The Biot-Savart LawMagnetic Sources

•Magnetic fields go around the wire – they are perpendicularto the direction of current•Magnetic fields are perpendicular to the separationbetween the wire and the point where you measure it

•Sounds like a cross product!

r

I

2

Id

rB

ds

ˆd s r

02

ˆ

4

I d

r

s rB

70 4 10 T m/A

•Permeability of free space•The Amp is defined to work out this way

Sample Problem

P

A loop of wire consists of two quarter circles of radii R and 2R, both centered at a point P, and connected with wires going radially from one to the other. If a current I flows in the loop, what is the magnetic field at point P?

R 2R

•Do one side at a time•First do one of the straight segments•ds and r-hat are parallel

•No contribution to the integral•Other straight segment is the same

dsr̂ˆ 0d s r

dsr̂

•Now do inner quarter loop

dsr̂

ˆˆd ds s r k

0inner 2

ˆ4

I ds

R

B k 02

2ˆ4 4

I R

R

k 0 ˆ8

I

R

k

•Outer loop opposite direction, similar

ds

r̂

0outer

ˆ16

I

R

B k

0 ˆ16

I

R

B k

I

Magnetic Field from a Finite Wire

r

I

ds

•Magnetic field from a finite straight wire:•Let a be the distance the point is from the wire•Let x be the horizontal separation

P

a

x

ˆ ˆx a r i j

2 2r x a

x1 x2

ˆd dxs i

2

1

03 22 2

ˆ ˆ ˆ

4

x

x

dx x aI

x a

i i jB

2

1

03 22 2

ˆ4

x

x

Ia dx

x a

k

0 2 1

2 2 2 22 1

ˆ

4

I x x

a x a x a

k

01 2cos cos

4

IB

a

21

11 2 2

1

cosx

x a

22 2 2

2

cosx

x a

Warning: My 2 differs from that of the book

Magnetic Field from a Wire

I

•Magnitude is found from the formula•Direction is found from the right hand rule

•Place thumb in direction of current flow•Fingers curl in direction of B-field

a

01 2cos cos

4

IB

a

21

•Infinite wire:•Angles are simple 1 2

1 2

0

cos cos 1

0

2

IB

a

Sample ProblemA regular hexagon whose center is a distance a = 1 cm from the nearest

side has current I = 4.00 A flowing around it. The current flows N = 500 times around. What is the total magnetic field at the center?

I

a60

60

•Draw in the two directions from the center to the corners of one segment•Top angle is one-sixth of a circle, or 60 degrees•Total angles in circle is 180, so other two angles are 60 each•Use formula to get magnetic field – right hand rule says up.

60 0 cos 60 cos 604

IB

a

0

4

I

a

•Multiply by all six side, and then by 500 cycles

0tot

6

4

INB

a

76 4 10 T m/A 4 A 500

4 0.01 m

0.120 T

Right Hand Rule for Loops

Force Between Parallel Wires

•One wire – infinite – creates a magnetic field•Other wire – finite or infinite – feels the force

I1

I2d

L

0 1

2

IB

d

2 2I F L B

0 1 2

2

I IF

L d

•Attractive if current is parallel, repulsive if anti-parallel

F

I

•If you curl your fingers in the direction the current flows, thumb points in direction of B-field inside the loop•Works for solenoids too (later)

Ampere’s Law (original recipe)•Suppose we have a wire coming out of the plane•Let’s integrate the magnetic field around a closed path

•There’s a funky new symbol for such an integral•Circle means “over a closed loop”

•The magnetic field is parallel to direction of integration

0

2

IB

a

dB s B ds 0 22

Ir

r

0I

•What if we pick a different path?

dB s cosBds I

0

2

Ird

r

0d I B s

ds

cosds rd

ds cosrd

•We have demonstrated this is true no matter what path you take•Wire need not even be straight infinite wire

•All that matters is that current passes through the closed Ampere loop

Announcements

ASSIGNMENTSDay Read Quiz HomeworkToday Sec. 30.4-30.5 Quiz 30b noneFriday Sec. 31.1-31.3 Quiz 31a Hwk. 30aMonday Sec. 31.4-31.6 Quiz 31b Hwk. 30b

2/29

Test Next WednesdayReview session here4:30-6:00 Monday

13 cm

12 cm

5 cm

Right angle

B

Total angle:270 + ( - )

Sorry About Today’s Reading Quiz

Understanding Ampere’s Law•If multiple currents flow through, add up all that are inside the loop•Use right-hand rule to determine if they count as + or –

•Curl fingers in direction of Ampere loop•If thumb points in direction of current, plus, otherwise minus

•The wire can be bent, the loop can be any shape, even non-planar

0d I B s

5 A2 A 1 A

4 A

7 A

0 3 Ad B s

We will later realize that this formula is imperfect

Using Ampere’s Law•Ampere’s Law can be used – rarely – to calculate magnetic fields•Need lots of symmetry – usually cylindrical

A wire of radius a has total current I distribu-ted uniformly across its cross-sectional area. Find the magnetic field

everywhere.I I

End-on view

•Draw an Ampere loop outside the wire – it contains all the current•Magnetic field is parallel to the direction of this loop, and constant around it•Use Ampere’s Law

•But we used a loop outside the wire, so we only have it for r > a

0I d B s B ds 2 rB0

2

IB

r

Using Ampere’s Law (2)•Now do it inside the wire•Ampere loop inside the wire does not contain all the current•The fraction is proportional to the area

End-on view

2 rB

2

2rI r

I a

a

2 2rI Ir a

0 rI d B s

0

2rI

Br

022

Ir

a

02

0

2

2

I r r aB

Ir a r a

Solenoids•Consider a planar loop of wire – any shape – with a current I going around it•Now, stack many, many such loops

•Treat spacing as very closely spaced•Assume stack is tall compared to size of loop

•Can show using symmetry that magnetic field is only in vertical direction•Can use Ampere’s Law to show that it is constant inside or outside the solenoid

00 I d B s 1B L 0 2B L 0 1 2B B

•But magnetic field at infinity must be zero

outside 0B

Field Inside a Solenoid•It remains only to calculate the magnetic field inside•We use Ampere’s law

•Recall, no significant B-field outside•Only the inside segment contributes

•There may be many (N) current loops within this Ampere loop•Let n = N/L be loops per unit length

0 totI d B s

L

inB L totI NI

0in

NIB

L

in 0B nI

•Works for any shape solenoid, not just cylindrical•For finite length solenoids, there are “end effects”•Real solenoids have each loop connected to the next, like a helix, so it’s just one long wire

Magnetic Flux•Magnetic flux is defined exactly the same way for magnetism as it was for electricity

ˆB dA B n

A cylindrical solenoid of radius 10 cm has length 50 cm and has 1000 turns of wire going around it. What is the

magnetic field inside it, and the magnetic flux through it, when a current of 2.00 A is passing through the wire?

0in

NIB

L

7 1000 2 A4 10 T m/A

0.5 m in 0.00503 TB

B BA 2B R 20.00503 T 0.1 m 4 21.579 10 T mB

A Tesla meter2 is also called a Weber (Wb)

Gauss’s Law for Magnetism

•Magnetic field lines always go in circles – there are no magnetic monopole sources•For closed surfaces, any flux in must go out somewhere else

ˆ 0B dA B n

Earth’s Magnetic Field•Deep inside the Earth there are currents flowing•These generate magnetic field lines

•Exiting near the south geographic pole•Entering near the north geographic pole

•Current loops near the surface of the Earth try to line up with B-field•Compass needles are really just loops of current (sort of)

•Compass needles point approximately north on most of the Earth

Equations for Test 2

End of material for Test 2

Resistors: Kirchoff’s Laws:Magnetic Forces:

1 2R R R

1 2

1 1 1

R R R

in outI I

loop

0 V q F E v B

I F L B

NT

A m

V N m C Units:

V IR I V P

dQI

dt

RC Circuits:RC

A C/s

With Capacitors:

V Q C I JA

ˆ 0B dA B n

Gauss’s Law for Magnetism