1 / 34

# the laws of biot-savart ampere - PowerPoint PPT Presentation

´. dl. I. Magnetism. The Laws of Biot-Savart & Ampere. Overview of Lecture. Fundamental Law for Calculating Magnetic Field Biot-Savart Law (“brute force”) Ampere’s Law (“high symmetry”) Example: Calculate Magnetic Field of ¥ Straight Wire from Biot-Savart Law from Ampere’s Law

Related searches for the laws of biot-savart ampere

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'the laws of biot-savart ampere' - omer

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

dl

I

Magnetism

The Laws of Biot-Savart & Ampere

• Fundamental Law for Calculating Magnetic Field

• Biot-Savart Law (“brute force”)

• Ampere’s Law (“high symmetry”)

• Example: Calculate Magnetic Field of ¥ Straight Wire

• from Biot-Savart Law

• from Ampere’s Law

• Calculate Force on Two Parallel Current-Carrying Conductors

Text Reference: Chapter 30.1-4

"High symmetry"

Calculation of Electric Field

• Two ways to calculate the Electric Field:

• Coulomb's Law:

• Gauss' Law

• What are the analogous equations for the Magnetic Field?

"Brute force"

I

"High symmetry"

Calculation of Magnetic Field

• Two ways to calculate the Magnetic Field:

• Biot-Savart Law:

• Ampere's Law

• These are the analogous equations for the Magnetic Field!

X

r

q

dB

I

Biot-Savart Law…bits and pieces

(~1819)

So, the magnetic field “circulates” around the wire

r

R

q

x

I

dx

Þ

Þ

\

Þ

Magnetic Field of ¥ Straight Wire

• Calculate field at point P using Biot-Savart Law:

• Rewrite in terms of R,q:

Which way is B?

r

R

q

x

I

dx

Þ

\

Magnetic Field of ¥ Straight Wire

R

(c) B = (m0I)/(2pR)

(b) B = (m0I)/(2R)

(a) B = 0

Lecture 14, ACT 1

• What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I?

r

Idx

R

(c) B = (m0I)/(2pR)

(b) B = (m0I)/(2R)

(a) B = 0

Lecture 14, ACT 1

• What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I?

• To calculate the magnetic field at the center, we must use the Biot-Savart Law:

• Two nice things about calculating B at the center of the loop:

• Idxis always perpendicular to r

• r is a constant (=R)

I

R

• Evaluate line integral in Ampere’s Law:

• Apply Ampere’s Law:

Þ

Magnetic Field of ¥ Straight Wire

• Calculate field at distance R from wire using Ampere's Law:

´

• Choose loop to be circle of radius R centered on the wire in a plane ^ to wire.

• Why?

• Magnitude of B is constant (fcn of R only)

• Direction of B is parallel to the path.

• Current enclosed by path =I

• Ampere's Law simplifies the calculation thanks to symmetry of the current! ( axial/cylindrical )

a

x

x

x

b

2I

x

x

I

x

x

x

(c) Bx(a) > 0

(b) Bx(a) = 0

(a) Bx(a) < 0

x

(c) Bx(b) > 0

(b) Bx(b) = 0

(a) Bx(b) < 0

Lecture 14, ACT 2

• A currentI flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current 2I in the -z direction.

• What is the magnetic field Bx(a) at point a, just outside the cylinder as shown?

• What is the magnetic field Bx(b) at point b, just inside the cylinder as shown?

a

x

x

x

b

2I

x

x

I

x

x

x

(c) Bx(a) > 0

(b) Bx(a) = 0

(a) Bx(a) < 0

x

B

x

B

I

B

B

Lecture 14, ACT 2

• A currentI flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current 2I in the -z direction.

• What is the magnetic field Bx(a) at point a, just outside the cylinder as shown?

• This situation has massive cylindrical symmetry!

• Applying Ampere’s Law, we see that the field at point a must just be the field from an infinite wire with current I flowing in the -z direction!

a

x

x

x

b

2I

x

x

I

x

x

x

(c) Bx(a) > 0

(b) Bx(a) = 0

(a) Bx(a) < 0

x

(c) Bx(b) > 0

(b) Bx(b) = 0

(a) Bx(b) < 0

• A currentI flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current 2I in the -z direction.

• What is the magnetic field Bx(a) at point a, just outside the cylinder as shown?

Lecture 14, ACT 2

What is the magnetic field Bx(b) at point b, just inside the cylinder as shown?

• Just inside the cylinder, the total current enclosed by the Ampere loop will be I in the +z direction!

• Therefore, the magnetic field at b will just be minus the magnetic field at a!!

´

Ib

d

L

Ia

Þ

Force on b =

Ib

d

L

Ia

´

Þ

Force on a =

F

Force on 2 ParallelCurrent-Carrying Conductors

• Calculate force on length L of wire b due to field of wire a:

• The field at b due to a is given by:

• Calculate force on length L of wire a due to field of wire b:

The field at a due to b is given by:

y

I

(a) Fx < 0

(c) Fx > 0

(b) Fx = 0

x

Lecture 14, ACT 3

• A current I flows in the positive y direction in an infinite wire; a current I also flows in the loop as shown in the diagram.

• What is Fx, net force on the loop in the x-direction?

Ftop

y

Fright

X

Fleft

I

Fbottom

(a) Fx < 0

(c) Fx > 0

(b) Fx = 0

x

• The direction of the magnetic field at the current loop is in the -z direction.

• The forces on the top and bottom segments of the loop DO indeed cancel!!

• A current I flows in the positive y direction in an infinite wire; a current I also flows in the loop as shown in the diagram.

• What is Fx, net force on the loop in the x-direction?

Lecture 14, ACT 3

• You may have remembered from a previous ACT that the net force on a current loop in a constant magnetic field is zero.

• However, the magnetic field produced by the infinite wire is not a constant field!!

• The forces on the left and right segments of the loop DO NOT cancel!!

• The left segment of the loop is in a larger magnetic field.

• Therefore, Fleft > Fright

x

dB

r

q

x

R

x

q

x

z

x

z

x

R

r

x

dB

x x x x x

x x x x x x x x

x x x x x x x x x

x x x x x x x x

x x x x x

x

x

x

x

x

a

r

Examples of Magnetic Field Calculations

• Calculate Magnetic Fields

• Inside a Long Straight Wire

• Infinite Current Sheet

• Solenoid

• Toroid

• Circular Loop

Text Reference: Chapter 30.1-5

Current “enclosed” by that path

´

I

Today is Ampere’s Law Day

"High symmetry"

x x x x x x x x

x x x x x x x x x

x x x x x x x x

x x x x x

a

r

Þ

Þ

B Field Insidea Long Wire

• Suppose a total current I flows through the wire of radius a into the screen as shown.

• Calculate B field as a fcn of r, the distance from the center of the wire.

• Bfield is only a fcn of rÞ take path to be circle of radius r:

• Current passing through circle:

• Ampere's Law:

B

r

B Field of aLong Wire

• Inside the wire: (r < a)

• Outside the wire: (r>a)

I

a

3a

2a

3a

1A

(a) BL(6a)< BR(6a)

(b) BL(6a)= BR(6a)

(c) BL(6a)> BR(6a)

• What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)?

1B

(a) BL(2a)< BR(2a)

(b) BL(2a)= BR(2a)

(c) BL(2a)> BR(2a)

Lecture 15, ACT 1

• Two cylindrical conductors each carry current I into the screen as shown.The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a and R=3a.

• What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)?

I

a

3a

2a

3a

1A

(a) BL(6a)< BR(6a)

(b) BL(6a)= BR(6a)

(c) BL(6a)> BR(6a)

Lecture 15, ACT 1

• Two cylindrical conductors each carry current I into the screen as shown.The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a and R=3a.

• What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)?

• Ampere’s Law can be used to find the field in both cases.

• The Amperian loop in each case is a circle of radius R=6a in the plane of the screen.

• The field in each case has cylindrical symmetry, being everywhere tangent to the circle.

• Therefore the field at R=6a depends only on the total current enclosed!!

• In each case, a total current I is enclosed.

I

a

3a

2a

3a

1A

(a) BL(6a)< BR(6a)

(b) BL(6a)= BR(6a)

(c) BL(6a)> BR(6a)

• What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)?

1B

(a) BL(2a)< BR(2a)

(b) BL(2a)= BR(2a)

(c) BL(2a)> BR(2a)

Lecture 15, ACT 1

• Two cylindrical conductors each carry current I into the screen as shown.The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a and R=3a.

• What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)?

• Once again, the field depends only on how much current is enclosed.

• For the LEFT conductor:

• For the RIGHT conductor:

x

x

x

x

x

x

x

x

x

x

x

x

x

constant

constant

\

Þ

B Field of ¥ Current Sheet

• Consider an ¥ sheet of current described by n wires/length each carrying current i into the screen as shown. Calculate the B field.

• What is the direction of the field?

• Symmetry Þ y direction!

• Calculate using Ampere's law using a square of side w:

• A solenoid is defined by a current I flowing through a wire which is wrapped n turns per unit length on a cylinder of radius a and length L.

a

B Field of a Solenoid

• A constant magnetic field can (in principle) be produced by an ¥ sheet of current. In practice, however, a constant magnetic field is often produced by a solenoid.

• Ifa <<L, the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law.

x

x

x

x

x

x

x

x

x

Þ

B Field of a ¥ Solenoid

• To calculate the B field of the ¥ solenoid using Ampere's Law, we need to justify the claim that the B field is 0 outside the solenoid.

• To do this, view the ¥ solenoid from the side as 2 ¥ current sheets.

• The fields are in the same direction in the region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid).

• Draw square path of side w:

x

x

x

x

x

x

x

x

x

r

x

x

x

x

x

x

x

B

Þ

Toroid

• Toroid defined by N total turns with current i.

• To find B inside, consider circle of radiusr, centered at the center of the toroid.

• B=0 outside toroid! (Consider integrating B on circle outside toroid)

Apply Ampere’s Law:

dB

r

q

R

q

z

z

R

r

dB

x

Circular Loop

• Circular loop of radius R carries current i. Calculate B along the axis of the loop:

• Magnitude of dB from element dl:

• What is the direction of the field?

• Symmetry Þ B in z-direction.

Þ

dB

r

q

R

q

z

z

R

r

dB

x

Circular Loop

• Note the form the field takes for z>>R:

Þ

• Expressed in terms of the magnetic moment:

note the typical dipole field behavior!

Þ

(c) Bz(A) > 0

(b) Bz(A) = 0

(a) Bz(A) < 0

• What is the magnetic field Bz(B) at point B, just to the right of the right loop?

2B

(c) Bz(B) > 0

(b) Bz(B) = 0

(a) Bz(B) < 0

Lecture 15, ACT 2

• Equal currentsI flow in identical circular loops as shown in the diagram. The loop on the right(left) carries current in the ccw(cw) direction as seen looking along the +z direction.

• What is the magnetic field Bz(A) at point A, the midpoint between the two loops?

Lecture 15, ACT 2

• Equal currentsI flow in identical circular loops as shown in the diagram. The loop on the right(left) carries current in the ccw(cw) direction as seen looking along the +z direction.

• What is the magnetic field Bz(A) at point A, the midpoint between the two loops?

(c) Bz(A) > 0

(b) Bz(A) = 0

(a) Bz(A) < 0

• The right current loop gives rise to Bz <0 at point A.

• The left current loop gives rise to Bz >0 at point A.

• From symmetry, the magnitudes of the fields must be equal.

• Therefore, B(A) = 0!

2B

Lecture 15, ACT 2

• Equal currentsI flow in identical circular loops as shown in the diagram. The loop on the right(left) carries current in the ccw(cw) direction as seen looking along the +z direction.

• What is the magnetic field Bz(A) at point A, the midpoint between the two loops?

(c) Bz(A) > 0

(b) Bz(A) = 0

(a) Bz(A) < 0

• What is the magnetic field Bz(B) at point B, just to the right of the right loop?

(c) Bz(B) > 0

(b) Bz(B) = 0

(a) Bz(B) < 0

• The signs of the fields from each loop are the same at B as they are at A!

• However, point B is closer to the right loop, so its field wins!

»

z

3

Circular Loop

R

B

z

0

0

z