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.

+

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.

Magnetic Field of a Moving Charged Particle

cross products of unit vectors

We are going to be doing lots of cross products of unit vectors.Here are some handy ways to do the cross product.

ˆˆ ˆ

x y z

x y z

i j k

C= det A A Av

B B B

C=A B

The “always works if you can do math” method

Example: ˆˆ ˆ

ˆ ˆ ˆ ˆ

i j k

k j = det 0 0 1 =i 0- -1 =i

0 -1 0

cross products of unit vectors

We are going to be doing lots of cross products of unit vectors.Here are some handy ways to do the cross product.

The right hand rule method

http://en.wikipedia.org/wiki/Cross_product

link to image

cross products of unit vectors

We are going to be doing lots of cross products of unit vectors.Here are some handy ways to do the cross product.

The “I learned this decades ago and forgot the name for it” method

i j k i j k

ˆˆ̂i j=k

cross products of unit vectors

i j k i j k

ˆˆ̂j i=-k

i j k i j k

ˆˆ ˆi k=-j

i j k i j k

ˆˆ ˆ ˆj k=- -i =i

This is example 28.1 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: this and the next 3 slides!

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.

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.

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

To calculate the magnetic force: 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.