Chapter 29: Magnetic Fields Due to Currents
PHY2049: Chapter 29
1
Law of Magnetism
ÎUnlike
pieces
the law of static electricity, comes in two
‹Piece
1: Effect of B field on moving charge
r
r
F = qv × B (Chapt. 28)
‹Piece
2: B field produced by current
Equivalent!
Biot-Savart Law
Ampere’s Law
Proof of equivalence not in the book (Requires vector
calculus and relies on the absence of magnetic monopoles)
Reminds you of similar equivalence between Coulomb’s law
and Gauss’ law
PHY2049: Chapter 29
2
ÎSources
Creating Magnetic Fields
of magnetic fields
‹Electric
current (moving charges)
‹Atomic orbits of electrons (angular momentum L > 0 only)
‹Internal “spin” of elementary particles (mostly electrons)
ÎMagnetic
field produced by current is fundamental
‹How
about field produced by a bar magnet? Bar magnet
← magnetic ions ← orbital motion and spin of electrons in
them ← they are microscopic currents
ÎThree
examples studied here
‹Long
wire
‹Wire loop
‹Solenoid
PHY2049: Chapter 29
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B Field Around Very Long Wire
ÎField
around wire is circular, intensity falls with
distance
‹Direction
given by RHR #2 (compass follows field lines)
µ 0i
B=
2π r
µ0 = 4π ×10−7
Right Hand Rule #2
Derived from Ampere’s law
PHY2049: Chapter 29
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ÎWhy
(continued)
does µ0 have such a simple value?
‹Magnetism
is inseparable from electricity. This allows
the units in electricity and magnetism (in particular
coulomb and tesla) to be chosen so that only one
constant, ε0, has a non-trivial value.
ÎThis
example illustrates important general
property of magnetic fields:
‹Magnetic
field lines have no
beginning/end, unlike electric
field lines.
PHY2049: Chapter 29
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Long Wire B Field Example
ÎI
= 500 A toward observer. Find B vs r
‹RHR
#2 ⇒ field is counterclockwise
µ 0i
B=
=
2π r
(
)
4π ×10−7 500
2π r
0.0001
=
r
‹r
= 1 mm
B = 0.10 T = 1000 gauss
‹r = 1 cm (~0.4”) B = 0.010 T = 100 gauss
‹r = 10 cm (~4”) B = 0.001 T = 10 gauss
PHY2049: Chapter 29
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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 = 4 mm
‹Direction
 µ0 I 
F = evB = ev 

2
π
r


−7

2
10
×
× 400 
−19
6
F = 1.6 × 10
5 × 10 


0.004


F = 1.6 × 10−14 N
(
)(
)
PHY2049: Chapter 29
v
I
7
Ampere’s Law First (Biot-Savart law later)
ÎTake
arbitrary closed path around set of currents
‹Let ienc
be total enclosed current (signs +/– according to
RHR #2)
‹Let B be magnetic field, and ds be differential length
along path
Not included
B ⋅ ds = µ0ienc
in ienc
‹ Direction of field due to each current
element obeys RHR #2
∫
ÎOnly currents inside path count!
‹ 5 currents inside path (included)
‹ 1 outside path (not included)
‹ This does not mean that current outside path
does not contribute to B (note similarity to Gauss’ law)
PHY2049: Chapter 29
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Ampere’s Law For Straight Wire
ÎLet’s
point P
try this for long wire. Find B at distance at
‹According
to RHR #2, B field has only azimuthal
component, no radial component
‹Draw circular path passing through P (radius r)
‹From symmetry, strength of B must be constant along
path
P
∫∫BB⋅ ⋅ddss = B ( 2π r ) = µ0i
r
µ 0i
B=
2π r
An easy derivation
PHY2049: Chapter 29
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Ampere’s Law: More Application
ÎFind
B vs r inside long wire, assuming uniform
current
‹Wire
radius R, total current i
‹Draw circular path of radius r
ÎKey
fact: enclosed current ∝ area
‹Inside
ienc
r2
=i 2
R
R
r
B(2πr ) = µ0ienc
µ0ir
µ0i
B=
2
B=
On surface
2πR
2π R
‹Outside: B = µ0 i (derived in previous slide)
2πr
PHY2049: Chapter 29
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Question 10
ÎFigure
shows the magnitude of B field inside and
outside four long wires. Current is uniformly
distributed in each wire. Which wire carries the
largest current?
‹(a)
1
‹(b) 1 and 2
‹(c) 1 and 3
‹(d) Insufficient info
1
B
2
ÎIn
which wire is the current
density the highest?
‹(a) 1
‹(b) 1 and 2
‹(c) 1 and 3
‹(d) Insufficient info
PHY2049: Chapter 29
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4
r
11
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
‹Must be the same and towards I2
‹Why? Newton’s third law
Or view from behind the screen.
(I1 is now on left, and I2 now on right.)
ÎMagnetic
I2
I2
I1
forces attract two parallel currents
I1
PHY2049: Chapter 29
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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
‹Must be the same and away from I2
I2
ÎMagnetic forces repel two antiparallel currents
I2
I1
I1
PHY2049: Chapter 29
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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
PHY2049: Chapter 29
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Question 6
ÎLong
wires, carrying equal currents, are parallel to
each other and equally spaced. In which
arrangement is the net force on the central wire the
largest?
‹(a)
a
‹(b) b
‹(c) c
‹(d) d
‹(e) Insufficient info
ÎWires
‹(a)
OUT IN OUT IN OUT
OUT OUT IN
are of equal lengths.
a
‹(b) b
‹(c) c
‹(d) d
IN
IN
OUT IN OUT OUT IN
IN OUT IN
PHY2049: Chapter 29
IN
IN
15