Assignment 17: Ampere's Law, magnetic materials

Due: 8:00am on Friday, March 16, 2012

Note: To understand how points are awarded, read your instructor's Grading Policy.

Ampère’s Law Explained

Learning Goal: To understand Ampère’s law and its application.

Ampère’s law is often written .

Part A

The integral on the left is

ANSWER:

the integral throughout the chosen volume.

the surface integral over the open surface.

the surface integral over the closed surface bounded by the loop.

the line integral along the closed loop.

the line integral from start to finish.

Correct

Part B

What physical property does the symbol represent?

ANSWER:

The current along the path in the same direction as the magnetic field

The current in the path in the opposite direction from the magnetic field

The total current passing through the loop in either direction

The net current through the loop

Correct

The positive direction of the line integral and the positive direction for the current are related by the right-hand rule: Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction

for and the direction of your thumb is the positive direction for the net current. Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)

Part C

The circle on the integral means that must be integrated

ANSWER:

over a circle or a sphere.

along any closed line that you choose.

along the path of a closed physical conductor.

over the surface bounded by the current-carrying wire.

Correct

Part D

Which of the following choices of path allow you to use Ampère’s law to find ?

a. The path must pass through the point .

b. The path must have enough symmetry so that is constant along large parts of it. c. The path must be a circle.

ANSWER:

a only

a and b

a and c

b and c

Correct

Part E

Ampère’s law can be used to find the magnetic field around a straight current-carrying wire. Is this statement true or false?

ANSWER:

true

false

Correct

In fact Ampère's law can be used to find the magnetic field inside a cylindrical conductor (i.e., at a radius less

than the radius of the wire, ). In this case is just that current inside , not the current inside (which is the total current in the wire).

Part F

Ampère’s law can be used to find the magnetic field at the center of a square loop carrying a constant current. Is this statement true or false?

ANSWER:

true

false

Correct

The key point is that to be able to use Ampère's law, the path along which you take the line integral of must

have sufficient symmetry to allow you to pull the magnitude of outside the integral. Whether the current distribution has symmetry is incidental.

Part G

Ampère’s law can be used to find the magnetic field at the center of a circle formed by a current-carrying conductor. Is this statement true or false?

ANSWER:

true

false

Correct

Part H

Ampère’s law can be used to find the magnetic field inside a toroid. (A toroid is a doughnut shape wound uniformly with many turns of wire.) Is this statement true or false?

ANSWER:

true

false

Correct

Therefore, though Ampère’s law holds quite generally, it is useful in finding the magnetic field only in some cases, when a suitable path through the point of interest exists, typically such that all other points on the path have the same magnetic field through them.

Magnetic Field of a Current-Carrying Wire

Find the magnetic field a distance from the center of a long wire that has radius and carries a uniform current

per unit area in the positive z direction.

Part A

First find the magnetic field, , outside the wire (i.e., when the distance is greater than ).

Hint A.1 Ampère's law with current density

Hint not displayed

Hint A.2 Find the direction of the field

Hint not displayed

Hint A.3 Find the left-hand side of Ampère's law

Hint not displayed

Hint A.4 Find the right-hand side of Ampère's law

Hint not displayed

Express in terms of the given parameters, the permeability constant , the variables , (the magnitude of

), , , and , and the corresponding unit vectors , , and . You may not need all these in your answer.

ANSWER:

= Correct

Part B

Now find the magnetic field inside the wire (i.e., when the distance is less than ).

Hint B.1 Establish the relationiship to Part A

Hint not displayed

Hint B.2 Integrate over the Ampèrean loop

Hint not displayed

Express in terms of the given parameters, the permeability constant , the distance from the center of

the wire, and the unit vectors , , and . You may not need all these in your answer.

ANSWER:

= Correct

Exercise 28.45

Coaxial Cable. A solid conductor with radius is supported by insulating disks on the axis of a conducting tube

with inner radius and outer radius in the following figure. The central conductor and tube carry equal currents

in opposite directions. The currents are distributed uniformly over the cross sections of each conductor.

Part A

Derive an expression for the magnitude of the magnetic field at points outside the central, solid conductor but

inside the tube ( < < ).

Express your answer in terms of the variables , , and .

ANSWER:

=

Correct

Part B

Derive an expression for the magnitude of the magnetic field at points outside the tube ( > ).

Express your answer in terms of the variables , , and .

ANSWER:

=

0 Correct

Current Sheet

Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current runs in the -y direction

through each wire. There are wires per unit length in the x direction.

Part A

Write an expression for , the magnetic field a distance above the xy plane of the sheet.

Use for the permeability of free space.

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Find

Hint not displayed

Hint A.3 Determine the direction of the magnetic field

Hint not displayed

Hint A.4 Magnitude of the magnetic field

Hint not displayed

Hint A.5 Evaluate

Hint not displayed

Express the magnetic field as a vector in terms of any or all of the following: , , , , , and the unit vectors ,

, and/or .

ANSWER:

= Correct

This equation is analogous to on either side of a infinitely charged sheet. The correspondence seems

more obvious if you set the current per unit length . Then the magnetic field you just calculated is

.

The electric field, though, points along the perpendicular to the surface.

Do you see why you had to pick the rean loop you used? That is, why would any other loop not have worked.

Did you notice that by using Ampère's law you could find the field by using a much simpler integral than Biot-Savart's law? The drawback is that you may not always be able to find a convenient loop in situations where the current distribution is more complicated.

Magnetic Field inside a Very Long Solenoid

Learning Goal: To apply Ampère's law to find the magnetic field inside an infinite solenoid. In this problem we will apply Ampère's law, written

,

to calculate the magnetic field inside a very long solenoid (only a relatively short segment of the solenoid is

shown in the pictures). The solenoid has length , diameter , and turns per unit length with each carrying

current . It is usual to assume that the component of the current along the z axis is negligible. (This may be assured by winding two layers of closely spaced wires that spiral in opposite directions.)

From symmetry considerations it is possible to show that far from the ends of the solenoid, the magnetic field is axial.

Part A

Which figure shows the loop that the must be used as the Ampèrean loop for finding for inside the solenoid?

Hint A.1 Choice of path for loop integral

Hint not displayed

ANSWER:

A

B

C

D

Correct

Part B

Assume that loop B (in the Part A figure) has length along . What is the loop integral in Ampère's law? Assume that the top end of the loop is very far from the solenoid (even though it may not look like it in the figure), so that the field there is assumed to be small and can be ignored.

Express your answer in terms of , , and other quantities given in the introduction.

ANSWER:

=

Correct

Part C

What physical property does the symbol represent?

ANSWER:

The current along the path in the same direction as the magnetic field

The current in the path in the opposite direction from the magnetic field

The total current passing through the Ampèrean loop in either direction

The net current through the Ampèrean loop

Correct

The positive direction of the line integral and the positive direction for the current are related by the right-hand rule: Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction

for and the direction of your thumb is the positive direction for the net current. Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)

Part D

What is , the current passing through the chosen loop?

Express your answer in terms of (the length of the Ampèrean loop along the axis of the solenoid) and other variables given in the introduction.

ANSWER:

=

Correct

Part E

Find , the z component of the magnetic field inside the solenoid where Ampère's law applies.

Express your answer in terms of , , , , and physical constants such as .

ANSWER:

= Correct

Part F

What is , the z component of the magnetic field outside the solenoid?

Hint F.1 Find the Ampèrean loop to use

Hint not displayed

ANSWER:

0

Correct

Part G

The magnetic field inside a solenoid can be found exactly using Ampère's law only if the solenoid is infinitely long. Otherwise, the Biot-Savart law must be used to find an exact answer. In practice, the field can be determined with very little error by using Ampère's law, as long as certain conditions hold that make the field similar to that in an infinitely long solenoid.

Which of the following conditions must hold to allow you to use Ampère's law to find a good approximation?

a. Consider only locations where the distance from the ends is many times .

b. Consider any location inside the solenoid, as long as is much larger than for the solenoid. c. Consider only locations along the axis of the solenoid.

Hint G.1 Implications of symmetry

Hint not displayed

Hint G.2 Off-axis field dependence

Hint not displayed

ANSWER:

a only

b only

c only

a and b

a and c

b and c

Correct

Magnetic Field inside a Toroid

A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy Slinky® with ends joined to make

a circle. Consider a toroid consisting of turns of a single wire with current flowing through it.

Consider the toroid to be lying in the r plane of a cylindrical coordinate system, with the z axis along the axis of

the toroid (pointing out of the screen). Let represent the angular position around the toroid, and let be the distance from the axis of the toroid.

For now, treat the toroid as ideal; that is, ignore the component of the current in the direction.

Part A

The magnetic field inside the toroid varies as a function of which parameters?

Hint A.1 Consider rotational symmetry

Hint not displayed

ANSWER:

only

only

both and

Correct

Part B

Inside the toroid, in which direction does the magnetic field point?

ANSWER:

(outward from the center)

(inward toward the center)

(clockwise)

(counterclockwise)

(out of the screen)

(into the screen)

Correct

Notice that the direction is antiparallel to the path shown by the Ampèrean loop in the figure. Also, by

definition, .

Part C

What is , the magnitude of the magnetic field inside the toroid and at a distance from the axis of the toroid?

Hint C.1 How to approach the problem

Hint not displayed

Hint C.2 Evaluate the line integral

Hint not displayed

Hint C.3 Find

Hint not displayed

Express the magnetic field in terms of , (the permeability of free space), , and .

ANSWER:

= Correct

Part D

In an ideal toroid, current would flow only in the and directions. The magnetic field in the central plane, outside of the coils of such a toroid, is zero. For the toroid shown in the figures however, this field is not quite zero. This is because in this problem, there is a single wire that is wrapped around a doughnut shape. This wire

must point somewhat in the direction, and thus the current must actually have a component in the direction.

Compute , the magnitude of the magnetic field in the center of the toroid, that is, on the z axis in the plane of

the toroid. Assume that the toroid has an overall radius of (the distance from the center of the toroid to the

middle of the wire loops) and that is large compared to the diameter of the individual turns of the toroid coils.

Note that whether the field points upward or downward depends on the direction of the current, that is, on whether the coil is wound clockwise or counterclockwise.

Hint D.1 Simplifying the problem

Hint not displayed

Express in terms of , , , , and the local diameter of the coils.

ANSWER:

=

Correct

This is the same expression that you would derive for the magnetic field at the center of a circular loop of

current-carrying wire. To see why this makes sense, imagine that the local diameter of the coils gets so small that it is negligible in comparison to the radius of the toroid. The wire makes one complete turn around the axis of the toroid. So, to a point in the center, the toroid looks like a simple current loop.

Magnetic Materials

Part A

You are given a material which produces no initial magnetic field when in free space. When it is placed in a region of uniform magnetic field, the material produces an additional internal magnetic field parallel to the original field. However, this induced magnetic field disappears when the external field is removed. What type of magnetism does this material exhibit?

ANSWER:

diamagnetism

paramagnetism

ferromagnetism

Correct

When a paramagnetic material is placed in a magnetic field, the field helps align the magnetic moments of the atoms. This produces a magnetic field in the material that is parallel to the applied field.

Part B

Once again, you are given an unknown material that initially generates no magnetic field. When this material is placed in a magnetic field, it produces a strong internal magnetic field, parallel to the external magnetic field. This field is found to remain even after the external magnetic field is removed. Your material is which of the following?

ANSWER:

diamagnetic

paramagnetic

ferromagnetic

Correct

Very good! Materials that exhibit a magnetic field even after an external magnetic field is removed are called ferromagnetic materials. Iron and nickel are the most common ferromagnetic elements, but the strongest permanent magnets are made from alloys that contain rare earth elements as well.

Part C

What type of magnetism is characteristic of most materials?

ANSWER:

ferromagnetism

paramagnetism

diamagnetism

no magnetism

Correct

Almost all materials exhibit diamagnetism to some degree, even materials that also exhibit paramagnetism or ferromagnetism. This is because a magnetic moment can be induced in most common atoms when the atom is placed in a magnetic field. This induced magnetic moment is in a direction opposite to the external magnetic field. The addition of all of these weak magnetic moments gives the material a very weak magnetic field overall. This field disappears when the external magnetic field is removed. The effect of diamagnetism is often masked in paramagnetic or ferromagnetic materials, whose constituent atoms or molecules have permanent magnetic moments and a strong tendency to align in the same direction as the external magnetic field.

In 28.54 "atomic currents" refers to that part of the magnetic field in the ferromagnetic material due to alignment of permanent magnetic dipoles as distinguished from the field due to electric current in the toroid wires

Exercise 28.54

A toroidal solenoid with turns of wire and a mean radius of carries a current of . The relative

permeability of the core is .

Part A

What is the magnetic field in the core?

ANSWER:

=

2.67×10−2 Correct

Part B

What part of the magnetic field is due to atomic currents?

ANSWER:

=

2.63×10−2 Correct

Exercise 28.56

The current in the windings of a toroidal solenoid is 2.900 . There are 500 turns and the mean radius is 28.00 . The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be

1.820 .

Part A

Calculate the relative permeability. Express your answer using four significant figures.

ANSWER:

=

1757 Correct

Part B

Calculate the magnetic susceptibility of the material that fills the toroid. Express your answer using four significant figures.

ANSWER:

=

1756 Correct

Problem 28.84

Long, straight conductors with square cross section, each carrying current , are laid side-by-side to form an infinite current sheet with current directed out of the plane of the page (see the figure

). A second infinite current sheet is a distance below the first and is parallel to it. The second sheet carries current into the plane of the page. Each sheet has conductors per unit length.

Part A

Calculate the magnitude of the net magnetic field at point P (above the upper sheet).

Express your answer in terms of the variables , , , and appropriate constants.

ANSWER:

=

0 Correct

Part B

Find the direction of the net magnetic field at point P (above the upper sheet).

ANSWER:

to the left

to the right

no field

Correct

Part C

Calculate the magnitude of the net magnetic field at point R (midway between the two sheets).

Express your answer in terms of the variables , , , and appropriate constants.

ANSWER:

=

Correct

Part D

Find the direction of the net magnetic field at point R (midway between the two sheets).

ANSWER:

to the left

to the right

no field

Correct

Part E

Calculate the magnitude of the net magnetic field at point S (below the lower sheet).

Express your answer in terms of the variables , , , and appropriate constants.

ANSWER:

=

0 Correct

Part F

Find the direction of the net magnetic field at point S (below the lower sheet).

ANSWER:

to the left

to the right

no field

Correct