b < r < a, and (d) r > a . (e) Test these expressions for all the special cases that occur to you. (f) Assume that a = 2.0 cm, b = 1.8 cm, c = 0.40 cm, and i = 120 A and plot the function B (r) over the range 0 < r < 3 cm .

Short Answer

Expert verified
  1. B=μ0ir2πc2.
  2. B=μ0i2πr
  3. B=μ0i2πr·a2-r2a2-b2
  4. B=0
  5. Tested the expressions for all the special cases.
  6. The graph is drawn below.

Step by step solution

01

Given

  • The radius of the central solid cylindrical part of the cablec=0.40cm=0.40×10-2m .
  • The inner radius of the outer hollow cylindrical part of the cable b=1.8cm=1.8×10-2m.
  • The outer radius of the outer hollow cylindrical part of the cablea=2cm=2.0×10-2m .
  • Current through inner and outer cylindrical part i = 120 A flowing in the opposite sense.
02

Understanding the concept

For this problem, we can use Ampere’s law by considering a suitable Ampere loop to find the magnetic field at the given point over the specified region. After that, we can find out the current enclosed in the ampere loop. Finally, we get the expression for the Bfield at the particular point.

Formula:

B·ds=μ0Ienclosed

03

(a) Derive expressions for B (r) with radial distance r  in the range r < c

For r < c,

B·ds=μ0IenclosedBds=μ0IenclosedB·2πr=μ0IenclosedB=μ0Ienclosed2πr

Now let’s findthecurrent enclosed in the ampere loop of radius r:

Ienclosed=πr2πc2i=r2c2i

So we get,

B=μ02πrr2c2=μ0ir2πc2

Hence,B=μ0ir2πc2.

04

(b) Derive expressions for B (r)  with radial distance r in the range  c < r < b

Forc < r < b

Now let’s find the current enclosed in the ampere loop of radius r:

B·ds=μ0IenclosedBds=μ0IenclosedB·2πr=μ0IenclosedB=μ0Ienclosed2πr

Ienclosed=i, so we getB=μ0i2πr

Hence,B=μ0i2πr

05

(c) Derive expressions for B (r) with radial distance r in the range b < r < a

Forb < r < a

B·ds=μ0IenclosedBds=μ0IenclosedB·2πr=μ0IenclosedB=μ0Ienclosed2πr

Let’s findthecurrent enclosed in the ampere loop of radiusr:

Ienclosed=i-ir2-b2a2-b2=a2-r2a2-b2iIenclosed=a2-r2a2-b2i

Hence using this, we get B=μ0i2πr·a2-r2a2-b2.

06

(d) Derive expressions for B (r) with radial distance r in the range r > a

For r > a

Current through the inner and outer cylindrical part is i flowing in the opposite direction, so the current enclosed in the Ampere loop of radius r

Ienclosed=0.0

Hence B=0.0.

07

(e) Test these expressions for all the special cases

The B (r)for the special cases:

We have

Br=μ0i2πr·a2-r2a2-b2

When a = 0, we get

Br=μ0i2πr·0-r20-b2=μ0i2πr·r2b2=μ0i2π·rb2

Whenr = b,we get

Br=μ0i2πr·a2-r2a2-b2=μ0i2πb·a2-b2a2-b2=μ0i2πb

When r = a, we get

Br=μ0i2πa·a2-a2a2-b2=μ0i2πa·0=0.0

When b = 0, we get

Br=μ0i2πr·a2-r2a2-0=μ0i2πr·1-r2a2

Hence, it is tested the expressions for all special cases.

08

(f) Plot the function B (r) over the range 0 < r < 3 cm

The plot of the function B (r) over the range 0 < r < 3 cm and i=120A:c=0.40cm=0.40×10-2m, b=1.8cm=1.8×10-2m,a=2cm=2.0×10-2m.

Hence the graph is drawn.

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