1 Exam 2 covers Ch. 27-33, Lecture, Discussion, HW, Lab Chapter 27: The Electric Field Chapter 29:...

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Exam 2 covers Ch. 27-33,Lecture, Discussion, HW, Lab

Chapter 27: The Electric Field Chapter 29: Electric potential & work Chapter 30: Electric potential & field

(exclude 30.7) Chapter 31: Current & Resistance Chapter 32: Fundamentals of Circuits

(exclude 32.8) Chapter 33: The Magnetic Field

(exclude 33.5-33.6, 33.9-10, & Hall effect)

Exam 2 is Tue. Oct. 27, 5:30-7 pm, 145 Birge

Tue. Oct. 27, 2009 Physics 208 Lecture 16 2

Law of Biot-Savart

Each short length of current produces contribution to magnetic field.r

I in plane of pageds

€

dr B =

μo

4π

Idr s × ˆ r

r2

B out of page

ds

dB

r

€

μo = 4π ×10−7 N / A2= permeability of free space

r = distance from current element

Field from very short section of current

€

dr s

Vector cross product

Tue. Oct. 27, 2009 Physics 208 Lecture 16 3

€

rC

€

rD

€

rC ×

r D

€

dr B =

μo

4π

Idr s × ˆ r

r2

€

dr B =

μoI

4π r2dr s × ˆ r

Short length of current

Unit vector toward point at which field is evaluated

Dist. to point at which field is evaluated

€

dr B

€

dr s

€

rr

Tue. Oct. 27, 2009 Physics 208 Lecture 16 4

Field from a circular loop

Each current element produce dB All contributions add as vectors Along axis, all

components cancelexcept for x-comp

Tue. Oct. 27, 2009 Physics 208 Lecture 16 5

Magnetic field from loop of current

Looks like magnetic dipole

Tue. Oct. 27, 2009 Physics 208 Lecture 16 6

Building a solenoid

Tue. Oct. 27, 2009 Physics 208 Lecture 16 7

Solenoid: many current loops

€

Bsolenoid =μoNI

L= μonI

Tue. Oct. 27, 2009 Physics 208 Lecture 16 8

Magnetic Force on a Current

S

N

I

Current

Magnetic field

Magnetic force

€

rF =

r I ×

r B L

€

qr v ×

r B Force on each charge

Force on length of wire

€

dr s

€

Idr s ×

r B

Force on straight section of wire, length L

Tue. Oct. 27, 2009 Physics 208 Lecture 16 9

Quick Quiz

A current I flows in a square loop of wire with side length L.

A constant B field points in the x-direction, perpendicular to the plane of the loop. What is the net force on the wire loop?

x

y

I

I

I

I

L

A. 4LB

B. 2LB

C. LB

D. 0

No force, but torque

Torque is Net torque can be nonzero even when

net force is zero.

Tue. Oct. 27, 2009 Physics 208 Lecture 16 10

€

rr ×

r F

Lever armForce

12/09/2002 U. Wisconsin, Physics 208, Fall 2006 11

Which of these loop orientations has the largest magnitude torque? Loops are identical apart from orientation.

(A) a (B) b (C) c

Question on torque

a bc

Quick QuizWhich of these different sized current loops has

the greatest torque from a uniform magnetic field to the right? All have same current.

Tue. Oct. 27, 2009 Physics 208 Lecture 16 12

L

W

L/2

2W

2L

W/2

A. B.

C. D. All same

€

rB

Tue. Oct. 27, 2009 Physics 208 Lecture 16 13

Torque on current loop

€

rτ =

rr ×

r F

€

rτ =2

l

2F sinθ

⎛

⎝ ⎜

⎞

⎠ ⎟

€

F = IBl ⇒ τ = AIBsinθ

€

A = l 2 =loop area

B

F

B

I

€

rr

I

F

Torque proportional to

• Loop area

• Current

• sinθ

Tue. Oct. 27, 2009 Physics 208 Lecture 16 14

Current loops & magnetic dipoles Current loop produces magnetic dipole field. Magnetic dipole moment:

€

rμ

€

rμ =IA

currentArea of loop

€

rμ

magnitude direction

In a uniform magnetic field

Magnetic field exerts torqueTorque rotates loop to align with

€

rτ =

rμ ×

rB ,

€

rμ

€

rB

€

rτ =

rμ

rB sinθ

Works for any shape planar loop

Tue. Oct. 27, 2009 Physics 208 Lecture 16 15

I

€

rμ =IA

€

rμ perpendicular to loop

Torque in uniform magnetic field

€

rτ =

rμ ×

rB ,

€

rτ =

rμ

rB sinθ

Potential energy of rotation:

€

U = −r μ ⋅

r B = −μBcosθ

Lowest energy aligned w/ magnetic field

Highest energy perpendicular to magnetic field

Tue. Oct. 27, 2009 Physics 208 Lecture 16 16

Magnetic flux Magnetic flux is defined

exactly as electric flux (Component of B surface) x (Area element)

€

ΦB = B • dA∫

zero flux Maximum flux

SI unit of magnetic flux is the Weber ( = 1 T-m2 )

Tue. Oct. 27, 2009 Physics 208 Lecture 16 17

Magnetic Flux Magnetic flux Φ through a surface:

(component of B-field surface) X (surface area) Proportional to

# B- field lines penetrating surface

Tue. Oct. 27, 2009 Physics 208 Lecture 16 18

Why perpendicular component? Suppose surface make angle surface normal

ΦB = BA cos ΦB =0 if B parallel A ΦB = BA (max) if B A

Flux SI units are T·m2=Weber

€

rB = B||

ˆ s + B⊥ˆ n

€

ˆ n

€

ˆ s

€

rA = A ˆ n

Component || surface

Component surface

Only component‘goes through’ surface

€

ΦM =r B •

r A

Tue. Oct. 27, 2009 Physics 208 Lecture 16 19

Total flux E not constant

add up small areas where it is constant

Surface not flat add up small areas

where it is ~ flat

€

δΦBi = BiδA icosθ =

r B i • δ

r A i

Add them all up:

€

ΦB =r B • d

r A

surface

∫

Tue. Oct. 27, 2009 Physics 208 Lecture 16 20

Magnetic flux

What is that magnetic flux through this surface?

A. Positive

B. Negative

C. Zero

Tue. Oct. 27, 2009 Physics 208 Lecture 16 21

Properties of flux lines Net magnetic flux through any closed

surface is always zero:

€

Φmagnetic = 0

No magnetic ‘charge’, so right-hand side=0 for mag.

Basic magnetic element is the dipole

€

Φelectric =Qenclosed

εo

For electric charges, and electric flux

Tue. Oct. 27, 2009 Physics 208 Lecture 16 22

Time-dependent fields Up to this point, have discussed only magnetic

and electric fields constant in time. E-fields arise from charges B-fields arise from moving charges (currents)

Faraday’s discovery

Another source of electric field Time-varying magnetic field creates

electric field

Tue. Oct. 27, 2009 Physics 208 Lecture 16 23

Measuring the induced field

A changing magnetic flux produces an EMF around the closed path.

How to measure this? Use a real loop of wire for the closed path.

The EMF corresponds to a current flow:

€

ε=IR