Today’s agenda: Magnetic Fields Due To A Moving Charged

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 calculate forces between current-carrying conductors.

*last week we studied the effects of magnetic fields on charges, today we learn how to produce magnetic fields

+

B

r

r

vˆμ .

4π

02

qv rB =r

0 is a constant, 0=4x10-7 T·m/A

Remember: r is unit vector from source point(the thing that causes the field) to the field point P (location where the field is being measured).

Magnetic Field of a Moving Charged Particle

• moving charge creates magnetic field

As in lecture 14: Motion with respect to what? You, the earth, the sun? Highly nontrivial, leads to Einstein’s theory of relativity.

P

Detour: cross products of unit vectors

• need lots of cross products of unit vectors

Work out determinant:

Example: ˆˆ ˆ

ˆ ˆ ˆ ˆ

i j kk j = det 0 0 1 = i 0 - -1 = i

0 -1 0

ˆˆ ˆi, j, k

Use right-hand rule:

Detour: cross products of unit vectors

i j k i j k

ˆˆ ˆi j=k

Cyclic property:

i j k i j k

ˆˆ ˆj i = -k

“forward” “backward”

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

v0

E

Homework Hint: this and the next 3 slides!



Electric force on proton 2:

1

r

x

y

z

2

v0

v0

• Electric field due to proton 1 at the position of proton 2:

ˆˆ

1

1 2 2q1 1 eE = r = j

4 r 4 r

r

FE

ˆ ˆ

2

E 1 2 21 e 1 eF = qE = e j= j

4 r 4 r

E• this electric field exerts a force

on proton 2

1

r

x

y

z

2

r

FE

Magnetic force on proton 2:

•magnetic field due to proton 1 at the position of proton 2

ˆ

1 1

1 2q v rB =

4 r

ˆ ˆ

0

1 2ev i jB =

4 r

ˆ

0

1 2evB = k

4 r

B1

v0

v0

1

r

x

y

z

2

r

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 2evF = ev i k

4 r

ˆ

2 20

B 2e vF = j

4 r

FB

v0

v0

What would proton 1 “feel?”

1

r

x

y

z

2

r

FE

B1

• both forces are in the +y direction• ratio of their magnitudes:

2 20

2B

2E

2

e v4 rF =

F 1 e4 r

FB

v0

v0 2B0

E

F = vF

Later we will find that

21=c

1

r

x

y

z

2

r

FE

B1

Thus

FB

v0

v0

20B2

E

vF =F c

If v0=106 m/s, then

26-5B

28E

10F = 1.11 10F 3 10

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.

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


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.”


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:


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.

current I in infinitesimal length of wire gives rise to magnetic field

dB

r

r

dl

Biot-Savart Law

ˆμ4π

02

I d rdB =r

I

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

Biot-Savart Law: magnetic field of a current elementd

dB

Derived, as in lecture 15, by summing contributions of all charges in wire element

Applying the Biot-Savart Law

I

ds

r

r

dBˆμ

4π

02

I ds rdB = r

μ θ4π

02

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.

P

ˆ ˆ

ds r = ds r sin

ˆ = ds sin because r =1

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.)

ˆμ4π

02

I ds rdB =r

ˆˆ θ

ds r = ds sin k

μ θ4π

02

I ds sindB =r

ds is an infinitesimal quantity in the direction of dx, soμ θ4π

02

I dx sindB =r

I

y

r

x

dBP

dsr

xz

a

L

2 2r = x +aθ asin =r

I

y

r

x

dBP

dsr

xz

μ θ4π

02

I dx sindB =r

a

μ μ4π 4π

0 03/23 2 2

I dx a I dx adB = =r x +a

μ4π

L/2 03/2-L/2 2 2

I dx aB =x +a

μ

4π L/20

3/2-L/2 2 2

I a dxB =x +a

L

I

y

r

x

dBP

dsr

xz

a

μ

4π L/20

3/2-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

dsr

xz

a

μ4π

01/22 2 2

I a 2L/2B =a L /4+a

μ4π0

1/22 2

I L 1B =a L /4+a

μπ

02 2

I L 1B =2 a L +4a

μπ0

2

2

I 1B =2 a 4a1+

L

L

I

y

r

x

dBP

dsr

xz

a μπ0

2

2

I 1B =2 a 4a1+

L

When L, μ .π0I B =

2 a

L

r = − x i+a j

r= √x2+a2

r= − x √x2+a2

i+ a √x2+a2

j

d s=dx i

Solution beginning with vector notation

d B=μo I4π

d s×rr 2

d B=μo I4π

dx i×( − x √x 2+a2

i+ a √x 2+a2

j)x2+a2

d B=μo I4π

adx(x2+a2)3/2

k

B=μo I a4π ∫ − L/2

L/2 dx(x2+a2)3/2

k

B=μo I a4π [ x

a2 √x2+a2k ]

− L /2

L /2

B=μo I a4π [ x

a2 √x2+a2k ]

− L/2

L/2

B=μo I a4π [ L /2

a2 √L2

4 +a2

− − L /2

a2 √L2

4 +a2 ]kB=

μo I a4π [ L

a2 √L2

4+a2] k

I

Br

Magnetic Field of a Long Straight Wire

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

μπ0 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!

link to image source

r is shortest (perpendicular) distance between field point and wire

Looking “down” along the wire:

I B• magnetic field is not constant

• at fixed distance r from wire, magnitude of field is constant (but vector magnetic field is not uniform).

• magnetic field direction is a tangent to imaginary circles around wire

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

I

A´

C

A

R

C´

O



I see three “parts” to the wire.

As usual, break the problem up into simpler parts.

A’ to A

A to C

C to C’


I

A´

C

A

R

C´

O



For segment A’ to A: dsr

ˆ ˆ 0

ds r = ds r sin =0

A'AdB =0

A'AB =0

ˆμ4π

02

I ds rdB =r


A´

C

A

R

C´

O



For segment C to C’: dsr

ˆ ˆ

ds r = ds r sin 180 =0

CC'dB =0

CC'B =0

ˆμ4π

02

I ds rdB =r

I

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 .

A´

C

A

R

C´

Important technique, handy for homework and exams:

The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to along those paths.

rO

dsr

dsr




For segment A to C: ˆ ˆ

ds r = ds r sin 90 =0

ˆμ4π

02

I ds rdB =r

A´

C

A

R

C´

O

Ir

ds

= ds 1 1 = ds

μ4π

0

AC AC 2I dsdB = dB =R

A´

C

A

R

C´

O

Ir

ds

If we use the standard xyz axes, the direction is -k.

Cross into . Direction is “into” the page, or , for all wire elements.

ds r

x

y

z

Direction of

dB

A´

C

A

R

C´

O

Ir

ds

μ4π

0AC 2

I dsdB =R

μ4π 0

AC AC 2arc arc

I dsB = dB =R

μ θ4π

0AC

IB =R

μ4π 0

AC 2arc

IB = dsR

The integral of ds is just the arc length; just use that if you already know it.

μ θ4π 0

AC 2arc

IB = R dR

μ θ ˆ4π

0

A'A AC CC'IB = B + B + B = - kRFinal answer:

A´

C

A

R

C´

O

Ir

ds

Important technique, handy for exams:

Along path AC, ds is perpendicular to .r

ˆ ˆ

ds r = ds r sin 90

ˆ

ds r = ds

Today’s agenda:


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

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

I1 I2

F12 F21

I1 I2

F12 F21

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

L

μ ˆˆπ0 2

2IB = k

2 d

12 1 1 2F = I L B

L2L1

B2

μ ˆˆπ

0 212 1

IF = I Lj k2 d

μ ˆπ

0 1 2

12I I LF = i

2 dforce per unit length of wire i: μ ˆ

π

0 1 212 I IF = i.

L 2 d

x

y

z

Force on wire 1 produced by wire 2

I1 I2

F12 F21

d

Lμ ˆπ

0 1

1IB = - k

2 d

21 2 2 1F = I L B

L2L1

B1μ ˆˆ

π

0 1

21 2IF = I Lj k

2 d

μ ˆπ

0 1 2

21I I LF = - i

2 d

The force per unit length of wire is μ ˆπ

0 1 221 I IF = - i.

L 2 d

B

x

y

z

Force on wire 2 produced by wire 1

Analogously:If currents are in opposite directions, force is repulsive.

I1 I2

F12 F21

d

L

L2L1

Official definition of the Ampere:

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

μπ

0 1 212 21

I I LF =F =2 d

ππ

-7-71 2

12 21 1 24 10 I I L LF =F = =2 10 I I

2 d d

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Today’s agenda: Magnetic Fields Due To A Moving Charged