Transcript Lect14

Biot-Savart’s Law
•
Rq
P
q
r
R
I
x
R
x
dB
q
z
q
dx
r
z
r
dB
Our Study of Magnetism
• Lorentz Force Eqn

  
F  qE  qv  B
mv
• Motion in a uniform B-field circular orbit R 
qB
• Forces on charges
moving in wires
• Magnetic dipole
• Today:
 

dF  Idl  B
 
  AI
  
  B
 
U    B
fundamentals of how currents
generate magnetic fields
• Next Time: Ampere’s Law  simplifies
calculation
Today...
• Biot-Savart Law for Calculating B-field
• Biot-Savart Law (“brute force”)
• Ampere’s Law
(“high symmetry”)
• B-field of an  Straight Wire
• Force on Two Parallel Current
Carrying Wires
• B-field of a Circular Loop
Last time:Potential Energy
of
Dipole
 
U    B

B

x
B
x

x
=0
 = B X
=0
U = -B
U=0
U = B
negative work
1
positive work
B
Lecture 14, Act 1
• A circular loop of radius R carries current I
as shown in the diagram. A constant
1A
magnetic field B exists in the +x direction.
Initially the loop is in the x-y plane.
– The coil will rotate to which of the
following positions?
y
y
w
a
z
1B
R
a
I b
x
(c) It will not rotate
a
b
B
w
(b)
(a)
y
b
z
What is the potential energy U0 of the loop in its initial position?
(a) U0 is minimum
(b) U0 is maximum
(c) neither
Lecture 14, Act 1
• A circular loop of radius R carries current I
as shown in the diagram. A constant
1A
magnetic field B exists in the +x direction.
Initially the loop is in the x-y plane.
– The coil will rotate to which of the
following positions?
y
y
w
a
z
•
•
•
•
R
a
I b
x
(c) It will not rotate
a
b
B
w
(b)
(a)
y
b
z
  
The coil will rotate if the torque on it is non-zero: τ  μ  B
The magnetic moment  is in +z direction.
Therefore the torque  is in the +y direction.
Therefore the loop will rotate as shown in (b).
Lecture 14, Act 1
• A circular loop of radius R carries current I
as shown in the diagram. A constant
magnetic field B exists in the +x direction.
1B
Initially the loop is in the x-y plane.
– What is the potential energy U0 of the
loop in its initial position?
y
B
R
I b
a
x
(a) U0 is minimum
(b) U0 is maximum
(c) neither
 
– The potential energy of the loop is given by: U   μ  B
– In its initial position, the loop’s magnetic moment vector points in
the +z direction, so initial potential energy is ZERO
– This does NOT mean that the potential energy is a minimum!!!
– When the loop is in the y-z plane and its magnetic moment points
in the same direction as the field, its potential energy is NEGATIVE
and is in fact the minimum.
– Since U0 is not minimum, the coil will rotate, converting potential
energy to kinetic energy!
Calculation of Electric Field
• Two Ways to calculate
– Coulomb’s Law
dq
dE  k 2 rˆ
r
"Brute force"
– Gauss’ Law
 
e 0  E  dS  q
"High symmetry"
What are the analogous equations for the
Magnetic Field?
Calculation of Magnetic Field
• Two Ways to calculate
μ0 I dl  rˆ
dB 
4π r 2
– Biot-Savart Law
(“Brute force”)
– Ampere’s Law
(“High symmetry”)
I
 
 B  dl  0 I
–AMPERIAN SURFACE/LOOP
These are the analogous equations
Biot-Savart Law...
...bits and pieces
μ0 I dl  rˆ μ0 I dl  r
dB 

2
4π r
4π r 3
dl
q
r
X
dB
I
N
 0  4 10
A2
7
The magnetic field “circulates”
around the wire
Use right-hand rule: thumb along I,
fingers curl in direction of B.
Preflight 14:
A current carrying wire (with no remarkable
symmetry) is oriented in the x-y plane. Points
A,B, & C lie in the same plane as the wire. The
z-axis points out of the screen.
5) In what direction is the magnetic field contribution from the segment dl at point
A. Check all non-zero components.
+x
-x
+y
-y
+z
-z
6) In what direction is the magnetic field contribution from the segment dl at point
B. Check all non-zero components.
+x
-x
+y
-y
+z
-z
7) In what direction is the magnetic field contribution from the segment dl at point
C. Check all non-zero components.
+x
-x
+y
-y
+z
-z
 
dB points in the direction of dl  r

 μ0 I dl  r
dB 
4π r 3
A: dl is to the right, and r is up  dB is out of the page
B: dl is to the right, and r is up and right  dB is out of the page
C: dl is to the right, and r is down and right  dB is into the page
Conclusion: at every point above the wire, dB is . Below the wire ,
dB is 
Preflight 14:
9) Would any of your answers for questions 5-7
change if we integrated dl over the whole wire?
NO!
Why or why not?

 μ0 I dl  r
dB 
4π r 3
 
dB points in the direction of dl  r
At point A: dl is to the right, and r is up  dB is out of the page
At point B: dl is to the right, and r is up and right  dB is out of the page
At point C: dl is to the right, and r is down and right  dB is into the page
For every point in the x-y plane and every piece of wire dl:
every dl and every r are always in the x-y plane
Since dB must be perpendicular to r and dl, dB is always in the ±z direction!
Conclusion:
At every point above the wire, the dB due to every piece dl is .
Below the wire, the dB due to every piece dl is 
B-field of  Straight Wire
• Calculate field at point P using
Biot-Savart Law:
 μ0 I dx  r
dB 
4π r 3
 (z )
Which way is B?
y P
q
r
R
q
dx
I
x

μ0 I (dx)r sin θ
3
4
π
r

B   dB  
This calculation appears in its entirety in the appendix
Its result is:
μ0 I
B
2πR
• Units of B are Tesla
• 1 Tesla = 104 Gauss
• Earth’s B-field ~ 0.5 G
Putting it all together
• We know that a current-carrying wire can experience
force from a B-field.
• We know that a a current-carrying wire produces a Bfield.
• Therefore: We expect one current-carrying wire to exert
a force on another current-carrying wire:
Ib
d
F
F
• Current goes together  wires come together
• Current goes opposite  wires go opposite
Ia
F on 2 Parallel CurrentCarrying Wires
• Calculate force on length L of wire b
due to field of wire a:
The field at b due to a is given by:
F •
Ib
d
Ia
μ0 I a
 
Magnitude
0 I a Ib L

Ba 
F

I
L

B

= b b
a
of
F
on
b
2d
2πd
• Calculate force on length L of wire a
due to field of wire b:
The field at a due to b is given by:
Ib
F
d
Ia
×
μ0 I b
Magnitude F  I L  B  0 I a Ib L

Bb 
= a a
b
2 d
of
F
on
a
2πd
2
Preflight 14:
2) Two slack wires are carrying current in opposite
directions. What will happen to the wires? They will:
a) attract
b) repel
c) twist due to torque
3) Now, two slack wires are carrying current in the same
direction. What will happen to the wires? They will:
a) attract
b) repel
c) twist due to torque
a) Find B due to one wire at the position of the other wire
b) Use F  IL  B to find the direction of F in each case
a: Point your thumb down, your fingers wrap in the direction of B around the
wire: The direction of B due to the left wire at the position of the right wire is
out of the screen.
b: I is up, B is out of the screen, so F is to the right.  the force is repulsive.
Lecture 14, Act 2A
• A current I flows in the +y direction in an
infinite wire; a current I also flows in the
loop as shown in the diagram.
– What is Fx, the net force on the loop
in the x-direction?
(a) Fx < 0
(b) Fx = 0
I
Ftop
y
Fleft
X
Fright
I
(c) Fx > 0
Fbottom
x
• You may have remembered a previous act in which the net
force on a current loop in a uniform B-field is zero, but the
B-field produced by an infinite wire is not uniform!
• Forces cancel on the top and bottom of the loop.
• Forces do not cancel on the left and right sides of the loop.
• The left segment is in a larger magnetic field than the right
• Therefore, Fleft > Fright
Lecture 14, Act 2B
I
What is the magnitude of the magnetic
field at the center of a loop of radius R,
carrying current I?
R
(a) B = 0
(b) B = (0I)/(2R)
(c) B = (0I)•(2R)
We can immediately guess the right answer must be (b).
(a) is wrong, because all parts of the loop contribute
to a B going into the paper at the center of the
loop.
(c) also cannot be correct—it has the wrong units:
B
Amp
meter
Lecture 14, Act 2
dl
I
But how do we calculate B?
r
We must use Biot-Savart Law to calculate the
magnetic field at the center:
R
 0 I dl  r
dB 
4 r 3
Two nice things about calculating B at the center of the loop:
• I dl is always perpendicular to r: dl  r  r into page
• r is constant (r = R)
μ0 I (dl ) R
B  ˜ dB  ˜
4 π R3
μ0 I

˜ dl
2
4πR
μ0 I
μ0 I

(2 π R) 
2
4πR
2R
Note: B 
Amp
meter
Circular Loop, in more detail
• Circular loop of radius R carries
current I. Calculate B along the
axis of the loop:
•
Rq
• Magnitude of dB from element dl:
0 I dl 0 I dl
dB =
=
4  r 2 4  z2 +R2
r
q
z
R
x
dB
z
r
dB
• What is the direction of the field?
• Symmetry  B in z-direction.
0 I dl
dB z =
cosq
2
2
4  z +R
cosq =
R
z2 +R2

0 I
R

Bz =
dl
2
2
3/2
4  (z +R )
Circular Loop, cont.
•
Rq
0 I
R

Bz =
dl
4  (z 2 +R2 ) 3/2
r
q
z
dl = 2 R
R



Bz =
0
z
r
x
IR2
Bz 
• Expressed in terms of the magnetic moment:
=

dB
2(z 2 +R2 ) 3/2
• Note the form the field takes for z >> R:
I R2
dB
0 
Bz 
2  z3
note the typical
dipole field
behavior!
0 IR2
2z 3
Circular Loop, anywhere on axis
Bz 
y=
0 IR 2
2 z  R
2
0 IR 2
Bz  z  R  
2z3

2 32
Expressed in terms of the
magnetic moment   I  R 2
f(x)
1
0 
Bz  z  R  
2 z 3
Bz
Note the typical 1/z3
dipole field behavior!
0
0
0
0
z
R
x=
x
3
3
Lecture 14, Act 3
• Equal currents I flow in identical circular
loops as shown in the diagram. The loop
3A
on the right (left) carries current in the
ccw (cw) direction as seen looking along
the +z direction.
I
o
I
x
B
A
– What is the magnetic field Bz(A) at point
A, the midpoint between the two loops?
x
(a) Bz(A) < 0
3B
(b) Bz(A) = 0
o
(c) Bz(A) > 0
• What is the magnetic field Bz(B) at point B, just to the right of the
right loop?
(a) Bz(B) < 0
(b) Bz(B) = 0
(c) Bz(B) > 0
z
Lecture 14, Act 3
• Equal currents I flow in identical circular
loops as shown in the diagram. The loop
3A
on the right (left) carries current in the
ccw (cw) direction as seen looking along
the +z direction.
I
o
I
x
B
A
– What is the magnetic field Bz(A) at point
A, the midpoint between the two loops?
x
(a) Bz(A) < 0
•
•
•
•
(b) Bz(A) = 0
o
(c) Bz(A) > 0
The right current loop gives rise to Bz <0 at point A.
The left current loop gives rise to Bz >0 at point A.
From symmetry, the magnitudes of the fields must be equal.
Therefore, B(A) = 0
z
Lecture 14, Act 3
• Equal currents I flow in identical circular
loops as shown in the diagram. The loop
on the right (left) carries current in the
3B
ccw (cw) direction as seen looking along
the +z direction.
I
o
I
x
B
A
– What is the magnetic field Bz(B) at point
B, just to the right of the right loop?
x
(a) Bz(B) < 0
(b) Bz(B) = 0
z
o
(c) Bz(B) > 0
• The signs of the fields from each loop are the same at B
as they are at A!
• However, point B is closer to the right loop, so its field wins!
Summary
• Biot-Savart Law for Calculating B-Fields
μ0 I dl  rˆ μ0 I dl  r
dB 

2
3
4π r
4π r
• Current-Carrying Wires Make B-Fields and
Are Affected by B-Fields
• Current Loop Produces a Dipole Field
Appendix A: B-field of  Straight Wire
y
• Calculate field at point P using
Biot-Savart Law:
 μ0 I dx  r
dB 
4π r 3
Which way is B?

+z
μ0 I (dx)r sin θ
3
4
π
r

• Rewrite in terms of R,q :
B   dB  
r
R
sin θ

R
tan θ 
x
1 

therefore, dx   R 
dθ 
2
 sin θ 
P
q
r
R
q
I
dx

x  R cot θ
dx(sin θ ) sin θdθ

2
r
R
x
Appendix A: B-field of  Straight Wire
P
π
μ0 I dθ
B
sin θ
4π R
0
q
r
q
dx
π
μ0 I
B
sin θdθ

4πR 0
therefore,
R

I
μ0 I
π
 cos θ 0
B
4πR
μ0 I
B
2πR
x