Announcements

Physics 2135 spreadsheets for all sections, with Exam 2 scores, will be available today on the Physics 2135 web site. You need your PIN to find your grade.

Physics 2135 Exam 2 will be returned in recitation Thursday. When you get the exam back, please check that points were added correctly. Review the course handbook and be sure to follow proper procedures before requesting a regrade. Get your regrade requests in on time! (They are due by next Thursday’s recitation.)On a separate sheet of paper, briefly explain the reason for your regrade request. This should be based on the work actually shown on paper, not what was in your head. Attach to the exam and turn it in by the end of your next Thursday’s recitation.

Preliminary exam average is about 75.0% (12 sections out of 12 reporting). Reasonable! Scores ranged from a low of 37 to a high of 200 (3 students). I will fill in the ??’s during the “live” lecture and in its “.ppt” file.

Announcements

Midterm grades include your first 3 labs. If you are missing one of them, this does impact your grade. Remember, the one lowest lab score is dropped, and there are no makeups.

Today’s agenda:

Magnetic Fields Due To A Moving Charged Particle.You must be able to calculate the magnetic field due to a moving charged particle.

Biot-Savart Law: Magnetic Field due to a Current Element. You must be able to use the Biot-Savart Law to calculate the magnetic field of a current-carrying conductor (for example: a long straight wire).

Force Between Current-Carrying Conductors.You must be able to begin with starting equations and calculate forces between current-carrying conductors.

We’ve been working with the effects of magnetic fields without considering where they come from. Today welearn about sources of magnetic fields.

Magnetic Field of a Moving Charged Particle

+

B

r̂

r

v

It is experimentally observed that a moving point charge q gives rise to a magnetic field

ˆμ.

4π

02

qv rB=

r

0 is a constant, and its value is 0=4x10-7 T·m/A

Let’s start with the magnetic field of a moving charged particle.

Remember: the direction of r is always from the source point (the thing that causes the field) to the field point (the location where the field is

being measured.

This is example 8.1 in your text.

Example: proton 1 has a speed v0 (v0<<c) and is moving along the x-axis in the +x direction. Proton 2 has the same speed and is moving parallel to the x-axis in the –x direction, at a distance r directly above the x-axis. Determine the electric and magnetic forces on proton 2 at the instant the protons pass closest to each other.

1

r

x

y

z

2

v0

v0The electric force is

ˆ

1 2

E 2

q q1F = r

4 rr̂

E

FE

ˆ

2

E 2

1 eF = j

4 r

Homework Hint: slides 6-9!

Alternative approach to calculating electric force. This is “better” because we use the concept of field to calculate both of electric and (later) magnetic forces.

1

r

x

y

z

2

v0

v0

At the position of proton 2 there is an electric field due to proton 1.

ˆˆ

1

1 2 2

q1 1 eE = r= j

4 r 4 r

r̂

FE

ˆ ˆ

2

E 1 2 2

1 e 1 eF =qE =e j= j

4 r 4 r

E

This electric field exerts a forceon proton 2.

1

r

x

y

z

2

r̂

FE

Also at the position of proton 2 there is a magnetic field due to proton 1.

ˆ

1 1

1 2

q v rB =

4 r

ˆ ˆ

0

1 2

ev i jB =

4 r

ˆ

0

1 2

evB = k

4 rB1

v0

v0

1

r

x

y

z

2

r̂

FE

B1

Proton 2 “feels” a magnetic force due to the magnetic field of proton 1.

B 2 2 1F =q v B

ˆˆ

0

B 0 2

evF =ev i k

4 r

ˆ

2 20

B 2

e vF = j

4 r

FB

v0

v0

What would proton 1 “feel?”Caution! Relativity overrules Newtonian mechanics!However, in this case, the force is “equal & opposite.”

1

r

x

y

z

2

r̂

FE

B1

Both forces are in the +y direction. The ratio of their magnitudes is

2 20

2B

2E

2

e v4 rF

=F 1 e

4 r

FB

v0

v0 2B0

E

F= v

F

Later we will find that

2

1=

c

1

r

x

y

z

2

r̂

FE

B1

Thus

FB

v0

v0

20B2

E

vF=

F c

If v0=106 m/s, then

26

-5B28

E

10F= 1.11 10

F 3 10

Don’t you feel sorry for the poor, weak magnetic force?What if you are a nanohuman, lounging on proton 1. You rightfully claim you are at rest. There is no magnetic field from your proton, and no magnetic force on 2.If you don’t like being confused,

close your eyesand cover your ears.

Or see here, here, and here for a hint about how to resolve the paradox.

What if you are a nanohuman, lounging on proton 1. You rightfully claim you are at rest. There is no magnetic field from your proton, and no magnetic force on 2.

Another nanohuman riding on proton 2 would say “I am at rest, so there is no magnetic force on my proton, even though there is a magnetic field from proton 1.”

What if you are a nanohuman, lounging on proton 1. You rightfully claim you are at rest. There is no magnetic field from your proton, and no magnetic force on 2.

Another nanohuman riding on proton 2 would say “I am at rest, so there is no magnetic force on my proton, even though there is a magnetic field from proton 1.”

This calculation says there is a magnetic field and force. Who is right? Take Physics 2305 (107) to learn the answer.

Today’s agenda:

Magnetic Fields Due To A Moving Charged Particle.You must be able to calculate the magnetic field due to a moving charged particle.

Biot-Savart Law: Magnetic Field Due to a Current Element. You must be able to use the Biot-Savart Law to calculate the magnetic field of a current-carrying conductor (for example: a long straight wire).

Force Between Current-Carrying Conductors.You must be able to begin with starting equations and calculate forces between current-carrying conductors.

From the equation for the magnetic field of a moving charged particle, it is “easy” to show that a current I in a little length dl of wire gives rise to a little bit of magnetic field.

dB

r̂

r

dl

The Biot-Savart Law

ˆμ

4π

02

I d rdB=

r

I

You may see the equation written using ̂r =r r .

If you like to be more precise in your language, substitute “an infinitesimal” for “a little length” and “a little bit of.”I often use ds instead of dl because the script l does not display very well.

Biot-Savart Law: magnetic field of a current element

Applying the Biot-Savart Law

I

ds

r

r̂

dBˆμ ˆ

4π

02

I ds r rdB= where r=

r r

μ θ

4π0

2

I ds sin dB=

r

B= dB

Homework Hint: if you have a tiny piece of a wire, just calculate dB; no need to integrate.

Example: calculate the magnetic field at point P due to a thin straight wire of length L carrying a current I. (P is on the perpendicular bisector of the wire at distance a.)

ˆμ

4π

02

I ds rdB=

r

ˆˆ θ

ds r=ds sin k

μ θ

4π0

2

I ds sindB=

r

ds is an infinitesimal quantity in the direction of dx, so

μ θ

4π0

2

I dx sindB=

r

I

y

r

x

dBP

ds

r̂

x

z

a

L

2 2r= x +aθa

sin =r

I

y

r

x

dBP

ds

r̂

x

z

μ θ

4π0

2

I dx sindB=

r

a

μ μ

4π 4π0 0

3/23 2 2

I dx a I dx adB= =

r x +a

μ

4πL/2

03/2-L/2 2 2

I dx aB=

x +a

μ

4π L/2

03/2-L/2 2 2

I a dxB=

x +a

L

I

y

r

x

dBP

ds

r̂

x

z

a

μ

4π L/2

03/2-L/2 2 2

I a dxB=

x +a

look integral up in tables, use the web,or use trig substitutions

3/2 1/22 2 2 2 2

dx x=

x +a a x +a

μ

4π

L/2

01/22 2 2

-L/2

I a xB=

a x +a

μ

4π

01/2 1/22 22 2 2 2

I a L/2 -L/2=

a L/2 +a a -L/2 +a

L

I

y

r

x

dBP

ds

r̂

x

z

a

μ

4π

01/22 2 2

I a 2L/2B=

a L /4+a

μ

4π0

1/22 2

I L 1B=

a L /4+a

μ

π0

2 2

I L 1B=

2 a L +4a

μ

π0

2

2

I 1B=

2 a 4a1+

L

I

y

r

x

dBP

ds

r̂

x

z

a μ

π0

2

2

I 1B=

2 a 4a1+

L

When L, μ

.π0I B=

2 a

μ

π0I B=

2 ror The r in this equation has a different

meaning than the r in the diagram!

I

B

r

Magnetic Field of a Long Straight Wire

We’ve just derived the equation for the magnetic field around a long, straight wire*

μ

π0 IB=

2 r

with a direction given by a “new” right-hand rule.

*Don’t use this equation unless you have a long, straight wire!

Looking “down” along the wire: I B

The magnetic field is not constant.

At a fixed distance r from the wire, the magnitude of the magnetic field is constant.

The magnetic field direction is always tangent to the imaginary circles drawn around the wire, and perpendicular to the radius “connecting” the wire and the point where the field is being calculated.

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two straight segments and a circular arc of radius R that subtends angle .

ds

r̂I

A´

C

A

R

Thanks to Dr. Waddill for the use of the diagram.

C´

Important technique, handy for exams:

The magnetic field due to wire segments A’A and CC’ is zero because ds is parallel to along these paths.

r̂

ds

r̂I

A´

C

A

R

Thanks to Dr. Waddill for the use of the diagram.

C´

Important technique, handy for exams:

Along path AC, ds is perpendicular to .

r̂

ˆˆ

ds r=-ds k

ˆ

ds r =ds

If we use the “usual” xyz axes.

ˆμ

4π

02

I ds rdB=

r

μ

4π0

2

I dsdB=

R

ds

r̂I

A´

C

A

R

Thanks to Dr. Waddill for the use of the diagram.

C´

μ

4π0

2

I dsdB=

R

μ

4π0

2

I dsB=

R

μ

4π 0

2

IB= ds

R

μθ

4π 0

2

IB= R d

R

μθ

4π 0IB= dR

μθ

4π0IB=R

Today’s agenda:

Magnetic Fields Due To A Moving Charged Particle.You must be able to calculate the magnetic field due to a moving charged particle.

Biot-Savart Law: Magnetic Field Due to a Current Element. You must be able to use the Biot-Savart Law to calculate the magnetic field of a current-carrying conductor (for example: a long straight wire).

Force Between Current-Carrying Conductors.You must be able to begin with starting equations and calculate forces between current-carrying conductors.

Magnetic Field of a Current-Carrying Wire

“Knowledge advances when Theory does battle with Real Data.”—Ian Redmount

It is experimentally observed that parallel wires exert forces on each other when current flows.

I1 I2

F12 F21

I1 I2

F12 F21

Skip ahead to slide 29.

The magnitude of the force depends on the two currents, the length of the wires, and the distance between them. μ

π0 1 2 I I L

F=2 d

I1 I2

F12 F21

d

L

The wires are electrically neutral, so this is not a Coulomb force.

We showed that a long straight wire carrying a current I gives rise to a magnetic field B at a distance r from the wire given byμ

π0 IB=

2 r

Remember, the direction of the field is given another (different) right-hand rule: grasp the wire and point the thumb of your right hand in the direction of I; your fingers curl around the wire and show the direction of the magnetic field.

The magnetic field of one wire exerts a force on a nearby current-carrying wire.

This is NOT a starting equation

I

B

r

Example: use the expression for B due to a current-carrying wire to calculate the force between two current-carrying wires.

I1 I2

F12

d

Lμ ˆˆπ0 2

2

IB = k

2 d

12 1 1 2F =I L B

L2L1

B2μ ˆˆπ

0 212 1

IF =I Lj k

2 d

x

y

μ ˆπ

0 1 2

12

I I LF = i

2 d

The force per unit length of wire is μ ˆπ

0 1 212 I IF

= i.L 2 d

Homework Hint: use this technique for any homework problems with long

parallel wires!

I1 I2

F12 F21

d

Lμ ˆπ

0 1

1

IB =- k

2 d

21 2 2 1F =I L B

L2L1

B1μ ˆˆπ

0 1

21 2

IF =I Lj k

2 d

x

y

μ ˆπ

0 1 2

21

I I LF =- i

2 d

The force per unit length of wire is μ ˆπ

0 1 221 I IF

=- i.L 2 d

If the currents in the wires are in the opposite direction, the force is repulsive.

I1 I2

F12 F21

d

L

L2L1y

I1 I2

F12 F21

d

L2L1The official definition of the Ampere: 1 A is the current that produces a force of 2x10-7 N force per meter of length between two long parallel wires placed 1 meter apart in empty space.

μ

π0 1 2

12 21

I I LF =F =

2 d

π

π

-7-71 2

12 21 1 2

4 10 I I L LF =F = =2 10 I I

2 d d