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 Grades are on WebAssign

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.

a F i b P

.

F

i

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 ^ Equal and opposite forces

F

= to

a

.

iaB

are exerted on the sides

a

.

No forces exerted on

b

since

i B

Since net force is zero, we can evaluate t about P (cm).

t =

Fb

sin q t t

F

= =

iaB iabB

sin q (torque) about any point. Evaluate it For N loops we mult by N =

NiabB

sin q =

NiAB

sin A=ab=area of loop q

Torque on a current loop t t t  = =

NiAB

sin q

NiA

= magnetic dipole moment = =  

B

 sin

B

q

n

ˆ q B

Torque tends to rotate loop until plane is ^ to B n

ˆ Must reverse current using commutator to keep loop turning.

q

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



qv xB

 y

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

1.

E=0, B=0 Observe spot on screen 2. Set E to some value and measure y the deflection y = qEL 2 2 mv 2 3. Now turn on B until spot returns to the original position qE = qvB v = E B 4 Solve for m q = B 2 L 2 2 yE This ratio was first measured by Thompson to be lighter than hydrogen by 1000 Show demo of CRT

Topic: Moving charges produce magnetic fields

r

ˆ q q 

B

=  0 4  r

q v

 

r

ˆ

r

2 1.

2.

3.

4.

v

Determined fom Experiment

v

2.

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

ˆ

r

Magnetic permeability  0 = 4   10  7

N A

2

Example: A point charge q = 1 mC (1x10 -3 C) moves in the x direction with v = 10 8 m/s. It misses a mosquito by 1 mm. What is the B field experienced by the mosquito?

r

ˆ 90 o 10 8 m/s

B

=  0 4 

v q r

2

B

= 10  7

N A

2  10  3

C

 10 8

s m

 1 10  6

m

2

B

= 10 4

T

Topic: A current produces a magnetic field Recall the

E

field of a charge distribution “Coulombs Law” d  E = kdq r 2 rˆ  To find the field of a current distribution use:

r

ˆ Biot-Savart Law  d B =  0 4   id s  rˆ r 2 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.

i P

B

=   0 4 

i dl r

2 

r

ˆ

d l

 q

r

ˆ R d l   rˆ is a vector coming out of the paper The angle between dl and r is constant and equal to 90 degrees.

dl

 sin q

r

ˆ =

dl

sin q ˆ = sin 90 = 1

B dl

=  

r

ˆ =

dB dl

ˆ =  0 4 

k

ˆ 

idl R

2 =  0 4 

k

ˆ

i R

2 

dl

=  0 4 

k

ˆ

i R

2 2 

R

Magnitude of B field at center of loop. Direction is out of paper.

i R

k

ˆ  B =  0 i 2 R kˆ Magnitude is  0 2

R i

P q

d l



r

ˆ = = = =

B

= 

dB

=  0 4 

k

ˆ 

idl R

2  0 4 

i R

2

l i l f

 =  0 4   0 4   0 4   0 2

R i i i R

2

R

2

k

ˆ

k

ˆ (

l f



l i

)

k

ˆ (2 

R

 0)

i R

2

l f l i

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 i

B

=  2 0

i R B B

= 4   10  7

N A

2 2 (.

05

m

) = 1 .

2  10  6  10 2

T

10

A B

= 1 .

2  10  4

T

= 1 .

2

Gauss

Direction is out of the page.

dl

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

i

d l



r

ˆ

B

=   0 4 

i dl r

2 

r

ˆ

B

= =  0 4  

dB

=  0 4 

i R

2 ˆ

S i R

2  Total length of arc is S.

r

ˆ q 0 R

dl r

ˆ where S is the arc length S = R q 0 q 0 is in radians (not degrees) P

dl



r

sin q =

rdl

sin = sin 0 = 0 q ˆ

B

=  0 4 

i R

q 0

k

ˆ q 0 = 100 deg=100 deg  6.28 rad 360 deg = 100  0.0174

= 1.74 rad Why is the contribution to the B field at P equal to zero from the straight section of wire?

Suppose you had the following loop .

Find magnetic field at center of arc length What is the magnitude and direction of B at the origin?

B

=  4  0

i R

2

S B

=  4  0

i R

q 0

B

=  4 0  (

i R

q 0 

i R

/ 2 q 0 ) =   2  0

i R

q 0

Next topic:

Ampere’s Law

Allows us to solve certain highly symmetric current problems for the magnetic field as Gauss’ Law did in electrostatics.

Gauss’s Law 

E



dA

=

q enc

 0 Ampere’s Law is  

B



d l

 =  0

I c

Charge enclosed by surface Current enclosed by the path

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).

i

d l

 r i By symmetry 

B d l

  

B



d l

 = 

Bdl

=

B



dl

=

B

2 

r

=  0

i B

=  0 2 

r i

Suppose i = 10 A  0 = R = 10 cm 4   10  7

N A

2

B B

= = 2 2   10  7  10 10  5  1 T 10

Show Fe fillings around a straight wire with current, current loop, and solenoid.

Rules for finding direction of B field from a current flowing in a wire B =  2  0 r i

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.



F ba

=

i b



L

x 

B a

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.

From experiment we find

B a



F ba

= =  2 0

i



d a i b



L

x =

i

(Given by Ampere’s Law) 

B b L

  2 0

i



d a a L



B

and directed towards wire a (wires are attracted).

F ba F ba L

Suppose one of the currents is

F

= =

L i a i

2

a

2

i b



i b

 

d



d

0 0

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

B



d l

 B 2  r = =  u 0 I c

Bdl

=

B

2 

r

0 

dl

=

B

2 

r

I C

I C B

= current enclosed by the circle whose radius is r = = 

r

  0

a

2

r

2

a

2

i

2

i

= 2

r a

1 2 2 

r i B

=  0

ir

2 

a

2 Example: Find field inside a solenoid. See next slide.

Solenoid



B

=  0

ni

n is the number of turns per meter

d l

 d a

d l

 B

d l

 c

d l

 b

First evaluate the right side, it’s easy

 

B



d l

 =  0

I C

I C is the total current enclosed by the path

I c

=

nhi

The number of loops of current; h is the length of one side.

Right side =

 0

nhi

d l



d l

 d a B

d l



Evaluate left side:

 

B



d l

 =

a



b B



dl



b



c B



dl



c



d B



dl



d



a B



dl

=

Bh

 0  0  0

Bh B

= =  0

nhi

 0

ni

n = the number of loops or turns per meter c

d l

 b

Toroid a b   

B



d l



Bdl

= =  0

I C

 0

Ni B B

2 

r

=  = 0 

Ni

2 

r

0

Ni

N is the total number of turns

d l

 a < r < b 

B



d l

 cos 0

o

=

Bdl

= 1 Tokamak Toroid at Princeton I = 73,000 Amps for 3 secs B = 5.2 T

Magnetic dipole inverse cube law

B

=  0 2  

z

3 P

.

z  Magnetic field lines for a bar magnet and a current loop are similar