Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo

Check Your Understanding

12.1

1.41 meters

12.2

μ 0 I 2 R μ 0 I 2 R

12.3

4 amps flowing out of the page

12.4

Both have a force per unit length of 9.23×10−12N/m9.23×10−12N/m

12.5

0.608 meters

12.6

In these cases the integrals around the Ampèrian loop are very difficult because there is no symmetry, so this method would not be useful.

12.7

a. 1.00382; b. 1.00015

12.8

a. 1.0×10−4T1.0×10−4T; b. 0.60 T; c. 6.0×1036.0×103

Conceptual Questions

1.

Biot-Savart law’s advantage is that it works with any magnetic field produced by a current loop. The disadvantage is that it can take a long time.

3.

If you were to go to the start of a line segment and calculate the angle θθ to be approximately 0°0°, the wire can be considered infinite. This judgment is based also on the precision you need in the result.

5.

You would make sure the currents flow perpendicular to one another.

7.

A magnetic field line gives the direction of the magnetic field at any point in space. The density of magnetic field lines indicates the strength of the magnetic field.

9.

The spring reduces in length since each coil will have a north pole-produced magnetic field next to a south pole of the next coil.

11.

Ampère’s law is valid for all closed paths, but it is not useful for calculating fields when the magnetic field produced lacks symmetry that can be exploited by a suitable choice of path.

13.

If there is no current inside the loop, there is no magnetic field (see Ampère’s law). Outside the pipe, there may be an enclosed current through the copper pipe, so the magnetic field may not be zero outside the pipe.

15.

The bar magnet will then become two magnets, each with their own north and south poles. There are no magnetic monopoles or single pole magnets.

Problems

17.

5.66 × 10 −5 T 5.66 × 10 −5 T

19.

B=μoI8(1a1b)B=μoI8(1a1b) out of the page

21.

a=2Rπa=2Rπ; the current in the wire to the right must flow up the page.

23.

20 A

25.

Both answers have the magnitude of magnetic field of 4.5×10−5T.4.5×10−5T.

27.

At P1, the net magnetic field is zero. At P2, B=3μoI8πaB=3μoI8πa into the page.

29.

The magnetic field is at a minimum at distance a from the top wire, or half-way between the wires.

31.

a. F/l=2×10−5N/mF/l=2×10−5N/m away from the other wire; b. F/l=2×10−5N/mF/l=2×10−5N/m toward the other wire

33.

B = μ o I 2 π a 2 b ( ( a 2 + b 2 ) i ^ + b ( a 2 b 2 ) j ^ ) B = μ o I 2 π a 2 b ( ( a 2 + b 2 ) i ^ + b ( a 2 b 2 ) j ^ )

35.

0.019 m

37.

N×6.28×10−5TN×6.28×10−5T

39.

B=μoIR2N((d2)2+R2)3/2B=μoIR2N((d2)2+R2)3/2

41.

a. μ0I;μ0I; b. 0; c. μ0I;μ0I; d. 0

43.

a. 3μ0I;3μ0I; b. 0; c. 7μ0I;7μ0I; d. −2μ0I−2μ0I

45.

at the radius R

47.
Graph shows the variation of B with r. B linearly increases with r until the point a. Then it starts to decreases proportionally to the inverse of r.
49.

B = 1.3 × 10 −2 T B = 1.3 × 10 −2 T

51.

roughly eight turns per cm

53.

B = 1 2 μ 0 n I B = 1 2 μ 0 n I

55.

0.0181 A

57.

0.0008 T

59.

317.31

61.

2.1×10−4A·m22.1×10−4A·m2
2.7A2.7A

63.

0.18 T

Additional Problems

65.

B = 1.4 × 10 −4 T B = 1.4 × 10 −4 T

67.

3.2×10−19N3.2×10−19N in an arc away from the wire

69.

a. above and below B=μ0j,B=μ0j, in the middle B=0;B=0; b. above and below B=0,B=0, in the middle B=μ0jB=μ0j

71.

d B B = d r r d B B = d r r

73.

a. 5026 turns; b. 0.00957 T

75.

B 1 ( x ) = μ 0 I R 2 2 ( R 2 + z 2 ) 3 / 2 B 1 ( x ) = μ 0 I R 2 2 ( R 2 + z 2 ) 3 / 2

77.

B = μ 0 σ ω 2 R B = μ 0 σ ω 2 R

79.

derivation

81.

derivation

83.

As the radial distance goes to infinity, the magnetic fields of each of these formulae go to zero.

85.

a. B=μ0I2πrB=μ0I2πr; b. B=μ0J0r23RB=μ0J0r23R

87.

B(r)=μ0NI/2πrB(r)=μ0NI/2πr

This figure shows a torus with the inner radius a and an outer radius b. A thin wire is wound evenly on the torus.

Challenge Problems

89.

B = μ 0 I 2 π a ln x + a x . B = μ 0 I 2 π a ln x + a x .

91.

a. B=μ0σω2[2h2+R2R2+h2−2h]B=μ0σω2[2h2+R2R2+h2−2h]; b. B=4.09×10−5T,B=4.09×10−5T, 82% of Earth’s magnetic field

Order a print copy

As an Amazon Associate we earn from qualifying purchases.

Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction
Citation information

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.