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´. 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

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the laws of biot savart ampere

´

dl

I

Magnetism

The Laws of Biot-Savart & Ampere

overview of lecture
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
  • Calculate Force on Two Parallel Current-Carrying Conductors

Text Reference: Chapter 30.1-4

calculation of electric field

"Brute force"

"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?
calculation of 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!
biot savart law bits and pieces

dl

X

r

q

dB

I

Biot-Savart Law…bits and pieces

(~1819)

So, the magnetic field “circulates” around the wire

magnetic field of straight wire

P

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?

lecture 14 act 1

I

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?
lecture 14 act 19

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)
magnetic field of straight wire10

dl

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 )
lecture 14 act 2

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

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?
lecture 14 act 212

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

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!
lecture 14 act 213

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?
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!!
force on 2 parallel current carrying conductors

F

´

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:

lecture 14 act 3

I

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?
lecture 14 act 316

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?
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
examples of magnetic field calculations

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

overview of lecture18
Overview of Lecture
  • Calculate Magnetic Fields
    • Inside a Long Straight Wire
    • Infinite Current Sheet
    • Solenoid
    • Toroid
    • Circular Loop

Text Reference: Chapter 30.1-5

slide19

Integral around a path … hopefully a simple one

Current “enclosed” by that path

´

I

Today is Ampere’s Law Day

"High symmetry"

b field inside a long wire

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

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 field of a long wire

a

B

r

B Field of aLong Wire
  • Inside the wire: (r < a)
  • Outside the wire: (r>a)
lecture 15 act 1

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)

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)?
lecture 15 act 123

I

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.
lecture 15 act 124

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)

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:
b field of current sheet

y

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:
b field of a solenoid

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

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.
b field of a solenoid27

x

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:
toroid

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:

circular loop

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

circular loop30

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

lecture 15 act 2

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

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 232

2A

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!
lecture 15 act 233

2A

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!
circular loop34

1

»

z

3

Circular Loop

R

B

z

0

0

z