Magnetic Force
ÎA
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
B
I is parallel to B, so
no magnetic force
I
PHY2054: Chapter 19
25
Torque on Current Loop
Î Consider
a
rectangular current loop
Forces in left, right branches = 0
‹ Forces in top/bottom branches cancel
‹ No net force! (true for any shape)
‹
Î But
b
there is a net torque!
Bottom side up, top side down (RHR)
‹ Rotates around horizontal axis
‹
τ = Fd = ( iBa ) b = iBab = iBA
Îμ
B
b
a
Plane normal is ⊥ B here
= NiA ⇒ “magnetic moment” (N turns)
True for any shape!!
‹ Direction of μ given by RHR
‹ Fingers curl around loop and thumb points
in direction of μ
‹
PHY2054: Chapter 19
26
General Treatment of Magnetic Moment, Torque
Îμ
= NiA is magnetic moment (with N turns)
‹ Direction
of μ given by RHR
θ
ÎTorque
depends on angle θ between μ and B
τ = μ B sin θ
PHY2054: Chapter 19
27
Torque Example
ÎA
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.
B
30°
2
2
Î μ = NiA = 3iπ r = 3 × 5 × 3.14 × ( 0.03 ) = 0.0339 A ⋅ m
2
Îτ
= μ B sin 30° = 0.0339 × 2.5 × 0.5 = 0.042 N ⋅ m
ÎRotation
always in direction to align μ with B field
PHY2054: Chapter 19
28
Trajectory in a Constant Magnetic Field
ÎA
charge q enters B field with velocity v perpendicular to
B. What path will q follow?
is always ⊥ velocity and ⊥ B
‹ Path will be a circle. F is the centripetal force needed to keep the
charge in its circular orbit. Let’s calculate radius R
‹ Force
x x x x x x x x x x x x x x
B
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 vx x
F
F
v
v
F
q
R
PHY2054: Chapter 19
29
Circular Motion of Positive Particle
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 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
v
B
F
q
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 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
2
mv
= qvB
R
mv
R=
qB
PHY2054: Chapter 19
30
Cosmic Ray Example
with energy 1 MeV move ⊥ earth B field of 0.5
Gauss or B = 5 × 10-5 T. Find radius & frequency of orbit.
ÎProtons
K=
1 mv 2
2
2K
⇒ v=
m
( )(
)
K = 106 1.6 × 10−19 =1.6 × 10−13 J
m = 1.67 × 10−27 kg
R = 2900 m
mv
2mK
R=
=
eB
eB
1
v
v
eB
f = =
=
=
T 2π R 2π ( mv / eB ) 2π m
f = 760 Hz
Frequency is independent of v!
PHY2054: Chapter 19
31
Helical Motion in B Field
ÎVelocity
of particle has 2 components
G G G
‹ v = v& + v⊥ (parallel to B and perp. to B)
‹ Only
v⊥ = v sinφ contributes to circular motion
‹ v|| = v cosφ is unchanged
ÎSo
the particle moves in a helical path
‹ v||
is the constant velocity along the B field
‹ v⊥ is the velocity around the circle
v||
v
v⊥
B
φ
mv⊥
R=
qB
PHY2054: Chapter 19
32
Helical Motion in Earth’s B Field
PHY2054: Chapter 19
33
Magnetic Field and Work
ÎMagnetic
force is always perpendicular to velocity
‹ Therefore
B field does Gno work! G
G
G
‹ Why? Because ΔK = F ⋅ Δx = F ⋅ ( v Δt ) = 0
ÎConsequences
‹ Kinetic
energy does not change
‹ Speed does not change
‹ Only direction changes
G
‹ Particle moves in a circle (if v ⊥
G
B)
PHY2054: Chapter 19
34
Magnetic Force
ÎTwo
particles of the same charge enter a magnetic field
with the same speed. Which one has the bigger mass?
‹A
‹B
‹ Both
masses are equal
‹ Cannot tell without more info
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 x x x x x x x
mv
R=
qB
x x x x x x x x x x x x
x x x x x x x x x x x x
A
Bigger mass means
bigger radius
PHY2054: Chapter 19
B
35
Mass Spectrometer
PHY2054: Chapter 19
36
Mass Spectrometer Operation
ions first enter a “velocity selector” where E ⊥ B
and values are adjusted to allow only undeflected particles
to enter mass spectrometer.
ÎPositive
‹ Balance
forces in selector ⇒ “select” v
qE = qvB
v= E/B
2
‹ Spectrometer:
Determine mass
from v and measured radius r
1
m1v
r1 =
qB
m2v
r2 =
qB
PHY2054: Chapter 19
37
Mass Spectrometer Example
ÎA
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?
eE = evB
B = E / v = 105 / 5 × 105 = 0.2T
‹ If
the electric field is suddenly turned off, what is the radius and
frequency of the circular orbit of the deuterons?
2
mv
mv
= evB ⇒ R =
=
R
eB
(
)( ) = 5.2 ×10
) ( 0.2)
3.34 × 10−27 5 × 105
(1.6 ×10
−19
−2
m
v
1
5 × 105
f = =
=
= 1.5 × 106 Hz
T 2π R ( 6.28 ) 5.2 × 10−2
(
)
PHY2054: Chapter 19
38
Quiz: Work and Energy
ÎA
charged particle enters a uniform magnetic field. What
happens to the kinetic energy of the particle?
‹ (1)
‹ (2)
‹ (3)
‹ (4)
‹ (5)
it
it
it
it
it
increases
decreases
stays the same
changes with the direction of the velocity
depends on the direction of the magnetic field
Magnetic field does no work, so K is constant
PHY2054: Chapter 19
39
Magnetic Force
ÎA
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
Forces cancel on
opposite sides of loop
B
z
y
x
PHY2054: Chapter 19
40
Hall Effect: Do + or – Charges Carry Current?
Î
Î
+ charges moving counterclockwise experience upward force
Upper plate at higher potential
Î
Î
– charges moving clockwise
experience upward force
Upper plate at lower potential
Equilibrium between magnetic (up) & electrostatic forces (down):
Fup = qvdrift B
Fdown = qEinduced
VH
=q
w
VH = vdrift Bw = "Hall voltage"
This type of experiment led to the discovery (E. Hall, 1879) that current
in conductors is carried by negative charges
41
PHY2054: Chapter 19
Electromagnetic Flowmeter
E
¾
¾
¾
¾
¾
Moving ions in the blood are deflected by magnetic force
Positive ions deflected down, negative ions deflected up
This separation of charge creates an electric field E pointing up
E field creates potential difference V = Ed between the electrodes
The velocity of blood flow is measured by v = E/B
PHY2054: Chapter 19
42
Creating Magnetic Fields
ÎSources
of magnetic fields
‹ Spin
of elementary particles (mostly electrons)
‹ Atomic orbits (L > 0 only)
‹ Moving charges (electric current)
ÎCurrents
generate the most intense magnetic fields
‹ Discovered
ÎThree
by Oersted in 1819 (deflection of compass needle)
examples studied here
‹ Long
wire
‹ Wire loop
‹ Solenoid
PHY2054: Chapter 19
43
B Field Around Very Long Wire
ÎField
around wire is circular, intensity falls with distance
‹ Direction
given by RHR (compass follows field lines)
μ 0i
B=
2π r
μ0 = 4π ×10−7
Right Hand Rule #2
μ0 = “Permeability of free space”
PHY2054: Chapter 19
44
Visual of B Field Around Wire
PHY2054: Chapter 19
45
B Field Example
ÎI
= 500 A toward observer. Find B vs r
‹ RHR
⇒ field is counterclockwise
μ i ( 4π ×10 ) 500 10
B=
=
=
−7
0
2π r
‹r
‹r
‹r
‹r
‹r
‹r
‹r
=
=
=
=
=
=
=
0.001 m
0.005 m
0.01 m
0.05 m
0.10 m
0.50 m
1.0 m
2π r
B
B
B
B
B
B
B
=
=
=
=
=
=
=
−4
r
0.10 T
0.02 T
0.010 T
0.002 T
0.001 T
0.0002 T
0.0001 T
=
=
=
=
=
=
=
1000 G
200 G
100 G
20 G
10 G
2G
1G
PHY2054: Chapter 19
46
Charged Particle Moving Near Wire
ÎWire
carries current of 400 A upwards
moving at v = 5 × 106 m/s downwards, 4 mm from wire
‹ Find magnitude and direction of force on proton
‹ Proton
ÎSolution
of force is to left, away from wire
‹ Magnitude of force at r = 0.004 m
‹ Direction
⎛ μ0 I ⎞
F = evB = ev ⎜
⎟
2
π
r
⎝
⎠
(
F = 1.6 × 10−19
)(
−7
⎛
2
10
×
× 400 ⎞
6
5 × 10 ⎜
⎟⎟
⎜
0.004
⎝
⎠
)
F = 1.6 × 10−14 N
PHY2054: Chapter 19
v
I
47
Ampere’s Law
ÎTake
arbitrary path around set of currents
be total enclosed current (+ up, − down)
‹ Let Bll be component of B along path
Not included
B& Δs = μ0ienc
in ienc
‹ Let ienc
∑
i
ÎOnly
currents inside path contribute!
‹5
currents inside path (included)
‹ 1 outside path (not included)
PHY2054: Chapter 19
48
Ampere’s Law For Straight Wire
ÎLet’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
∑ B&Δs = B ( 2π r ) = μ0i
P
i
μ 0i
B=
2π r
ÎAn
r
easy derivation
PHY2054: Chapter 19
49
Useful Application of Ampere’s Law
ÎFind
B field inside long wire, assuming uniform current
‹ Wire
radius R, total current i
‹ Find B at radius r = R/2
ÎKey
ienc
∑ B&Δs = μ0ienc
i
fact: enclosed current ∝ area
⎛ π r2 ⎞ i
Aenc
= i×
= i×⎜
⎟⎟ =
2
⎜
Atot
⎝πR ⎠ 4
R
r = R/2
r
R⎞
i
⎛
∑ i B&Δs =B ⎜⎝ 2π 2 ⎟⎠ = μ0 4
1 μ 0i
μ0i
B=
B=
On surface
2 2π R
2π R
PHY2054: Chapter 19
50
Ampere’s Law (cont)
problems: use Ampere’s law to solve for B at any r
‹ Wire radius R, total current i
∑ B&Δs = μ0ienc
ÎSame
i
ienc
⎛ π r2 ⎞
Aenc
r2
= i×
= i×⎜
⎟⎟ = i 2
2
⎜
Atot
R
⎝πR ⎠
=i
(r ≤ R)
R
r
(r ≥ R)
⎛ r2 ⎞
∑ i B&Δs =B ( 2π r ) = μ0i ⎜⎜ R 2 ⎟⎟ or μ0i
⎝
⎠
μ0i
μ0i r
B=
B=
(r
≤
R)
2π r
2π R R
PHY2054: Chapter 19
r≥R
51
Force Between Two Parallel Currents
ÎForce
on I2 from I1
μ0 I1I 2
⎛ μ0 I1 ⎞
F2 = I 2 B1L = I 2 ⎜
L=
L
⎟
2π r
⎝ 2π r ⎠
‹ RHR
⇒ Force towards I1
ÎForce
on I1 from I2
μ0 I1I 2
⎛ μ0 I 2 ⎞
F1 = I1B2 L = I1 ⎜
L=
L
⎟
2π r
⎝ 2π r ⎠
‹ RHR
⇒ Force towards I2
ÎMagnetic
I2
I2
I1
forces attract two parallel currents
I1
PHY2054: Chapter 19
52
Force Between Two Anti-Parallel Currents
ÎForce
on I2 from I1
μ0 I1I 2
⎛ μ0 I1 ⎞
F2 = I 2 B1L = I 2 ⎜
L=
L
⎟
2π r
⎝ 2π r ⎠
‹ RHR
⇒ Force away from I1
ÎForce
on I1 from I2
μ0 I1I 2
⎛ μ0 I 2 ⎞
F1 = I1B2 L = I1 ⎜
L=
L
⎟
2π r
⎝ 2π r ⎠
‹ RHR
⇒ Force away from I2
ÎMagnetic
I2
I2
I1
forces repel two antiparallel currents
I1
PHY2054: Chapter 19
53
Parallel Currents (cont.)
ÎLook
B
at them edge on to see B fields more clearly
B
2
Antiparallel: repel
1
F
2
1
F
B
2
1
B
Parallel: attract
F
1
2
F
PHY2054: Chapter 19
54
B Field @ Center of Circular Current Loop
ÎRadius
B=
R and current i: find B field at center of loop
μ 0i
2R
‹ Direction:
From calculus
RHR #3 (see picture)
ÎIf
N turns close together
N μ 0i
B=
2R
PHY2054: Chapter 19
55
Current Loop Example
Îi
= 500 A, r = 5 cm, N=20
B=N
μ 0i
2r
=
( 20 ) ( 4π ×10−7 ) 500
2 × 0.05
= 1.26T
PHY2054: Chapter 19
56
B Field of Solenoid
ÎFormula
found from Ampere’s law
i = current
‹ n = turns / meter
‹
B = μ0in
B ~ constant inside solenoid
‹ B ~ zero outside solenoid
‹ Most accurate when L R
‹
ÎExample:
‹
i = 100A, n = 10 turns/cm
n = 1000 turns / m
(
)
( )
B = 4π ×10−7 (100 ) 103 = 0.13T
PHY2054: Chapter 19
57