Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question:In Figure, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a=20cm. The currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3, and each wire carries 20 A. In unit-vector notation, what is the net magnetic field at the square’s center?

Figure 29-89 is an idealized schematic drawing of a rail gun. Projectile Psits between two wide rails of circular cross section; a source of current sends current through the rails and through the(conducting) projectile (a fuse is not used). (a) Let wbe the distance between the rails, Rthe radius of each rail, and i the current. Show that the force on the projectile is directed to the right along the rails and is given approximately byF=i2μ02π·ln(w+RR)

(b) If the projectile starts from the left end of the rails at rest, find the speed vat which it is expelled at the right. Assume that I = 450 kA, w = 12 mm, R = 6.7 cm, L = 4.0 m, and the projectile mass is 10 g.

In Figure, point P is at perpendicular distance R=2.00cmfrom a very long straight wire carrying a current. The magnetic field Bset up at point Pis due to contributions from all the identical current length elements idsalong the wire. What is the distanceto the element making (a) The greatest contribution to field Band (b) 10.0% of the greatest contribution?

A long wire is known to have a radius greater than4.00mmand to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is0.28mTat a point4.0mmfrom the axis of the wire, and0.20mTat a point 10 mm from the axis of the wire. What is the radius of the wire?

Question: Two long straight thin wires with current lie against an equally long plastic cylinder, at radius R=20.0cmfrom the cylinder’s central axis.

Figure 29-58ashows, in cross section, the cylinder and wire 1 but not wire 2. With wire 2 fixed in place, wire 1 is moved around the cylinder, from angle localid="1663154367897" θ1=0°to angle localid="1663154390159" θ1=180°, through the first and second quadrants of the xycoordinate system. The net magnetic field Bat the center of the cylinder is measured as a function of θ1. Figure 29-58b gives the x component Bxof that field as a function of θ1(the vertical scale is set by Bxs=6.0μT), and Fig. 29-58c gives the y component(the vertical scale is set by Bys=4.0μT). (a) At what angle θ2 is wire 2 located? What are the (b) size and (c) direction (into or out of the page) of the current in wire 1 and the (d) size and (e) direction of the current in wire 2?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free