Lesson 8 Ampère’s Law and Differential Operators

Lesson 8 Ampère’s Law and Differential Operators

Section 1 Visualizing Ampère’s Law

Amperian Loop An Amperian loop is Amperian loops include:
any closed loop Amperian loops include: a circle a square a rubber band Amperian loops do not include: a balloon a piece of string (with two ends)

The Field Contour of a Wire
Take a wire with current coming out of the screen.

The field contour is made of half-planes centered on the wire.

We draw arrows on each plane pointing in the direction of the magnetic field.

Amperian Loops We draw an Amperian loop around the wire.

Amperian Loops We wish to count the “net number” of field lines pierced by the Amperian loop.

Amperian Loops First, we put an arrow on the loop in an overall counterclockwise direction.

Counting Surfaces Pierced
To count the net number of surfaces pierced by Amperian loop, we add +1 when the loop is “in the direction’ of the plane and −1 when it is “opposite the direction” of the plane. +1 +1 +1 +1 +1 −1 +1 +1 +1 −1 +1 +1 −1 +1 +1 +1 +1 +1 −1 +1 +1 +1 +1 +1

Note there are “+1” appears 20 times and “-1” appears 4 times. +1 +1 +1 +1 +1 −1 +1 +1 +1 −1 +1 +1 −1 +1 +1 +1 +1 +1 −1 +1 +1 +1 +1 +1

The net number of surfaces pierced by the Amperian loop us therefore +16. +1 +1 +1 +1 +1 −1 +1 +1 +1 −1 +1 +1 −1 +1 +1 +1 +1 +1 −1 +1 +1 +1 +1 +1

Other Amperian Loops What is the net number of surfaces pierced by each of these Amperian loops?

Other Amperian Loops The net numbers of surfaces pierced by each of these loops is 16.

Other Amperian Loops What is the net number of surfaces pierced by this Amperian loop?

Other Amperian Loops This time the net number of surfaces pierced by the loop is 0. Why?

Other Wires. This is the same loop we saw earlier, but now only 8 surfaces are pierced, since there are only 8 surfaces extending outward from the wire.

Other Wires. There are 8 surfaces coming from the wire because the current through the wire is half as much as it was before.

Ampère’s Law The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the loop.

Sign Convention Always traverse the Amperian loop in a (generally) counterclockwise direction. If the number of surfaces pierced N>0, the current comes out of the screen. If N<0, the current goes into the screen.

Section 2 Cylindrically Symmetric Current Density

Current Density There are three kinds of charge density (ρ,σ,λ)
There is one kind of current density (current/unit area)

Current Density The current passing through a small gate of area ΔA is

Current Density The total current passing through the wire is the sum of the current passing through all small gates. small gates

Cylindrically Symmetric Current Distribution
The current density, j, can vary with r only. Below, we assume that the current density is greatest near the axis of the wire.

Outside the distribution, the field contour is composed of surfaces that are half planes, uniformly spaced.

Inside the distribution, it is difficult to draw perpendicular surfaces, as some surfaces die out as we move inward. – We need to draw many, many surfaces to keep them equally spaced as we move inward.

But we do know that if we draw enough surfaces, the distribution of the surfaces will be uniform, even inside the wire.

Let’s draw a circular Amperian loop at radius r. r

Now we split the wire into two parts – the part outside the Amperian loop and the part inside the Amperian loop. r r

The total electric field at r will be the sum of the electric fields from the two parts of the wire. r r

Inside a Hollow Wire The total number of perpendicular surfaces pierced by the Amperian loop is zero because there is no current passing through it. r

How can we get zero net surfaces?
1. We could have all the surfaces pierced twice, one in the positive sense and one in the negative… … but this violates symmetry!

2. We could have some surfaces oriented one way and some the other… … but this violates symmetry, too!

3. Or we could just have no surfaces at all inside the hollow wire. This is the only way it can be done!

The Magnetic Field inside a Hollow Wire
If the current distribution has cylindrical symmetry, the magnetic field inside a hollow wire must be zero.

Since the magnetic field inside a hollow wire is zero, the total magnetic field at a distance r from the center of a solid wire is the field of the “core,” the part of the wire within the Amperian loop. r r

Outside the core, the magnetic field is the same as that of a thin wire that has the same current as the total current passing through the Amperian loop. r

Inside a cylindrically symmetric current distribution, the magnetic field is: r

Section 3 Uniform Current Density

Example: Uniform Current Distribution
A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ?

A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? r

A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? r The current density is uniform, so:

A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? r Therefore:

Section 4 The Line Integral

Line Integral We know that the magnetic field is stronger where the perpendicular surfaces are closer together. Therefore, the number of surfaces pierced is Let

Line Integral Therefore, the number of surfaces pierced is
Λ is called the “line integral” The line integral is proportional to the number of contours pierced. Λ=Bℓ if ℓ is a section of a field line and B is constant on ℓ.

Line Integral Λ is called the line integral because it is more generally given by the expression

Line Integral Note that this is similar to expression for work you learned in mechanics. Work is the line integral of force along the path an object follows.

Line Integral The line integral is a way of measuring the number of contour surfaces pierced by a line segment.

Line Integral Roughly speaking, it is a measure of how much field lines along a path.

Ampère’s Law and the Line Integral
The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the Amperian loop.

Ampère’s Law and the Line Integral
The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the Amperian loop. Therefore:

Class 23 Today, we will use Ampere’s law to find the magnetic fields
inside and outside a long, straight wire with radial charge density of a plane of wires of a solenoid of a torus

Section 5 Applying Ampère’s Law

Ampère’s Law – the Practical Version

The number of surfaces pierced

The number of surfaces pierced The magnetic field on the Amperian loop – must be a constant over the whole loop.

The number of surfaces pierced The length of a the closed Amperian loop (or the part where the field is non-zero). The magnetic field on the Amperian loop – must be a constant over the whole loop.

The total current passing through the Amperian loop. The number of surfaces pierced The length of a the closed Amperian loop (or the part where the field is non-zero). The magnetic field on the Amperian loop – must be a constant over the whole loop.

This is NOT the line integral This is the line integral

Section 6 Ampère’s Law and Cylindrical Wires

A Typical Problem A wire of radius R has current density Find the magnetic field inside the wire. “Inside the wire” means at some point P at a radius r < R. P r R

Choosing the Amperian Loop
What shape of Amperian Loop should we choose for current traveling through a cylindrical wire?

Choosing the Amperian Loop
Choose a circular Amperian loop. r

A wire of radius R has current density
A wire of radius R has current density Find the magnetic field inside the wire. What is the correct expression for the ℓ in the line integral?

Integrating the Current Density
How do we do slice the region inside the Amperian loop to integrate the current density?

Integrating the Current Density
We slice the wire into rings. The current though each ring is dI = j dA

A wire of radius R has current density Find the magnetic field inside the wire. What is the correct expression for ?

A wire of radius R has current density Find the magnetic field inside the wire.

Another Problem A wire of radius R has current density Find the magnetic field outside the wire. “Outside the wire” means at some point P at a radius r > R. P r R

A wire of radius R has current density Find the magnetic field outside the wire

Section 7 Other Applications of Ampère’s Law

Ampere’s Law – Plane of Wires
The magnetic field from each wire forms circular loops around the wire. Therefore is perpendicular to . B1 1 2 3 P

Consider the magnetic field from three wires. B2 B3 B1 d 1 2 3 P

B2 B3 B1 d When we add the magnetic fields from each wire, the vector sum points upward. 1 2 3 P

Now let’s look at the line integral of the magnetic field around the dotted Amperian loop.

Now let’s look at the line integral of the magnetic field around the dotted Amperian loop. d

The bottom part of the Amperian loop pierces no contour surfaces – so the line integral here is zero. d We also know but the B field and the path are perpendicular on this segment so Λ=0 for this part of the path. Λ=0 for this top of the path, too.

Conclusion: The line integral over the top and bottom segments of the Amperian loop is zero because the magnetic field is perpendicular to the path. d

The line integral over the right side of the Amperian loop is d

The line integral over the left side of the Amperian loop is d

The total line integral is: d The enclosed current is the number of wires in the loop times the current in each wire:

d n is the number of wires per unit length.

Ampere’s Law – Two Planes of Wires
What would the field be like if there were two planes of wires with currents in opposite directions?

Field lines

Contour surfaces What can you conclude about the magnetic field?

Field lines of the right plane

Field lines of the left plane

The field of the both planes

What would the field be like if there were two planes of wires with currents in opposite directions? d 1 2 3

Ampere’s Law – Solenoid
A solenoid is similar to two planes of wires. The magnetic field inside this solenoid points downward. The magnetic field outside the solenoid is zero. 1 2 3 d

Ampere’s Law – Solenoid
1 2 3 d

Right-hand Rule #3 The direction of the magnetic field
inside a solenoid is given by a right-hand rule: Circle the fingers of your right hand in the direction of the current. The magnetic field is in the direction of your thumb.

Ampere’s Law – Torus r A torus is like a solenoid wrapped around a
doughnut-shaped core. The magnetic field inside forms circular loops. The magnetic field outside is zero.

Ampere’s Law – Torus r

Ampere’s Law – Torus r The wires on the inside of the torus are closer
together, so the field is stronger there.

Class 24 Today, we will use direct integration to find
electric fields of charged rods and loops electric potentials of charged rods and loops magnetic fields of current-carrying wire segments and loop segments (Biot-Savart law)

Section 8 Finding Fields by Direct Integration

Geometry for Extended Objects
origin to source P origin to P source to P

The Basic Laws The electric field and potential of a small charge dq:
The magnetic field of a current i in a small length of wire :

Origin of the Basic Laws
Electric field and potential for slowly moving point charges – Coulomb’s Law:

Electric field and potential for dq.
The Basic Laws – for dq Electric field and potential for dq.

Remember the geometry from Lesson 2 motion of source U θ head line thread x y P tail line S T ray line ψ

An expression for the magnetic field of a point charge:
Origin of the Basic Laws An expression for the magnetic field of a point charge:

An expression for the magnetic field of a point charge (moving slowly):

Magnetic field for dq A little sleight of hand, but it’s the same as a more formal proof.

The Biot-Savart Law The formula for the magnetic field of a wire segment A current i passes through the wire segment The length of the wire segment is dℓ. The direction of dℓ is the direction of the current. is the vector from the origin to a field point. is the vector from the segment (the origin) to a field point.

Equations for Extended Objects
P

Now, let’s work some problems…

A Charged Rod y P x +L −L The rod has a linear charge density
Find the electric field at P.

A Charged Rod y P x −L +L

A Charged Rod P y −L +L 1. Find

A Charged Rod P −L +L Find Choose a slice and find

A Charged Rod P +L −L Find Choose a slice and find
Be sure to put primes on the slice variables!

A Charged Rod P +L −L Find Choose a slice and find
Find the length and charge of the slice.

A Charged Rod Now just substitute! Find Choose a slice and find

A Charged Rod Find Choose a slice and find

A Charged Rod I won’t expect you to evaluate these integrals!

A Charged Rod y P x +L −L The rod has a linear charge density
Now find the electric potential at P.

A Charged Rod Find Choose a slice and find

A Charged Rod You don’t need to evaluate this integral, either!

Current in a Wire Segment
y P i x −L +L Current i travels to the left along a segment of wire. Find the magnetic field at P.

P y i −L +L 1. Find

P i −L +L Find Choose a slice and find

P i −L +L Find Choose a slice and find Be sure to put primes on the slice variables!

P i −L +L Find Choose a slice and find

Remember:

Find Choose a slice and find

A Charged Loop Segment y x a P The rod has a linear charge density
Find the electric field at P.

A Charged Loop Segment y x a P Find

A Charged Loop Segment y x a P Find Slice and find

A Charged Loop Segment y x a P Find Slice and find
Find the charge of the slice.

A Charged Loop Segment Find Slice and find
Find the charge of the slice.

A Charged Loop Segment Here we only have to integrate sines and cosines, so the integrals are easy!

A Segment of a Current Loop
i y x a P Current i travels counterclockwise along a segment of a loop of wire. Find the magnetic field at P.

i y x a P Find

i y x a P Find Slice and find

Find Slice and find

This is a really easy integral this time!

Class 25 Today, we will: learn the definition of divergence in terms of flux. learn the definition of curl in terms of the line integral. find the gradient, divergence, and curl in terms of derivatives (differential operators) write Gauss’s laws and Ampere’s law in differential form work several sample problems

Section 9 Gauss’s Law and Divergence

Gauss’s Law The net number of electric field lines passing through a Gaussian surface is proportional to the charge enclosed.

Gauss’s Law This is true no matter how small the Gaussian surface is. But the number of field lines gets smaller as the volume gets smaller.

Define divergence to be

Divergence Divergence is a scalar field – a scalar defined at every point in space – that tells us if diverging (or converging) field lines are being produced at that point. The larger the divergence, the more field lines are produced.

Divergence and Gauss’s Law
In a very small volume, charge density is nearly constant. (That is, until we get to the atomic scale where we can start seeing protons and electrons.)

In a very small volume, charge density is nearly constant.

Gauss’s Law in Differential Form

Electrical charge produces field lines that tend to spread from (or converge to) a point in space. Divergence is a measure of how much field lines spread from (+) or converge to (-) a point of space. It is a measurement of “spreadingness.”

Section 10 Ampère’s Law and Curl

Ampère’s Law The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the loop.

Ampere’s Law This is true no matter how small the Amperian loop is. But the number of surfaces pierced gets smaller as the area of the loop gets smaller.

Curl Take a point in space and a line in the x direction passing through the point. The x-component of the curl (around the line) is defined to be: P

Curl Take a point in space and a line in the y direction passing through the point. The y-component of the curl (around the line) is defined to be: P

Curl Take a point in space and a line in the z direction passing through the point. The z-component of the curl (around the line) is defined to be: P

Curl Take a point in space and a line in the x direction passing through the point. The curl (around the line) is defined by:

Curl Curl is a vector field, a vector defined a every point in space. The x component of curl tells us if something at the point is producing field lines that loop about a line going in the x direction and passing through the point.

In a very small area, current density is nearly constant.
Curl and Ampère’s Law In a very small area, current density is nearly constant.

In a very small volume, the density is nearly a constant.
Curl and Ampère’s Law In a very small volume, the density is nearly a constant.

Curl and Ampère’s Law More generally:
The curl points in the direction of the current at any point in space.

Curl and Ampère’s Law More generally:
Curl is a measure of how much field lines ccw (+) or cw to (-) around the direction of the current. It is a measurement of “loopiness.”

Curl and Ampère’s Law Electrical current produces magnetic field lines that form loops around the path of the moving charges.

Section 11 Differential Operators

The Gradient The gradient is a three-dimensional generalization of a slope (derivative). The gradient tells us the direction a scalar field increases the most rapidly and how much it changes per unit distance.

In terms of derivatives, the gradient is:

Electric Field and Electric Potential

The Divergence Operator
We can show that another way of representing divergence is in terms of derivatives.

We can also express the curl in terms of derivatives.
The Curl Operator We can also express the curl in terms of derivatives.

Gauss’s Laws and Ampère’s Law in Differential Form

Some Problems In a region of space, the electric potential is given by the expression Find the electric field.

Some Problems In a region of space, the electric potential is given by the expression Find the charge density.

Some Problems In a region of space, the magnetic field is given by the expression Find β in terms of α.

Some Problems

Some Problems In a region of space, the magnetic field is given by the expression Find the current density (magnitude and direction).

Some Problems

Lesson 8 Ampère’s Law and Differential Operators

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