Lecture 9 Magnetic Fields due to Currents Ch. 30• Cartoon - Shows magnetic field around a long current carrying wire and a loop of wire

• Opening Demo - Iron filings showing B fields around wires with currents

• Warm-up problem

• Topics

– Magnetic field produced by a moving charge

– Magnetic fields produced by currents. Big Bite as an example.

– Using Biot-Savart Law to calculate magnetic fields produced by currents.

– Examples: Field at center of loop of wire, at center of circular arc of wire, at center ofsegment of wire.

– Amperes’ Law : Analogous to Gauss’ L:aw in electrostatics, Useful in symmetric cases.

– Infinitely long straight wire of radius a. Find B outside and inside wire.

– Solenoid and Toroid Find B field.

– Forces between current carrying wires or parallel moving charges

• Demos

– Torque on a current loop(galvanometer)

– Iron filings showing B fields around wires with currents.

– Compass needle near current carrying wire

– Big Bite as an example of using a magnet as a research tool.

– Force between parallel wires carrying identical currents.

Reminders

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Quiz tomorrow

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Magnetic Fields due to Currents

Torque on a coil in a magnetic field demo– left over from last time

• So far we have used permanent magnets as our source of magnetic

field. Historically this is how it started.

• In early decades of the last century, it was learned that moving charges

and electric currents produced magnetic fields.

• How do you find the Magnetic field due to a moving point charge?

• How do you find the Magnetic field due to a current?

– Biot-Savart Law – direct integration

– Ampere’s Law – uses symmetry

• Examples and Demos

Torques on current loops

Electric motors operate by connecting a coil in a magnetic field to a current supply,which produces a torque on the coil causing it to rotate.

P

F

ia

bF

B

B

Above is a rectangular loop of wire of sides a and b carrying current i.

B is in the plane of the loop and ̂ to a.

Equal and opposite forces are exerted on the sides a.No forces exerted on b since

Since net force is zero, we can evaluate t (torque) about any point. Evaluate itabout P (cm).

iaBF =

Bi

! = NiabBsin" = NiABsin"

i

.

A=ab=area of loop

! = Fbsin"

F = iaB

! = iabBsin"

For N loops we mult by N

Torque on a current loop

! = NiABsin"

µ = NiA = magnetic dipole moment

! = µBsin"

! = µ # B

n̂

!

B

!n̂

Must reverse current using

commutator to keep loop

turning.

Torque !!tends to rotate loop until plane is " to B

( parallel to B).n̂

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.html#c4

Split ring

commutator +

-

Electron moving with speed v in a crossed electric

and magnetic field in a cathode ray tube.

�

! F = qE + q

! v x! B

y

Discovery of the electron by J.J. Thompson in 1897

2

2

mv2

qELy =

1. E=0, B=0 Observe spot on screen

2. Set E to some value and measure y the deflection

3. Now turn on B until spot returns to the original position

B

Ev

qvBqE

=

=

4 Solve foryE2

LB

q

m 22

=This ratio was first measured by Thompson

to be lighter than hydrogen by 1000

Show demo of CRT

Topic: Moving charges produce magnetic fields

q

r

"r̂

2

0 ˆ

4 r

rvqB

!=

!!

"

µ

2

70 104

A

N!"= #µ

Determined fom

Experiment

!v

1. Magnitude of B is proportional to q, , and 1/r2.

2. B is zero along the line of motion and proportional to sin at other

points.

3. The direction is given by the RHR rotating into

4. Magnetic permeability

r̂

!v

!v

Example:

A point charge q = 1 mC (1x10-3C) moves in the x direction with

v = 108 m/s. It misses a mosquito by 1 mm. What is the B field

experienced by the mosquito?

108 m/s90o

r̂2

0

4 r

vqB

!

µ=

26

83

2

7

10

1101010

mCB

s

m

A

N

!

!!"""=

TB4

10=

Recall the E field of a charge distribution

“Coulombs Law”

r̂r

kdqEd

2=

!

To find the field of a current distribution use:B

!

2

0

r

r̂sid

4Bd

!

"

µ=

!!

Biot-Savart Law

Topic: A current produces a magnetic field

r̂

This Law is found from experiment

Find B field at center of loop of wire lying in a plane with radius R

and total current i flowing in it.

B =µ0

4!"i d!l # r̂

r2

R

i

r̂ld !

!is a vector coming out of the paper The angle

between dl and r is constant and equal to 90

degrees.

!B = d

!B! =

µ0

4"k̂idl

R2! =

µ0

4"k̂i

R2dl! =

µ0

4"k̂i

R22"R

k̂R2

iB

0µ=!

Magnitude of B field at center of

loop. Direction is out of paper.

R k̂i

d!l ! r̂ = dl sin" k̂

sin" = sin90 = 1

d!l ! r̂ = dl k̂

r̂

ld!

P

"

Magnitude is µ0i

2R

Direction is k̂

!B = d

!B! =

µ0

4"k̂idl

R2!

=µ0

4"

i

R2k̂ dlli

l f

! =µ0

4"

i

R2k̂ l

li

l f

=µ0

4"

i

R2k̂(l f # li )

=µ0

4"

i

R2k̂(2"R # 0)

=µ0i

2Rk̂

r̂

ld!

P

"

Integration Details

Example

Loop of wire of radius R = 5 cm and current i = 10 A. What is B at the

center? Magnitude and direction

R

iB

2

0µ=

)05(.2

10104

2

7

m

AB

A

N!"= #

TB26

10102.1 !"=#

GaussTB 2.1102.14

=!=" Direction is out of

the page.

i

What is the B field at the center of a segment or circular

arc of wire?

"0 R

ild!

r̂

Total length of arc is S.

Why is the contribution to the B field at P equal to zero from

the straight section of wire?

P

!0 = 100 deg=100 deg "6.28 rad

360 deg= 100 " 0.0174 = 1.74 rad

!B = d

!B! =

µ0

4"

i

R2k̂ dl!

=µ0

4"

i

R2

k̂ S

!B =

µ0

4!

i

R "

0k̂

where S is the arc length S = R"0

"0 is in radians (not degrees)

B =µ0

4!"i d!l # r̂

r2

d!l !!r = rdl sin" k̂

sin" = sin0 = 0

d!l " r̂

d!l " r̂

Find magnetic field at center of arc length

0

0

4!

"

µ

R

iB =

Suppose you had the following loop.

What is the magnitude and direction of B at the origin?

SR

iB

2

0

4!

µ=

00

000

2)

2/(

4!

"

µ!!

"

µ

R

i

R

i

R

iB #=#=

Next topic: Ampere’s Law

Allows us to solve certain highly symmetric current problems for the

magnetic field as Gauss’ Law did in electrostatics.

Ampere’s Law is cIldB 0µ=!"!! Current enclosed by the path

!E"! "d!A =

qenc

#0

Gauss’s Law Charge enclosed by surface

Example: Use Ampere’s Law to find B near a very long, straight

wire. B is independent of position along the wire and only depends

on the distance from the wire (symmetry).

rld!

i i

By symmetry ldB!!

irBdlBBdlldB 02 µ! ====" # ##!!

r

iB

!

µ

2

0

=Suppose i = 10 A

R = 10 cmT 102

10

10102

5

1

7

!

!

!

"=

""=

B

B

Show Fe fillings around a straight wire with current,

current loop, and solenoid.

2

70 104

A

N!"= #µ

Rules for finding direction of B field from a current

flowing in a wire

r2

iB

0

!

µ=

Force between two current carrying wires

Find the force due to the current element of the first wire and the magnetic

field of the second wire. Integrate over the length of both wires. This will give

the force between the two wires.

abbaBLiF!"!

x =

Force between two current carrying wires

Find the force due to the current element of the first wire and the magnetic

field of the second wire. Integrate over the length of both wires. This will give

the force between the two wires.

d

iB

a

a!

µ

2

0

=

abbaBLiF!"!

x =

d

iLi

a

b!

µ

2

0

=

d

ii

L

F

d

iiLF

baba

ba

ba

!

µ!

µ

2

2

0

0

=

=

�

! L !" B and directed towards

wire a (wires are attracted).

Suppose one of the currents is

in the opposite direction? What

direction is ?baF!

From experiment we find

(Given by Ampere’s Law)

Example: Find magnetic field inside a long, thick wire of radius a

rBdlBBdlldB

r

!!

2

2

0

===" # ##!!

cIur2B0

=!

IC = current enclosed by the circle whose radius is r

ia

ri

a

rIC

2

2

2

2

==!

!

ri

a

rB

!µ

2

1

2

2

0= 2

0

2 a

irB

!

µ= Example: Find field inside a

solenoid. See next slide.

Solenoid

�

B = µ0ni

n is the number of turns per meter

ld!

ld!

ld!

ld!

Ba

d

b

c

CIldB 0µ=!"!!

First evaluate the right side, it’s easy

IC is the total current enclosed by the path

The number of loops of current; h is the length of

one side.

Right side =

nhiIc=

nhi0

µ

Evaluate left side:

!! ! !! "+"+"+"="a

d

b

a

c

b

d

c

dlBdlBdlBdlBldB!!

ld!

ld!

ld!

ld!

B

a

d

b

c

n = the number of loops or turns per meter

000 +++= Bh

Bh = µ0nhi

B = µ0ni

Toroid! =" CIldB 0µ

!!

! = NiBdl 0µ

NirB 02 µ! =

r

NiB

!

µ

2

0

=

a < r < b

a

b ld!

BdlldB =!

!!

10cos =o

Tokamak Toroid at Princeton

I = 73,000 Amps for 3 secs

B = 5.2 T

N is the total

number of turns

Magnetic dipole inverse cube law

�

! B =

µ0

2!

" µ

z3

z

.P

Magnetic field lines for a bar magnet

and a current loop are similar