Sources of magnetic field Lots of details so far on how Magnetic - - PDF document

Lots of details so far on how Magnetic Fields exert forces. But, where did those Magnetic Fields come from? B-fields are created by moving charges (currents). This seems quite opposite to what we might expect!

Sources of magnetic field

Magnetic Field created by a single moving charged particle at a position

2

ˆ 4 r r v q B × = v v π µ

Constant µ0 = 1.257 x 10-6 Tesla meter/Amp. is a vector in the direction of , but with magnitude = 1 (often referred to as a unit vector). direction is from charge to where you are evaluating B !

Field created by a moving charge

r r r ˆ r ˆ r r

Field created by a moving charge

A new Right Hand Rule… Thumb in direction of the motion, and fingers then curl in direction of the B-field.

2

ˆ 4 r r v q B × = v v π µ

2

ˆ 4 r r v q B × = v v π µ

Clicker Question What is the direction of the B- field at point P indicated? A)Up and to the Left B)Down and to the Left C)Out of the Page D)Into the Page E)None of the Above Point P r

Example: Forces between two moving protons z r ev r r v e B ˆ 4 ˆ 4

2 2

π µ π µ = × = v v

y r e FE ˆ 4 1

2 2

πε = r

( )

y r v e z x r ev v e z r ev x v e B v e FB ˆ 4 ˆ ˆ 4 ) ( ˆ 4 ) ˆ ( ) (

2 2 2 2 2

π µ π µ π µ = × − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × − = × − = r r r

Now the B-field created by many moving charges (i.e. current flowing in a wire). I

2 2 2

ˆ 4 ˆ ) ( 4 ˆ 4 r r l d I r r v nqAdl r r v dQ B d

d d

× = × = × = r r r v π µ π µ π µ

dl X r dB Biot-Savart first discovered this experimentally.

Magnetic field of a current element

I

2

ˆ 4 r r l d I B d × = r v π µ

Magnetic field of a current element

2

ˆ 4 r r L d I B d × = v v π µ

∫ ∫

× = =

2

ˆ 4 r r L d I B d Btot v v v π µ

This can be a very difficult integral to evaluate.

Magnetic field of a current-carrying wire

2

ˆ 4 r r L d I B d × = v v π µ page) the into ( sin 4

2

θ π µ r dy I B d = v

2 2 2

x y r + =

2 2

/ / sin y x x r x + = = θ Now a little geometry

Magnetic field from an infinite straight wire

( )

2 / 3 2 2

4 y x xdy I B d + = π µ v

∫ ∫

+∞ ∞ −

+ = =

2 / 3 2 2

) ( 4 y x xdy I B d B π µ v v

page) the (into 2 x I B π µ = v

( )

2 / 3 2 2

4 y x xdy I B d + = π µ v

Magnetic field from an infinite straight wire (cont.)

2 | | R I B π µ = v

This is a key result! B-field a perpendicular distance x away from an infinite (or very long) wire.

Magnetic field from an infinite straight wire (cont.)

Clicker Question i A long wire has a current moving as shown. What is the direction of the B-field created by the wire just below the wire? B=? A)Into the Page B)Out of the Page C)To the right D)Down E)None of the Above

2

ˆ 4 r r L d i B d × = v v π µ

L r

Clicker Question i A long wire has a current moving as shown. What is the direction of the B-field created by the wire just above the wire? B=? A)Into the Page B)Out of the Page C)To the right D)Down E)None of the Above

2

ˆ 4 r r L d i B d × = v v π µ

L r Power line has 500 Amps going through it. What is the B-field strength 15 meters below

• n the ground?

T m A A Tm R I B

6

• 6

10 7 . 6 ) 15 ( 2 ) 500 )( / 10 (1.26 2

−

× = × = = π π µ

Example with numbers

I I’

• Think of the Red Wire as creating

a B-field.

• Then think of that B-field creating

a force on the moving charges (current) in the Blue Wire.

Interaction between two current carrying wires

The two wires may exert forces on each other through Magnetic Interactions.

Interaction between two current carrying wires

B L I F v v v × = ' r r I B ˆ 2 π µ = r r L II LB I F π µ 2 ' ' = =

Clicker Question I I What is the direction of the Force acting on the Blue Wire? A)Up B)Right C)Left D)Into the Page E)Out of the Page

B L I F v v v × =

i1 i2 The B-field from the Red Wire at the location of the Blue Wire is into the page. X

left) ( 2

1 2 1 2 2

R i L i B L i F π µ = × = v v v

Then the Force on the Blue Wire is to the left.

page) the into ( 2

1 1

R i B π µ = v

R

Clicker Question i i What is the direction of the Force acting on the Red Wire? A)Up B)Right C)Left D)Into the Page E)Out of the Page

B L I F v v v × =

i1 i2 The B-field from the Blue Wire at the location of the Red Wire is out of the page. .

right) ( 2

2 1 2 1 1

R i L i B L i F π µ = × = v v v

Then the Force on the Red Wire is to the right.

page) the

• f
• ut

( 2

2 2

R i B π µ = v

R

i i Wires with parallel currents attract each other. i Wires with antiparallel currents repel each other. Try following the procedure we just outlined to confirm this for yourselves. i What happens if we flip the direction of one current?

Interaction between current-carrying wires

What about the B-field at the center of a circular loop of wire

• f radius a and current I?

) ( 4 4

2 2 2

a x dl I r dl I dB + = = π µ π µ

2 / 3 2 2

) ( 4 cos a x adl I dB dBx + = = π µ θ

Field through a circular loop

∫ ∫ ∫

+ = + = = dl a x Ia a x adl I dB B

x x 2 / 3 2 2 2 / 3 2 2

) ( 4 ) ( 4 π µ π µ

2 / 3 2 2 2

) ( 2 a x Ia Bx + = µ

Field through a circular loop

Which point A or B has the larger magnitude Magnetic Field? A B C : The B-field is the same at A and B.

A B

I I

Answer: Case B has the larger magnetic field. Use the Biot-Savart Law to get the directions of the B-field due to the two semi-circular portions of the loop. In A the two fields oppose each other; in B they add. Clicker Question

2

ˆ 4 r r l d I B d × = r v π µ

Clicker Question

2 / 3 2 2 2

) ( 2 a x Ia B + = µ

In the limit as x >> a, the expression for the B-field becomes? A) B) C)

3 2

2 x Ia B µ =

a I B 2 µ =

) ( 2

2 2 2

a x Ia B + = µ

a

I x B=?

x x Ia B ˆ 2

3 2

µ = v x x IA x x a I B ˆ 2 ˆ 2

3 3 2

π µ π π µ = =

a

I x B=? * A = area of loop

A I v v = µ

Magnetic Dipole Moment

3

2 x B µ π µ v v =

Magnetic Dipole moment

A I v v = µ

3

2 x B µ π µ v v =

Any current loop looks like a Magnetic Dipole far away. The B-field drops as the distance^3 And depends on the Magnetic Dipole Moment.

Magnetic Dipole moment

A square loop of side length a of wire carrying current I is in a uniform magnetic field B. The loop is perpendicular to B (B

• ut of the page). What is the magnitude of the net force on

the wire?

I B

A: IaB B: 4IaB C: 2IaB D: 0 E: None of these Clicker Question

B L I F v v v × =

Clicker Question

r B Bz = ⋅ $, where B = B(y) = A y

D B C A B stronger B weaker E: net force is zero

The same loop is now in a non-uniform field. where A is a constant. The direction of the net force is?

y x

In a uniform B-field, regardless of the orientation between the B and the Magnetic Moment of the loop µ, the net force is always zero.

However, that does not mean the net torque is zero!

Uniform fields produce zero net force!!!

µ If µ is not parallel to B, then there is a net torque.

∑ ×

= F r v v v τ ] sin ) 2 / [( 2 ) ( θ τ LB I L B L I r

∑

= × × = v v v v θ τ sin

2B

IL = v B A I v v v × = τ B v v v × = µ τ

Magnetic torque

µ

B v v v × = µ τ

Torque wants to twist the loop so that µ and B align.

The DC motor

Clicker Question i1 i2 Two loops of wire have current going around in the same direction. The forces between the loops is: A)Attractive B)Repulsive C)Net force is zero.

Magnetic Field Lines and Flux

Gauss’s Law for Magnetic Fields

Magnetic monopoles (so far) do not exist!!! There are no sources of magnetic flux

∫

= ⋅ A d B r r

=>Flux through a closed surface:

Is there something like Gauss’ Law for Electric Fields that helps us solve for Magnetic Fields in a simpler way for cases with nice symmetries?

Ampere’s law

∫

= ⋅

loop thru

I l d B µ v v

For any closed imaginary loop where the current is constant, the above relation is true.

Clicker Question

∫

= ⋅

loop thru

I l d B µ v v

We need a sign convention for I(thru). Place the fingers of your right hand around the imaginary loop and your thumb points in the direction of positive I(thru). What is I(thru) in the below case where all three wires have 5 A? A) I(thru) = +15 Amps B) I(thru) = +5 Amps C) I(thru) = -5 Amps D) None of the above X X i Place an imaginary circular loop of radius R around the wire.

∫

⋅

loop

l d B v v

Vector dot product!

Ampere’s law for a straight wire

Clicker Question Look at the previous wire from the view where the wire current is coming out of the page. The Ampere circular loop is drawn in blue. What is the relation of the vector dl that

• ne integrates around the loop as shown

and the Magnetic Field vector? A)Parallel B)Anti – Parallel C)Perpendicular D)Depends where in the loop for dl i dl i Thus, this is a special symmetry case where all around the loop the two vectors are parallel.

l d B v v ||

In fact, also due to symmetry, we know that the magnitude of the B-field is also constant.

loop the around constant | | B v

i

a B l d B l d B

loop loop

π 2 | | | | v v v v v = = ⋅

∫ ∫

Where a is the radius of the loop.

∫

= ⋅

loop thru

I l d B µ v v

I a B 2 | | µ π = v a I B π µ 2 | | = v

Ampere’s law for a straight wire

b r J = constant

A long straight copper wire has radius b and carries a constant current of magnitude i. The current density of magnitude J=i/(πb2) is uniform throughout the wire. What is the current contained in the circular loop £, with radius r < b, centered on the wire's center as shown? Clicker Question A) B) C) D) None of these. b r i

2 2

b r i

3 3

b r i

A)B ∝ r B) B = constant C) B ∝ 1/r D) B ∝ 1/r2 E) None of these

b r J = constant

A long straight copper wire has radius b and carries a constant current of magnitude i. The current density of magnitude J=i/(πb2) is uniform throughout the wire. How does the magnitude of the B-field a distance r < b from the center of the wire depend on r? Clicker Question

∫

= ⋅

loop thru

i l d B µ v v

b r J = constant

∫

= ⋅

loop thru

i l d B µ v v

For r < b:

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =

2 2

2 | | b r i r B µ π v

2 0 2

| | b r i B π µ = v

For r >= b:

r i B π µ 2 | | = v

|B| r b

Clicker Question We have determined that for the closed loop as drawn below the integral on the left is -10 Tesla meters. What is the direction of the current in the red wire shown below? A) Into the Page B) Out of the Page C) Left D) Up E) Down

∫

= ⋅

loop thru

i l d B µ v v

A single wire tightly coiled up into loops. Since it is a single wire, the current magnitude is the same in all parts of the coil.

Field of a solenoid

What is the B-field from a solenoid of N turns and length L? Length = L Number of Turns = N n = N/L Current Direction

Field of a solenoid

Side View X X X X X

∫

= ⋅

loop thru

I l d B µ v v ) ( | |

0 nlI

l Binside µ = v

l Uniform B Inside X X X Zero B Outside

I L N nI Binside | | µ µ = = v

B-Field Inside a Solenoid # loops in length l

Field of a solenoid

Three long straight solenoids all of length L and all with the same (large) number of closely-packed turns N, all with the same current i, have different cross-sections as shown.

A B C

Which solenoid has the largest field |B| at its center? A) A B) B C) C D) All three have the same magnitude magnetic field Clicker Question Answers: The field is the same magnitude and uniform for all three solenoids. The field within a solenoid is B = µni. This depends only on the current i and the turns per length n. This formula does not depend on either the cross-sectional shape

• f the solenoid or the position within the solenoid.

Toroidal solenoid

Using Ampere’s Law:

I r N B NI rB π µ µ π 2 2 = → =

If the “thickness” of the core is very small:

nI B n r N 2 µ π = → ≈

Projectile

Conducting Gas

I X B

B L I F v v v × =

How does a rail gun work?

Rail Guns are by far the most spectacular type of electromagnetic accelerators ever developed. They hold the record for fastest object accelerated of a significant mass, for the 16000m/s firing of a .1 gram object by Sandia National Research Laboratories' 6mm Hypervelocity Launcher, and they can also propel objects of very sizeable masses to equally impressive velocities, such as in the picture, where Maxwell Laboratories' 32 MegaJoule gun fires a 1.6 kilogram projectile at 3300m/s (that's 9 MegaJoules of kinetic energy!) at Green Farm research facility.

2

ˆ 4 r r l d I B d × = r v π µ

2

ˆ 4 r r v q B × = v v π µ

Moving charges (currents) create B-fields. Where are the moving charges?

Natural magnetism: Where is the B-field coming from?

Atoms have moving electric charges.

Natural magnetism: Where is the B-field coming from?

5Tesla 2 B ≈ = R I µ v

Consider a single electron moving in a circular orbit around the nucleus.

A m s m C πR ev I

4 10 6 19

10 6 . 7 ) 10 ( 2 ) / 10 3 )( 10 6 . 1 ( 2

− − −

× = × × = = π

What is the B-field at the center (at the nucleus)? Electron in orbit, but current in

• pposite direction.

Natural magnetism: Where is the B-field coming from?

In most materials all the electrons in orbit have random

• rientations.

Superposition of B-field vectors over many atoms gives B=0

Natural magnetism

In Ferromagnetic materials (Fe, Ni, Cr, some alloys containing these metals too), the atomic currents can all

• rient the same way, making a net B-field.

B B B B

Ferromagnetism

Sometimes the material is fragmented into many domains (top) and is thus unmagnetized. Domains that are magnetized in the direction of the field grow larger.

Ferromagnetism

In all permanent magnets (just like a current loop) the B-field points away from one end (North) and towards the other end (South). Very important convention. Thus, these magnets act just like a current loop! i

Permanent magnets

What is the current direction in this loop? And which side

• f the loop is the north pole?
• A. Current clockwise; north pole on top
• B. Current counterclockwise, north pole on top

C.Current clockwise; north pole on bottom D.Current counterclockwise, north pole on bottom i Clicker Question Atoms have a Magnetic Dipole Moments from the orbit of the electrons.

A I v v =

• rbit)

( µ

But, amazingly electrons themselves also have a Magnetic Dipole Moment.

spin)

• r

electron ( µ v

Electron “spin”

Is the electron really a fundamental point-like particle? Or does it have little charges moving around inside it? Or could the electron be a rotating sphere of charge?

We know from the world’s best microscopes (particle physics accelerators) that the electron is less than 10-18 meters across. Thus, knowing the Magnetic Dipole Moment, we can calculate how fast the outside of the sphere must be rotating.

• nd

meters/sec 10 5 . 2

14

× = v

Electron “spin”

light)

• f

(speed / 10 3 m/s 10 5 . 2

8 14

s m v × >> × =

This velocity is faster than the speed of light, and thus not allowed in Einstein’s Theory of Relativity. Thus, it may be that electrons just have an intrinsic property

• f Magnetic Dipole Moment.

This is not understood at any more fundamental level today.

Electron “spin”

It turns out in permanent magnets, the atoms have many electrons in orbits with different orientations. Thus, it is often the intrinsic magnetic dipole moment of the electrons that is important and tends to be aligned into domains. i Clicker Question A permanent bar magnet is broken in half. Do the pieces attract or repel?

1 2

A: Attract B: Repel C: Neither! There is no net force. Answer: They attract. The bar magnet can be thought of as a coil of current. Parallel currents attract. The currents on the ends are parallel so there is an attraction. i i

Clicker Question A permanent bar magnet is broken in half. The two pieces are interchanged, keeping their orientations fixed, as shown

• below. Do the pieces attract or repel?

1 2 2 1

A: Attract B: Repel C: Neither! There is no net force. Answer: They attract. The bar magnet can be thought of as a coil of current. Parallel currents attract. When the magnet is broken and rearranged, the currents on the ends are parallel so there is an attraction. Opposite Poles attract. Similar Poles repel. This is not new physics. It is identical to the physics of

• ur current loops.

Now one can see why we can never break a magnet and get just one pole!