# Lecture 9 Magnetic Fields due to Currents Ch. 2006. 7. 17.¢  Torques on current loops Electric motors

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• 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 of segment 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

Ungraded problem solutions are on website?

UVa email activated

Quiz tomorrow

Class in afternoon 1:00 PM

• 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

i a

b F

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 it about 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

!

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

qEL y =

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

E v

qvBqE

=

=

4 Solve for yE2

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

rv qB

! =

! !

"

µ

2

7 0 104

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

! 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/s 90o

r̂ 2

0

4 r

v qB

!

µ =

26

83

2

7

10

1 101010

m CB

s

m

A

N

!

!! """=

TB 4

10=

• Recall the E field of a charge distribution

“Coulombs Law”

r̂ r

kdq Ed

2 =

!

To find the field of a current distribution use:B !

2

0

r

r̂sid

4 Bd

!

"

µ =

! !

Biot-Savart Law

Topic: A current produces a magnetic field

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

r 2

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" k̂ idl

R 2! =

µ0

4" k̂ i

R 2 dl! =

µ0

4" k̂ i

R 2 2"R

k̂ R2

i B

0µ = !

Magnitude of B field at center of

loop. Direction is out of paper.

R k̂ i

d ! l ! r̂ = dl sin" k̂

sin" = sin90 = 1

d ! l ! r̂ = dl k̂

ld !

P

"

Magnitude is µ0i

2R

Direction is k̂

• ! B = d

! B! =

µ0

4" k̂ idl

R 2!

= µ0

4"

i

R 2 k̂ dl li

l f

! = µ0

4"

i

R 2 k̂ l

li

l f

= µ0

4"

i

R 2 k̂(l f # li )

= µ0

4"

i

R 2 k̂(2"R # 0)

= µ 0 i

2R k̂

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

i B

2

0µ =

)05(.2

10 104

2

7

m

A B

A

N! "= #

TB 26

10102.1 !"= #

GaussTB 2.1102.1 4

=!= " 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

i ld !

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

R 2 k̂ dl!

= µ0

4"

i

R 2

k̂ S

! B =

µ0

4!

i

R "

0 k̂

where S is the arc length S = R"0

"0 is in radians (not degrees)

B = µ0

4!" i d ! l # r̂

r 2

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

i B =

Suppose you had the following loop.

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

S R

i B

2

0

4!

µ =

0 0

00 0

2 )

2/ (

4 !

"

µ !!

"

µ

R

i

R

i

R

i B #=#=

• 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

i B

!

µ

2

0

= Suppose i = 10 A

R = 10 cm T 102

10

10102

5

1

7

!

!

!

"=

"" =

B

B

Show Fe fillings around a straight wire with current,

current loop, and solenoid.

2

7 0 104

A

N! "= #µ

• Rules for finding direction of B field from a current

flowing in a wire

r2

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