PHY2054: Chapter 19 25

Magnetic Force A vertical wire carries a current and is in a vertical magnetic field. What is the direction of the force on the wire?

(a) left (b) right (c) zero (d) into the page(e) out of the page

I

B

I is parallel to B, sono magnetic force

PHY2054: Chapter 19 26

a

a

bb

Torque on Current LoopConsider rectangular current loop

Forces in left, right branches = 0 Forces in top/bottom branches cancelNo net force! (true for any shape)

But there is a net torque!Bottom side up, top side down (RHR)Rotates around horizontal axis

μ = NiA ⇒ “magnetic moment” (N turns)True for any shape!!Direction of μ given by RHRFingers curl around loop and thumb points in direction of μ

B

( )Fd iBa b iBab iBAτ = = = =Plane normal is ⊥ B here

PHY2054: Chapter 19 27

General Treatment of Magnetic Moment, Torque

μ = NiA is magnetic moment (with N turns)Direction of μ given by RHR

Torque depends on angle θ between μ and B

sinBτ μ θ=

θ

PHY2054: Chapter 19 28

Torque ExampleA 3-turn circular loop of radius 3 cm carries 5A current in a B field of 2.5 T. Loop is tilted 30° to B field.

Rotation always in direction to align μ with B field

30°

( )22 23 3 5 3.14 0.03 0.0339A mNiA i rμ π= = = × × × = ⋅

sin30 0.0339 2.5 0.5 0.042 N mBτ μ= ° = × × = ⋅

B

PHY2054: Chapter 19 29

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x

Trajectory in a Constant Magnetic FieldA charge q enters B field with velocity v perpendicular to B. What path will q follow?

Force is always ⊥ velocity and ⊥ BPath will be a circle. F is the centripetal force needed to keep the charge in its circular orbit. Let’s calculate radius R

FFv

R

v

B

qF

v

PHY2054: Chapter 19 30

x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x

Circular Motion of Positive Particle

BqF

v

2mv qvBR

=mvRqB

=

PHY2054: Chapter 19 31

Cosmic Ray ExampleProtons with energy 1 MeV move ⊥ earth B field of 0.5 Gauss or B = 5 × 10-5 T. Find radius & frequency of orbit.

212

2KK mv vm

= ⇒ =

2mv mKReB eB

= =

( )( )6 19 13

27

10 1.6 10 =1.6 10 J

1.67 10 kg

K

m

− −

−

= × ×

= ×

( )1

2 2 / 2v v eBf

T R mv eB mπ π π= = = = 760Hzf =

2900mR =

Frequency is independent of v!

PHY2054: Chapter 19 32

Helical Motion in B FieldVelocity of particle has 2 components

(parallel to B and perp. to B)Only v⊥ = v sinφ contributes to circular motionv|| = v cosφ is unchanged

So the particle moves in a helical pathv|| is the constant velocity along the B fieldv⊥ is the velocity around the circle

v v v⊥= +

mvRqB

⊥=

Bv

φ

v⊥

v||

PHY2054: Chapter 19 33

Helical Motion in Earth’s B Field

PHY2054: Chapter 19 34

Magnetic Field and WorkMagnetic force is always perpendicular to velocity

Therefore B field does no work!Why? Because

ConsequencesKinetic energy does not changeSpeed does not changeOnly direction changesParticle moves in a circle (if )

( ) 0K F x F v tΔ = ⋅Δ = ⋅ Δ =

v B⊥

PHY2054: Chapter 19 35

Magnetic Force Two particles of the same charge enter a magnetic field with the same speed. Which one has the bigger mass?

ABBoth masses are equalCannot tell without more info

x x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xA B

mvRqB

=

Bigger mass meansbigger radius

PHY2054: Chapter 19 36

Mass Spectrometer

PHY2054: Chapter 19 37

Mass Spectrometer OperationPositive ions first enter a “velocity selector” where E ⊥ B and values are adjusted to allow only undeflected particles to enter mass spectrometer.

Balance forces in selector ⇒ “select” v

Spectrometer: Determine massfrom v and measured radius r

/qE qvBv E B

==

1

2

11

22

m vrqBm vrqB

=

=

PHY2054: Chapter 19 38

Mass Spectrometer Example A beam of deuterons travels right at v = 5 x 105 m/s

What value of B would make deuterons go undeflected through a region where E = 100,000 V/m pointing up vertically?

If the electric field is suddenly turned off, what is the radius and frequency of the circular orbit of the deuterons?

5 5/ 10 /5 10 0.2T

eE evB

B E v

=

= = × =

( )( )5

62

1 5 10 1.5 10 Hz2 6.28 5.2 10

vfT Rπ −

×= = = = ×

×

( )( )( )( )

27 522

19

3.34 10 5 105.2 10 m

1.6 10 0.2mv mvevB R

R eB

−−

−

× ×= ⇒ = = = ×

×

PHY2054: Chapter 19 39

Quiz: Work and Energy A charged particle enters a uniform magnetic field. What happens to the kinetic energy of the particle?

(1) it increases (2) it decreases(3) it stays the same(4) it changes with the direction of the velocity(5) it depends on the direction of the magnetic field

Magnetic field does no work, so K is constant

PHY2054: Chapter 19 40

Magnetic Force A rectangular current loop is in a uniform magnetic field. What direction is the net force on the loop?

(a) +x (b) +y (c) zero (d) –x(e) –y

B

x

z

y

Forces cancel onopposite sides of loop

PHY2054: Chapter 19 41

Hall Effect: Do + or – Charges Carry Current?

+ charges moving counter-clockwise experience upward force

Upper plate at higher potential

– charges moving clockwise experience upward force

Upper plate at lower potential

Equilibrium between magnetic (up) & electrostatic forces (down):

This type of experiment led to the discovery (E. Hall, 1879) that current in conductors is carried by negative charges

up driftF qv B= down inducedHVF qE qw

= =

drift "Hall voltage"HV v Bw= =

PHY2054: Chapter 19 42

Electromagnetic Flowmeter

Moving ions in the blood are deflected by magnetic forcePositive ions deflected down, negative ions deflected upThis separation of charge creates an electric field E pointing upE field creates potential difference V = Ed between the electrodesThe velocity of blood flow is measured by v = E/B

E

PHY2054: Chapter 19 43

Creating Magnetic FieldsSources of magnetic fields

Spin of elementary particles (mostly electrons)Atomic orbits (L > 0 only)Moving charges (electric current)

Currents generate the most intense magnetic fieldsDiscovered by Oersted in 1819 (deflection of compass needle)

Three examples studied hereLong wireWire loopSolenoid

PHY2054: Chapter 19 44

B Field Around Very Long WireField around wire is circular, intensity falls with distance

Direction given by RHR (compass follows field lines)

02

iBr

μπ

=

70 4 10μ π −= ×

Right Hand Rule #2

μ0 = “Permeability of free space”

PHY2054: Chapter 19 45

Visual of B Field Around Wire

PHY2054: Chapter 19 46

B Field ExampleI = 500 A toward observer. Find B vs r

RHR ⇒ field is counterclockwise

r = 0.001 m B = 0.10 T = 1000 Gr = 0.005 m B = 0.02 T = 200 Gr = 0.01 m B = 0.010 T = 100 Gr = 0.05 m B = 0.002 T = 20 Gr = 0.10 m B = 0.001 T = 10 Gr = 0.50 m B = 0.0002 T = 2 Gr = 1.0 m B = 0.0001 T = 1 G

( )7 40

4 10 500 102 2

iBr r r

πμπ π

− −×= = =

PHY2054: Chapter 19 47

Charged Particle Moving Near WireWire carries current of 400 A upwards

Proton moving at v = 5 × 106 m/s downwards, 4 mm from wireFind magnitude and direction of force on proton

SolutionDirection of force is to left, away from wireMagnitude of force at r = 0.004 m

Iv

02

IF evB evr

μπ

⎛ ⎞= = ⎜ ⎟⎝ ⎠

( )( )7

19 6 2 10 4001.6 10 5 100.004

F−

− ⎛ ⎞× ×= × × ⎜ ⎟⎜ ⎟

⎝ ⎠141.6 10 NF −= ×

PHY2054: Chapter 19 48

Ampere’s LawTake arbitrary path around set of currents

Let ienc be total enclosed current (+ up, − down)Let Bll be component of B along path

Only currents inside path contribute!5 currents inside path (included)1 outside path (not included)

0 enci

B s iμΔ =∑Not included

in ienc

PHY2054: Chapter 19 49

Ampere’s Law For Straight WireLet’s try this for long wire. Find B at distance at point P

Use circular path passing through P (center at wire, radius r)From symmetry, B field must be circular

An easy derivation

( ) 0

0

2

2

iB s B r i

iBr

π μ

μπ

Δ = =

=

∑

r

P

PHY2054: Chapter 19 50

Useful Application of Ampere’s LawFind B field inside long wire, assuming uniform current

Wire radius R, total current iFind B at radius r = R/2

Key fact: enclosed current ∝ area

2enc

enc 2tot 4

A r ii i iA R

ππ

⎛ ⎞= × = × =⎜ ⎟⎜ ⎟

⎝ ⎠

0

0

22 4

12 2

iR iB s B

iBR

π μ

μπ

⎛ ⎞Δ = =⎜ ⎟⎝ ⎠

=

∑

r

R

02

iBR

μπ

= On surface

0 enci

B s iμΔ =∑

r = R/2

PHY2054: Chapter 19 51

Ampere’s Law (cont)Same problems: use Ampere’s law to solve for B at any r

Wire radius R, total current i

2 2enc

enc 2 2tot

A r ri i i iA R R

i

ππ

⎛ ⎞= × = × =⎜ ⎟⎜ ⎟

⎝ ⎠=

( )2

0 02

0

2 or

2

irB s B r i iR

i rBR R

π μ μ

μπ

⎛ ⎞Δ = = ⎜ ⎟⎜ ⎟

⎝ ⎠

=

∑

r

R

02

iBr

μπ

=

0 enci

B s iμΔ =∑

(r ≤ R)

r ≥ R

(r ≥ R)

(r ≤ R)

PHY2054: Chapter 19 52

Force Between Two Parallel CurrentsForce on I2 from I1

RHR ⇒ Force towards I1

Force on I1 from I2

RHR ⇒ Force towards I2

Magnetic forces attract two parallel currents

I1I2

0 1 0 1 22 2 1 2 2 2

I I IF I B L I L Lr r

μ μπ π

⎛ ⎞= = =⎜ ⎟⎝ ⎠

I1I2

0 2 0 1 21 1 2 1 2 2

I I IF I B L I L Lr r

μ μπ π

⎛ ⎞= = =⎜ ⎟⎝ ⎠

PHY2054: Chapter 19 53

Force Between Two Anti-Parallel CurrentsForce on I2 from I1

RHR ⇒ Force away from I1

Force on I1 from I2

RHR ⇒ Force away from I2

Magnetic forces repel two antiparallel currents

I1I2

I1I2

0 1 0 1 22 2 1 2 2 2

I I IF I B L I L Lr r

μ μπ π

⎛ ⎞= = =⎜ ⎟⎝ ⎠

0 2 0 1 21 1 2 1 2 2

I I IF I B L I L Lr r

μ μπ π

⎛ ⎞= = =⎜ ⎟⎝ ⎠

PHY2054: Chapter 19 54

Parallel Currents (cont.)Look at them edge on to see B fields more clearly

Antiparallel: repel

FF

Parallel: attract

F F

B

BB

B

2 1

2

2

2

1

11

PHY2054: Chapter 19 55

B Field @ Center of Circular Current LoopRadius R and current i: find B field at center of loop

Direction: RHR #3 (see picture)

If N turns close together

02

iBR

μ=

02

N iBRμ

=

From calculus

PHY2054: Chapter 19 56

Current Loop Examplei = 500 A, r = 5 cm, N=20

( )( )70

20 4 10 5001.26T

2 2 0.05iB N

r

πμ−×

= = =×

PHY2054: Chapter 19 57

B Field of SolenoidFormula found from Ampere’s law

i = currentn = turns / meter

B ~ constant inside solenoidB ~ zero outside solenoidMost accurate when

Example: i = 100A, n = 10 turns/cmn = 1000 turns / m

0B inμ=

( )( )( )7 34 10 100 10 0.13TB π −= × =

L R