Physics 2112 Unit 15

Physics 2112

Unit 15

Today’s Concept:

Ampere’s Law

=• enclosedoIdB

Unit 15, Slide 1

Electricity & Magnetism Lecture 15, Slide 2

R

IB o

2=

We know for an infinite current

carrying wire

Ampere’s Law

IRB o = 2

IdB o=•

But what is 2R? Circumference of circle!


Ampere’s Law

ENCLo IdB =•

Any closed loop

Current enclosed by that closed

loop

Checkpoint 1A


Two loops are placed near identical current carrying wires as

shown in Case 1 and Case 2 below.

For which loop is ∫B·dl greater?

A. Case 1

B. Case 2

C. The integral is the same for

both

Checkpoint 1B


Two loops are placed near identical current carrying wires as

shown in Case 1 and Case 2 below.


A. Case 1

B. Case 2


both

Checkpoint 1C


Two loops are placed near current carrying wires as shown in

Case 1 and Case 2 below. In both cases the direction of the

current in the two wires are opposite to each other.


A. Case 1

B. Case 2


both

Example 15.1 (B field from a thick wire)

Unit 15, Slide 7

A wire with a radius of r=1cm has a

uniform current of 1A flowing

through it.

What is the B field 1 meter from the

center of the wire?

What is the B field 0.5cm from the

center of the wire?

Example 15.1 (graph)

Unit 15, Slide 8

A wire with a radius

of r=1cm has a

uniform current of 1A

flowing through it.

X

X X X

X X X X X

X

X XX

X XXXXB

r

CheckPoint 2A


XAn infinitely long hollow

conducting tube carries

current I in the direction

shown.

What is the direction of the magnetic

field inside the tube?

A. clockwise

B. counterclockwise

C. radially inward to the center

D. radially outward from the center

E. the magnetic field is zero

Why is that?


XX

Example 15.3 (Pipe of current)

An infinitely long cylindrical shell carries a uniformly distributed current of 5A out of the screen. The inner radius is a=4cm and outer radius=8cm

What is B at r = 2cm?What is B at r = 6cm?What is B at r = 16cm?

y

x

b

aI


Example 15.3 (Pipe of current)

➢ Conceptual AnalysisComplete cylindrical symmetry (can only

depend on r) can use Ampere’s law to calculate B

➢ Strategic Analysis

Calculate B for the three regions separately:

1) r < a

2) a < r < b

3) r > b


For circular path concentric with shell.

encoIdB =•

encoIdB =

Question


y

xba

Ir

encoIdB =•

0

so 0=B

What does |B| look like for r < a?

A B C

y

x

Question


ba

Ir

B

dl

A B C

What does |B| look like for r < b?

LHS: rBdBBddB 2===•

RHS: IIenclosed =

r

IB o

2=

Example Problem


y

x

What is the current density j (Amp/m2) in the conductor?

ba

I

A) B) C) 2b

Ij

=

22 ab

Ij

+=

22 ab

Ij

−=

j = I / area 22 abarea −=

22 ab

Ij

−=

Example Problemy

xba

Ir

What does |B| look like for a < r < b ?

A B C

Starts at 0 and increases almost linearly

)()(

2 22

22ar

ab

IrB o −

−=


encoIdB =•

enco jArB = 2

)(

)(

2 22

22

ab

ar

r

IB o

−

−=

An infinitely long cylindrical shell with inner radius a and outer radius b carries a uniformly distributed current I out of the screen.

Sketch |B| as a function of r.

Follow upy

x

b

aI


Follow-Up


y

x

b

a I

Add an infinite wire along the z axiscarrying current I0. (I is out of screen.)

What must be true about I0 such that there is some value of r, a < r < b, such that B(r) = 0 ?

I0

A) |I0| > |I| AND I0 into screen

B) |I0| > |I| AND I0 out of screen

C) |I0| < |I| AND I0 into screen

D) |I0| < |I| AND I0 out of screen

E) There is no current I0 that can produce B = 0 there

X

Example 15.2 (Two thick wires)

Unit 15, Slide 19

X

Two thick cables both of

radius R and length L carry

a current, I, side-by-side as

shown to the left. The left

cable has the current into

the screen and the right

cable has the current out of

the screen.

A B C

What is the magnetic field at points A, B, C and D?

- A is at the center of the left cable

- B is a distance R/2 from the center of the left cable

- C is at the point where the cables touch.

- D is the bottom of the left cable

D

The Plan

Unit 15, Slide 20

XA B

C

What is the magnetic field at

points A, B, C and D?

D

➢ Conceptual Analysisuse Ampere’s law to calculate B from each cable separately

➢ Strategic AnalysisNote direction of B from each cable at each point.

Add B like vectors

B Field in Solenoid

Unit 15, Slide 21

X X X X

. ...

~cancel out

add up

B Field in Solenoid

Unit 15, Slide 22

. ...

X X XX

. ...

X X XX

encoIdB =•

L

ILnLB ox ****000 =+++

nIB ox =n = turns per unit length (Ideal Solenoid L>>>r)

Unit 15, Slide 23

Solenoid → Bar Magnet

Use the right hand rule and curl your fingers along the direction of the current.

CheckPoint 2B


A current carrying wire is wrapped around cardboard tube

as shown below.

In which direction does the

magnetic field point inside the

tube?

A. left

B. right

C. up

D. down

E. out of the screen

F. into the screen

Unit 15, Slide 25

Example 15.3 (B from Solenoid)

A long thin solenoid consists of

500 turns of wire carrying 1.5A. It

has a length of 25cm and a

radius of 1cm.

What is the magnetic field in the exact center of the coil?

What is the contribution to this field from the 250th coil

(the center coil)?

Make Sense?

Unit 15, Slide 26

. ...

X X XX

. ...

X X XX

Note:

• B250/Btot = 0.025 (1/500 = 0.002)

• So closest coil contributes over 10 tens more than

average coil. Make sense?

Unit 15, Slide 27

Ampere’s Law wrong?

X X XX

X X XX

Consider 𝑩 ∙ 𝑑റ𝑙 inside

the solenoid.

No current in enclosed

and yet there is a B

field there.

X X XX

X X XX

Forces on Loop?

Unit 15, Slide 28

X X XX

X X XX Consider a single loop of

the solenoid.

X X XX

Current moving in B field

caused by other loops.

What do the forces on the

loop look like?

Zero (no

forces)

(A) (B) (C) (D)

Unit 15, Slide 29

When does “Loop” → “Solenoid”?


http://www.hep.uiuc.edu/home/mats/flash/BFP.html

Unit 15, Slide 31

The mathematical way for indicating that you’re

going to add up the strength of an electric field

in a closed loop, is

• ldE

What is this integral equal to?

A. oqencl

B. qencl/eo

C. Iencl/eo

D. zero

E. the value varies from situation to situation

Unit 15, Slide 32

The mathematical way for indicating that you’re

going to add up the strength of an magnetic

field in a closed surface, is

• AdB

What is this integral equal to?

A. oIencl

B. Iencl/eo

C. Iencl/eo

D. zero

E. the value varies from situation to situation

A hint…..

Unit 15, Slide 33

=•surface

AdB 0

ENCLo

loop

IldB =•

0=•loop

ldE

o

enc

surface

QAdE

e== 0

Physics 2112 Unit 15

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