Physics 2112 Unit 14 · of current carrying wire. ds G. ... Unit 14, Slide 8 4 10 7 Tm / A 0 P Su...

Today’s Concept:

What Causes Magnetic Fields

Physics 2112

Unit 14

2

0ˆ

4 r

rsdIBd

=

Unit 14, Slide 1

Unit 14, Slide 2

2

0ˆ

4 r

rsdIBd

=

2

12

120ˆ

4 r

rvqB

=

Biot-Savart Law

B field from one

moving charge

avgqnAvI =But remember from

previous slides

B field from tiny

of current

carrying wire.

sd

Compare to Electric Fields

Unit 14, Slide 3

2

12

120ˆ

4 r

rvqB

=

2

12

12ˆ

4

1

r

rqE

o=

In the same direction as r12 Perpendicular to r12

v out of the

screen

ATm /104 7

0

−=

x

z

BF

qv

Remembering Directions: The Right Hand Rule

BvqF

=

Unit 12, Slide 4

This method

always

works!

A

BAC

=

A B

C

AB

C

Another way of looking at it…..

Unit 14, Slide 5

Works when something is

going in a circle

What is the magnetic field a distance yo away from a infinitely long wire of current I?

Example 14.1 (Infinite wire of current)

2

0ˆ

4 r

rsdIBd

=

Unit 14, Slide 6

➢ Conceptual Plan

➢ Strategic AnalysisDone in prelecture in detail

Integrate

(Similar to E field for infinite line of charge)

Use Biot-Savart Law

Unit 14, Slide 7

Main Idea

2

0ˆ

4 r

rsdIB

=

)(

)sin(*

4 22

0

oyx

dxI

+=

f

Q

r̂

sd

f

sd

rx

y

Current I OUTr

Magnitude:B

Example 14.1 (answer)

r

IB

2

0=•

Unit 14, Slide 8

ATm /104 7

0

−=

Remember:

rE

o

2=

xy

z

BF

qv

Remembering Directions: The Right Hand Rule

Unit 12, Slide 9

vB

F

2

12

120ˆ

4 r

rvqB

=

BvqF

=

v

B

F

A long straight wire is carrying current from left to right. Two identical charges are moving with equal speed. Compare the magnitude of the force on charge a moving directly to the right, to the magnitude of the force on charge b moving up and to the right at the instant shown (i.e. same distance from the wire).

A) |Fa| > |Fb|

B) |Fa| = |Fb|

C) |Fa| < |Fb|

•B

v

I

v

(a)

r r

(b)

Forces are in different directions (careful with all the angles!)

•F

F

Same q, |v|, B and q (=90)

Question

Unit 14, Slide 10

qsinqvBF =

BvqF

=

Two long wires carry opposite current

What is the direction of the magnetic field above, and midway between the two wires carrying current – at the point marked “X”?

A) Left B) Right C) Up D) Down E) Zero

B

x

Question

Unit 14, Slide 11

x

Example 14.3 (From Loop)

Unit 14, Slide 12

yo

A current, I, flows clockwise

through a circular loop of wire.

The loop has a radius a.

What is the magnetic field at a

point P a distance yo above

the plane of the loop in the

center?a

P

x

x

Q

BcosQ

Q

Question

Unit 14, Slide 13

yo

When doing this integral, how

much of ds is perpendicular to r?

A) all of it

B) ds sinQ

C) ds cosQ

D) ds/sinQ

a

P

x

x

Q

BcosQ

Q

r

ds

Q

Force Between Current-Carrying Wires

BLIF

= 212

12122

Id

LIF o

=

Unit 14, Slide 14

I1 r

IB

2

10=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 X X X X

X X X X X X X X X X

. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . .

I2

F

I towards us • •

Another I towards usF

Conclusion: Currents in same direction attract!

•

I towards us

Another I away from us

F

Conclusion: Currents in opposite direction repel!

dB

Bd

Force Between Current-Carrying Wires

BLIF

= 212 12122

Id

LIF o

=

Unit 14, Slide 15

Example 14.4 (Two Current Carrying Wires)

Two current carrying wires a 10cm apart for a length for 50cm. Wire 1 carries 5A and Wire 2 carries 10A with both current to the left.

What is the magnitude and direction of the force on wire 2due to wire 1?

Unit 14, Slide 16

I1 = 5A

I2 = 10A

10cm

50cm

CheckPoint 1A

What is the direction of the force on wire 2 due to wire 1?

A) Up B) Down C) Into Screen D) Out of screen E) Zero

XB

F

Unit 14, Slide 17

What we thought…..

What is the direction of the force on wire 2 due to wire 1?

A) Up B) Down C) Into Screen D) Out of screen E) Zero

XB

F

Unit 14, Slide 18

A) The current through wire 2 creates a magnetic field pointing out of the screen at wire 1. Using the RHR, the force on wire 1 must be up.

B) One right-hand rule shows that the B field is into the page, so another right-hand rule shows that the force points down.

CheckPoint 1B

What is the direction of the torque on wire 2 due to wire 1?

Uniform force at every segment of wire

No torque about any axis

Unit 14, Slide 19

A) Up B) Down C) Into Screen D) Out of screen E) Zero

CheckPoint 3A

What is the direction of the force on wire 2 due to wire 1?

A) Up B) Down C) Into Screen

D) Out of screen E) Zero

Unit 14, Slide 20

CheckPoint 3B

What is the direction of the torque on wire 2 due to wire 1?

A) Up B) Down C) Into Screen

D) Out of screen E) Zero

Unit 14, Slide 21

LET’S DRAW A PICTURE!

B

B

IF out of screen

Net Force is Zero!

I F into screen

rr

Consider Force on Symmetric Segments

Electricity & Magnetism Lecture 14, Slide 22

But the torque is not!!!

CheckPoint 3B

What is the direction of the torque on wire 2 due to wire 1?

A) Up B) Down C) Into Screen

D) Out of screen E) Zero

Unit 14, Slide 23

Checkpoint 2: Force on a loop

A. the forces are in opposite directions

B. the net forces are the same.

C. the net force on the loop is greater than the net force on the wire segment

D. the net force on the loop is smaller than the net force on the wire segment

E. there is no net force on the loop

A current carrying loop of width a and

length b is placed near a current carrying

wire.

How does the net force on the loop

compare to the net force on a single wire

segment of length a carrying the same

amount of current placed at the same

distance from the wire?

Checkpoint question

Current flows in a loop as shown in

the diagram at the right. The direction

is such that someone standing at

point a and looking toward point b

would see the current flow clockwise.

What is the orientation of the

magnetic field produced by the loop at

points a and b on the axis?

(A)

(B)

(C)

(D)

B on axis from Current Loop

I

Resulting B Field

Current in Wire

Electricity & Magnetism Lecture 14, Slide 26

What about Off-Axis ?

Biot-Savart Works, but need to do numerically

See Simulation!

Unit 14, Slide 27

Two identical loops are hung next to each other. Current flows in the same direction in both.

The loops will:

A) Attract each other B) Repel each other C) There is no force between them

Two Current Loops

Unit 14, Slide 28

Two ways to see this:

1) Like currents attract

N S N S

2) Look like bar magnets

1. ANY CROSS PRODUCT

2. Direction of Magnetic Moment

Thumb: Magnetic Moment

Fingers: Current in Loop

3. Direction of Magnetic Field from WireThumb: Current

Fingers: Magnetic Field

Right Hand Rule Review

BvqF

= BLIF

=

Fr

= B

=

2

0ˆ

4 r

rsdIBd

=

Unit 14, Slide 29

Remembering difference between two formulas

Unit 14, Slide 30

Which one of these situations will

have bigger B field at point P?

I

P

a

aP

I

A) Situation 1 B) Situation 2

C) They will be equal

Remembering difference between two formulas

Unit 14, Slide 31

Which one of these equation will

have bigger B field at a distance a ?

A) equation 1 B) equation 2

C) They will be equal

sin 𝛼 ± sin 𝛽 = 2 sin1

2𝛼 ± 𝛽 cos

1

2𝛼 ∓ 𝛽

𝝁𝒐𝑰

𝟐𝝅𝒂

𝝁𝒐𝑰

𝟐𝒂

Remembering difference between two formulas

Unit 14, Slide 32

𝝁𝒐𝑰

𝟐𝝅𝒂

𝝁𝒐𝑰

𝟐𝒂

I

P

a

aP

I

➢ Conceptual AnalysisEach wire creates a magnetic field at P

B from infinite wire: B = 0I / 2r

Total magnetic field at P obtained from superposition

y

x

.

z

I1 = 1A

Front view Side view

➢ Strategic AnalysisCalculate B at P from each wire separately

Total B = vector sum of individual B fields

I2 = 1A

4cm

4cm

y

P

3cm

Example 14.2

Two parallel horizontal wires are located in the vertical (x,y) plane as shown. Each wire carries a current of I = 1A flowing in the directions shown.

What is the B field at point P?

Unit 14, Slide 33

What is the direction of Bat P?

.

z

y

P

..

z

y

P

.

z

y

P

.90o

Question

Electricity & Magnetism Lecture 14, Slide 34

y

x

.

z

I1 = 1A

Front view Side view

I2 = 1A

4cm

4cm

y

P

3cm

A B C D

z

P

y

Question

Electricity & Magnetism Lecture 14, Slide 35

What is the magnitude of B at P produced by the top current I1? (0 = 4 x 10−7 T−m/A)

A) 4.0 x 10−6 T B) 5.0 x 10−6 T C) 6.7 x 10−6 T

What is r?r = distance from wire axis to P

3cmz

y

.

4cm r

Two parallel horizontal wires are located in the vertical (x,y) plane as shown. Each wire carries a current of I = 1A flowing in the directions shown.

y

x

.

z

I1 = 1A

Front view Side view

I2 = 1A

4cm

4cm

y

P

3cm

r

IB

2

0=

y

x

.

z

I1 = 1A

Front view Side view

I2 = 1A

4cm

4cm

y

P

3cm

Example 14.2

Two parallel horizontal wires are located in the vertical (x,y)

plane as shown. Each wire carries a current of I = 1A

flowing in the directions shown.

What is the B field at point P?

Unit 14, Slide 36

What is the direction of the magnetic field at point P,

which is at the center of a semicircular loop of wire

carrying a current I as shown?

A. Goes in

B. Goes out

C. Goes left

D. Goes right

E. There is no magnetic field at point P.

P

Question

If I = 6A, what is the

magnitude of the

magnetic field at point

P?

Example 14.5 (Curved Loop of Wire)

P

20cm

12cmConceptual Plan

Strategic Analysis

Integrate both loops

Note straight sections cancel out.

2

0ˆ

4 r

rsdIBd

=

Use Biot-Savart Law

All of the current loops below carry the same

current I. Rate them according to the magnetic

field at the red dot, from largest to smallest.

A. A>B>C

B. A>C>B

C. B=C>A

D. C=B=A

E. C>B>A

Question

Good News!!!!!

Unit 14, Slide 41

Remember how we used Gauss’

Law to avoid doing integral in E

field?

We got similar law for B fields!