# Lecture 9 Magnetic Fields due to Currents Chp. 30 Cartoon - Shows magnetic field around a long...

### Transcript of Lecture 9 Magnetic Fields due to Currents Chp. 30 Cartoon - Shows magnetic field around a long...

Lecture 9 Magnetic Fields due to Currents Chp. 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• Physlet • Topics

– Magnetic field produced by a moving charge – Magnetic fields produced by currents. BigBite 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– BigBite as an example of using a magnet as a research tool.– Force between parallel wires carrying identical currents.

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

Crossed magnetic and electric fields

€

rF = qE + q

r v x

r B

y

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

€

y =qEL2

2mv 2

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 oriiginal position

€

qE = qvB

v = E /B

4 Solve for

€

m

q=

B2L2

2yE

This ratio was first measured by Thompson to be lighter than hydrogen by 1000

Show demo of CRT

Hall Effect Crossed fields to measure charge carrier density n

V=Ed

€

eE = evd B

vd =E

B=

V

Bd

€

vd=i

neA

€

n =Bid

VeA

Topic

A moving charge produces a magnetic field.q

r

vr̂

2

0 ˆ

4 r

rvqB

×=

vv

πμ

27

0 104A

N−×= πμ

1. Magnitude of B is proportional to q, v, 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 v into . r̂

Example: A point charge q = 1 mC 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/s90or̂

2

0

4 r

vqB

πμ

=

2683

27

10

1101010

mCB

s

m

A

N−

−− ×××=

TB 410=

To find the E field of a charge distribution use:

Use:

2

ˆ

r

rkdqEd =r

To find the field of a current distribution use:Br

Note that vdqdt

dsdqds

dt

dqids

v===

2

0 ˆ

4 r

rsidBd

×=

vv

πμ

Topic: Biot – Savart Law

Use to find B field at the center of a loop of wire. 2

0 ˆ

4 r

rlidB

×=∫

v

πμ

R

i Loop of wire lying in a plane. It has radius R and total current i flowing in it.

First find rld ˆ×v

rld ˆ×v

is a vector coming out of the paper at the same angle anywhere on the circle. The angle is constant.

r̂

r̂

ldv

ldv

RR

idl

R

i

R

idldBB π

πμ

πμ

πμ

2444 2

0

2

0

2

0==== ∫∫∫

R

iB

2

0μ=

kR

iB ˆ

2

0μ=

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

R k̂i

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−×= π

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?

∫= dlR

iB

2

0

4π

μ

0 R

ildv

r̂Total length of arc is S.

SR

iB

2

0

4πμ

= where S is the arc length S = R0

0 is in radians (not degrees)

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

the straight section of wire?

P

Find magnetic field at center of arc length

00

4

πμ

Ri

B=

0

i

ii

Radius R

Radius R/2

Suppose you had the following loop.

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

SR

iB

2

0

4πμ

=

€

B =μ0

4π(

i

Rθ0 −

i

R /2θ0) = −

μ0

2π

i

Rθ0

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=⋅∫vv 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).

rldv

i i

By symmetry ldBvv

r

iB

πμ2

0=

Suppose i = 10 A

R = 10 cm

€

B = 2 ×10−7 ×10 ×10−1

B = 2 ×10−7 T

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

27

0 104A

N−×= πμ

Rules for finding direction of B field from a current flowing in a wire

r

iB

o

πμ2

=

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

aa

πμ2

0=

€

vF ba = i

b

r L x

v B

a

€

=ibL μ 0ia

2πd

€

Fba = Liaibμ 0

2πd

Fba /L =iaibμ 0

2πd

€

rL ⊥

v B and directed towards

wire a (wires are attracted).

Suppose one of the currents is in the opposite direction? What direction is ?baF

v

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

Cross-sectional view

Path of integral dlar

CIurB 02 =πIC = current enclosed by the path

ia

ri

a

rIC

2

2

2

2

==ππ

ri

a

rB

πμ

21

2

2

0= 2

0

2 a

irB

πμ

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

Solenoid

€

B = μ0nin is the number of turns per meter

ldv

ldv

ldv

ldv

B

a

d

b

c

CIldB 0μ=⋅∫vv

First evaluate the right side, it’s easy

IC is the total current enclosed by the path

IC = nhi

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

Right side = μ0nhi

Evaluate left side:

∫∫ ∫ ∫∫ ⋅+⋅+⋅+⋅=⋅a

d

b

a

c

b

d

c

dlBdlBdlBdlBldBvv

ldv

ldv

ldv

ldv

B

a

d

b

c

Bh = μ0nhi

= Bh + 0 + 0 + 0

B = μ0ni n = the number of loops or turns per meter

Toroid∫ =⋅ CIldB 0μ

vv

∫ = NiBdl 0μ

NirB 02 μπ =

r

NiB

πμ20

=

a < r < b

a

b ldv

BdlldB =⋅vv

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

€

rB =

μ0

2π

v μ

z3

z

Warm up Problem Set 9Warm up set 9 Due 8:00 am Thursday 1. HRW6 30.TB.02. [119973] A "coulomb" is:

one ampere per second an abbreviation for a certain combination of kilogram, meter and second the quantity of charge which will exert a force of 1 N on a similar charge at a distance of 1 m the amount of current in each of two long parallel wires separated by 1 m, which produces a force of 2 10-7 N per meter the amount of charge which flows past a point in one second when the current is 1 A

2. HRW6 30.TB.03. [119974] Electrons are going around a circle in a counterclockwise direction as shown. At the center of the circle they produce a magnetic field that is:

to the left into the page out of the page zero to the right

3. HRW6 30.TB.07. [119978] Lines of the magnetic field produced by a long straight wire carrying a current:

are circles concentric with the wire leave the wire radially are in the direction of the current are lines similar to those produced by a bar magnet are opposite to the direction of the current