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Physics 121 Lecture 10
Sources of Magnetic Fields (Currents)
SJ Chapter 30, Sec. 1 - 5
• Magnetic fields are due to currents
• The Biot-Savart Law
• Calculating field at the centers of current loops
• Magnetic dipole moment
• Field due to a long wire
• Force between two parallel wires carrying currents
• Ampere’s Law
• Solenoids and toroids
• Summary
Fall 2008
Last time: a magnetic field exerts a force on
moving charges and currents
• Magnets only come in pairs of N
and S poles (no monopoles).
N S N S NS
• Magnetic field exerts a force on
moving charges (i.e. on currents).
• The force is perpendicular to both
the field and the velocity (i.e. it
uses the use cross product). The
magnetic force can not change a
FB q v B
particle’s speed or KE
• A charged particle moving in a
uniform magnetic field moves in a
circle or a spiral. The magnetic
force can not change a particle’s
speed or KE
2 qB
c
c m
• Because moving charges are a
current, a wire in a magnetic field
feels a force also using cross
product. This force is responsible
for the motor effect.
FB iL B
m B
• Magnetic dipole moment,
• torque, and potential energy
ˆ
Ni A n
Um B Fall 2008
Magnetic fields are due to currents
Oersted - 1820: Magnetic compass deflected by current
Magnetic fields are due to currents (free charges & in wires)
In fact, currents are the
only way to create
magnetic fields
Fall 2008
The Biot-Savart Law (1820)
• Same basic role as Coulomb’s Law: magnetic field due to a source
• Source strength measured by “current-length” i ds
• Fall off with distance as inverse-square of distance B
• New “permeability” constant 0 i s x
• Direction of B field depends on a cross-product ( another RH rule) r
use RH rule for current
segments: thumb along ids differential addition to field at P due
curled fingers show B to distant source ids
P’
0 ids r unit
ˆ vector
dB
dB
(out of page) along r
4 r 2
10-7 exactly
“permeability” 0 = 4x10-7 T.m/A.
Find total field B by integrating over the
whole current region (need lots of symmetry)
The magnetic field lines are circles wrapped around a ids sin( )
wire. Use the right hand rule to find the direction
dB 0
4 r2
Fall 2008
Magnetic field due to a long, straight wire
• Wire is infinitely long, current i flows up
0 ids r
• ˆ
Find B at point P dB
• dB is into page at P for ds anywhere along wire 4 r 2
• Magnitude of ids X r = ids sin()
• Integrate dB using Biot Savart Law from –infinity to +infinity
0 sin( )ds
B dB
4 R 2 s 2
i
• Denominator has same sign for s + and –
R
• Substitute: sin() sin( ) sin() - sin(-)
1/ 2
R 2
s2
0 ds
B i R
2 0 [R 2 s2 ]3 / 2
• From integral ds s
0
1
1
1 - 0
tables: [R 2 s 2 ]3 / 2 R 2 [R 2 s 2 ]1 / 2 0 R2
RIGHT HAND RULE FOR A WIRE
0i
B
2R
R is distance
perpendicular to FIELD LINES ARE CIRCLES
wire through P Fall OR
THEY DO NOT BEGIN 2008 END
Direction of Magnetic Field
Which sketch below shows the correct direction of the
magnetic field, B, at the point P?
B into B into B into
page B page page
B
i i i i i
P P P P P
A B C D E
Fall 2008
Magnetic field at the center of a current arc
0 ids r
• Circular arc, radius R is constant ˆ
• Find B at center, point C dB
• f is included arc angle, not the cross product angle
4 r 2
• Angle for the cross product is always 900
• dB at center C is up out of the paper
• ds = Rdf’ i
0 ds 0 d' ids
dB i 2 i B
4 R 4 R R
R
• integrate on arc angle f’ from 0 to f
f Right hand rule
0i 0i
df' 4R f
for wire segments
B f in radians
4R 0
• For a full loop of current - f = 2 radians:
0i
B B B (loop)
Thumb points along the
2R
current. Curled fingers
Another RH rule (for loops): show direction of B
Curl fingers along current, thumb shows direction of B at center
Fall 2008
?? What would formula be for f = 45o, 180o, 100 radians ??
Examples:
FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE
P
0
dB i ds sin() 0
4r 2
ˆ
ds r 0
B AT CENTER OF A HALF LOOP, RADIUS = r
0i 0i
B
4r 4r
into page
B AT CENTER OF TWO HALF LOOPS
0i 0i
OPPOSITE B 2x
CURRENTS 4r 2r
same as closed loop
0i 0i
PARALLEL B - 0
CURRENTS 4r 4r
into out of
page page
Fall 2008
Magnetic Field from Loops
The three loops below have the same current. Rank
them in terms of the magnitude of magnetic field at the
point shown, greatest first.
A. I, II, III.
B. III, I, II.
C. II, I, III.
D. III, II, I.
E. II, III, I.
I. II. III.
0i
B f f in radians Hint: consider radius, direction, arc angle
4R Fall 2008
Magnetic Field lines When there are two
near a straight wire parallel wires carrying
current, the magnetic field
from one causes a force on
the other.
i up
i Fa,b ibLb Ba
. When the currents are parallel,
0ia the force is attractive.
Ba . When the currents are anti-
2R parallel, the force is repulsive.
Fall 2008
Magnitude of the force between two long parallel wires
L
• Third Law says: F12 = - F21
i1 • Use result for B due to infinitely long wire
0i1 due to 1 at 2
x x x x x
B1
r x x x x x 2r into page via RH rule
x x x x x • Evaluate F12 = force on 2 due to field of 1
x x x x x i2L is normal to B
i2 | F1,2 | i2L2B1 Force is toward 1
End View 0 i1i2
F1,2 L F21= - F12
2 d
i1
• Attractive force for parallel currents
i2 • Repulsive force for opposed currents
Example: Two parallel wires 1 cm apart |i1| = |12| = 100 A.
2x10 -7 x 100 x 100
F/L force per unit length 0.2 N / m
.01
F 0.2 N for L 1 m
Fall 2008
Forces on Parallel Currents
Which of the four situations below results in the greatest force
to the right on the central conductor?
F greatest?
A. I.
B. II.
C. III. I.
D. IV.
E. Cannot II.
determine.
III.
IV.
Hints: Sketch the field directions. What is the net field midway
between two wires with parallel currents?
Which way is the net force?
Ftot i L Btot
Fall 2008
Ampere’s Law
• Another way to find B, given the current
• Derivable from Biot-Savart Law
• But B is inside an integral usable only for high symmetry (like Gauss’ Law)
• An “Amperian loop” is a
ienc= net current
B ds 0ienc
closed path of any shape
• Add up (integrate) components passing through the
of B along the loop (path) loop
To find B, you have to be able to do the integral, then solve for B
Picture for applications:
• Only the tangential component of
B contributes to the dot product
• Current outside the loop (i3) doesn’t
contribute
• Another version of RH rule:
- curl fingers along loop
- thumb shows + direction for net current
Fall 2008
Example: Magnetic field outside a long
straight wire carrying current
0i
We already used the Biot-Savart Law to show that: B
2r
Now show it again more simply, using
Ampere’s Law:
B ds 0ienc
Draw an Amperian loop of any shape, but since
the magnetic field goes in circles around a
wire, choose a circular loop (of radius r).
B and ds are then parallel, and B is constant
everywhere on the Amperian path
B ds B2r 0ienc
The integration is simple.
ienc is the total current.
Solve for B to get our earlier expression.
0i
B outside wire
2r Fall 2008
Magnetic field inside a long straight wire
carrying current
Using Ampere’s Law, we can even calculate B
inside the wire. B ds 0ienc
Assume the current density is uniform across the
wire. J = i/A. Current is evenly distributed
over the cross-section of the wire and must be
cylindrically symmetric.
Again draw a circular Amperian loop around the
axis, of radius r < R.
The enclosed current is less than the total
current i, because some is outside the Amperian
loop. The amount enclosed is
r 2
ienc i
R 2 B
Apply Ampere’s Law: ~r
~1/r
r2
B ds B2r 0ienc 0i R 2 R r
i Outside wire (r>R) it looks like an
B 0 2 r r R inside wire infinitely thin wire (previous expression)
2R Inside: B grows linearly Fall 2008
Counting currents within Amperian Loops
For the Amperian paths shown, rank the value of B ds
along each path, taking direction into account and putting the
more positive ahead of less positive values. All of the wires are
carrying the same current.
A. I, II, III, IV, V.
I.
B. II, III, IV, I, V.
II.
C. III, V, IV, II, I.
D. IV, V, III, II, I. III.
E. I, II, III, V, IV.
IV.
V.
B ds 0ienc
Fall 2008
Another Ampere’s Law example
Why use COAXIAL CABLE for CATV and other applications?
Find B inside and outside of the cable
Cross section:
Amperian shield wire
loop 1 current i into
sketch
Amperian
loop 2 center wire
current i out of
sketch
Inside – use Amperian loop 1:
0i Outer shield does not
B ds 0i Bx2r B
2r
affect field inside
Reminds one of Gauss’s Law
Outside – use Amperian loop 2:
Zero field outside due to opposed
currents + radial symmetry
B ds 0ienc 0 Losses and interference suppressed
Fall 2008
Solenoids: strengthen field by using many loops
L
cancellation
d
n # coils / unit length N/L
strong uniform
Approximation: field is constant inside and field in center
zero outside (just like capacitor)
“Long solenoid” d << L
IDEAL SOLENOID WITH AMPERIAN LOOP
• outside B = 0, no contribution from path c-d
• B perpendicular to ds of a-d and b-c
• inside B is uniform and parallel to ds on a-b
B ds Bh 0ienc 0inh
B 0in
ideal solenoid
only section that has non-zero
Fall 2008
contribution
Toroids: bend a long solenoid into a circle
Find the B field inside
Draw an Amperian loop parallel to the field, with
radius r
The coil has a total of N turns
LINES The Amperian loop encloses current Ni.
OF
B ARE
CIRCLES B ds B2r 0ienc 0iN
0iN
B inside toroid B0 outside
2 r
• N times the result for a long wire
• Depends on r
• Also same result as for long solenoid
N
n (turns/uni t length) B 0in
AMPERIAN LOOP IS 2 r
A CIRCLE ALONG B Fall 2008
B on z-axis of a current loop (dipole)
• Find B at a point P on the z axis of the current loop?
• We use the Biot-Savart Law directly
i ds r
dB 0
4 r 3
0 i ds cos() R
dBz dB|| dB cos() cos
4 R 2 z 2 r
dB cancels by symmetry
0 iR r R2 z2
dB z ds
4 (R z )
2 2 3/2
• Integrate on f around the current loop, noting
that the field is perpendicular to r.
• By symmetry, the perpendicular part of B cancels
around the loop - only the parallel part survives.
0 iR 0 iR 2 2
Bz dBz
4 (R 2 z 2 )3 / 2 4 (R 2 z 2 )3 / 2 0
ds df
as before recall dipole
0iR 2
i
B(z) B(z 0) 0 moment
2(R z )
2 2 3/2
2R NiA iR2 Fall 2008
B field far from the dipole (current loop)
Fall 2008
Summary: Lecture 10 Chapter 29 – Magnetic Fields from Currents
Fall 2008
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