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The Laws of Biot-Savart & Ampere

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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

"Brute force"

"High symmetry"

- 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"

- Two ways to calculate the Magnetic Field:
- Biot-Savart Law:

- Ampere's Law

- These are the analogous equations for the Magnetic Field!

dl

X

r

q

dB

I

(~1819)

So, the magnetic field “circulates” around the wire

P

r

R

q

x

I

dx

Þ

Þ

\

Þ

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

- Rewrite in terms of R,q:

Which way is B?

P

r

R

q

x

I

dx

Þ

\

I

R

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

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

(a) B = 0

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

I

r

Idx

R

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

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

(a) B = 0

- 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)

dl

I

R

- Evaluate line integral in Ampere’s Law:

- Apply Ampere’s Law:

Þ

- 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 )

y

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?

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

y

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

- 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!

y

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?

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!!

F

´

Ib

d

L

Ia

Þ

Force on b =

Ib

d

L

Ia

´

Þ

Force on a =

F

- 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:

I

y

I

(a) Fx < 0

(c) Fx > 0

(b) Fx = 0

x

- 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?

I

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?

- 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

•

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

Integral around a path … hopefully a simple one

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

x x x x x

a

r

Þ

Þ

- 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:

a

B

r

- Inside the wire: (r < a)

- Outside the wire: (r>a)

I

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)

- 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

I

a

3a

2a

3a

1A

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

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

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

- 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

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)

- 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:

y

x

x

x

x

x

x

x

x

x

x

x

x

x

constant

constant

•

•

\

Þ

- 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:

L

- 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

- 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

x

•

•

•

•

•

Þ

- 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 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 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

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

Þ

- Expressed in terms of the magnetic moment:

note the typical dipole field behavior!

Þ

2A

(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

- 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?

2A

- 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!

2A

2B

- 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!

1

»

z

3

R

B

z

0

0

z