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
34
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
35
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
36
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
37
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
38
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
PHY2054: Chapter 19
39
Visual of B Field Around Wire
PHY2054: Chapter 19
40
B Field Example
ÎI
= 500 A toward observer. Find B
‹ RHR
⇒ field is counterclockwise
μ i ( 4π ×10 ) 500 0.0001
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
=
=
=
=
=
=
=
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
41
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
42
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
43
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
44
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
Aenc
= i×
Atot
⎛ π ( R / 2 )2 ⎞ i
⎟=
= i×⎜
2
⎜ πR
⎟ 4
⎝
⎠
R
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
45
Ampere’s Law (cont)
ÎSame
problems: use Ampere’s law to solve for B at any r
‹ Wire
ienc
radius R, total current i
Aenc
= i×
Atot
⎛ π r2 ⎞
r2
= i×⎜
=i 2
⎜ π R 2 ⎟⎟
R
⎝
⎠
∑ B&Δs = μ0ienc
i
⎛ r2 ⎞
∑ i B&Δs =B ( 2π r ) = μ0i ⎜⎜ R 2 ⎟⎟
⎝
⎠
μ0i r
r≤R
B=
2π R R
μ0i
B=
2π r
R
r
r≥R
PHY2054: Chapter 19
46
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
47
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
48
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
49
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
50
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
51
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
52