Right-Hand-Rules: Get ready to Click

 
  
F  qv  B F  IL  B
 
 
  r F   B
 0 I d s  rˆ
dB 
4 r 2
Phys 122 Lecture 20
A wire bent in the shape shown below
carries a current in the direction marked.
What is the direction of the net magnetic
field at point A?
A. Toward us.
B. Away from us.
C. To the left.
D. To the right.
E. Field is zero.
A wire bent in the shape shown below
carries a current in the direction marked.
What is the direction of the net magnetic
field at point A?
A. Toward us.
B. Away from us.
C. To the left.
D. To the right.
E. Field is zero.
Last class we observed wire with radius
R is closer and thus “wins” but it is also
shorter. The two exactly cancel out:
B1  R / R
B2  R/2R
!
L1
L2
C

     B 


L3
A)
B)
C)
D)
E)
Clicker
rail
S
V
R
A conducting rail of length L1 rests
on the top of the circuit loop as
shown. It is free to move. A
uniform magnetic field exists in the
box of dimension L2 by L3.
When switch s is closed, which way
does the rail move (if at all) ?
Left
Right
Rotates clockwise
Rotates counterclockwise
Does not move
• Current through rail is down
• B is pointed toward you
• IL x B is to the LEFT
L1
L2
C

     B 


L3
A)
B)
C)
D)
E)
Clicker
rail
S
V
R
A conducting rail of length L1 rests
on the top of the circuit loop as
shown. It is free to move. A
uniform magnetic field exists in the
box of dimension L2 by L3.
What is the magnitude of the force
on the rail RC seconds after the
switch has been closed?
0.37(V/R)
0.63(VL2B)/R
0.37(VL1B)/R
0.37(VL2B)/R
Help, or My answer wasn’t listed 
•
•
•
•
Force on segment is F = IL2B (it’s at right angle)
I(t) = (V/R) exp(-t/RC)
I(RC) = 0.37V/R
F = 0.37(VL2B)/R
Clicker
• A loop of wire is formed in this circuit
as shown on the right of the drawing.
• We label the direction of positive
current through the loop, +I, as shown
I
2V0
• What is the direction of the current
and the magnetic moment?
A)
B)
C)
D)
I is > 0
I is < 0
I is > 0
I is < 0
&
&
&
&
R
 is out of the page
 is out of the page
 is into the page
 is into the page
1. -2V + 5V + IR = 0
2. I = - 3V/R;  clockwise through that loop
3. RHR follow clockwise current   into page
5V0
Clicker
z
• Consider the loop of current shown,
which is located in a uniform vertical
magnetic field.
• About which axis might this loop rotate?
A)
B)
C)
D)
x
y
z
It will not rotate
1. Magnetic moment  is into page (+y)
2. Torque on loop is  x B (+x direction)
3. Loop rotates around direction of torque
•  i.e., around the x axis
y
x
B
The lightness of phys122B
•
For the physics is hard... And full of vectors... Students Snow
• Magnetic field has too many formulas. >_< –LingLi
• I am going to make a big and powerful railgun with
my knowledge of amperes law, solenoids, and the
right hand rule. science is so amazing! -Andrew
• What is the name of the first electricity detective?
Sherlock Ohms. -Aman(Plx don't hate).
The Rest of
Today is Ampere’s Law Day
“Well, the good news is most of this makes sense…” – Clifford-
"High symmetry"
 B  dl   I
0
Integral around a path …
hopefully a simple one
Current “enclosed” by that path

I
I just want to make
sure I can correctly
apply Ampere's Law,
and also know when
to use the Right Hand
Rule. -Jonah
Infinite current‐carrying wire
 
LHS:  B  d    Bd  B  d  B  2R
RHS:
I enclosed  I
o I
B
2R
General Case
Checkpoint Summary: Not bad
CheckPoint 2
Ienclosed  I
Ienclosed  I
CheckPoint 4
CheckPoint 6
CheckPoint 8
Cylindrical Symmetry
X
X
X
X
Enclosed Current = 0
Check cancellations
Finite wire
B
Can’t use Ampere’s
Law for a finite wire
Need to include the
“rest of the circuit”…
Read textbook.
o I
2R
Can we assume uniform current?
Yes (for DC)
In electrostatics, only the superfluous charges get to the
surface. The negative electrons that cancel the positive nuclei
are still inside the metal. They form negative charge density,
which is cancelled by the positive charge density of the
nuclei. In statics, these densities are uniform. These
distributed electrons later form the DC current. The electrons
do repel each other, but this is balanced by the positive
nuclei, so the electrons are not pushed out to the surface, but
can stay distributed in the volume.
Only true for constant current.
For AC current flows on a “skin” with depth inversely
proportional to sqrt().
From Prelecture: B Field of a Long Wire
• Inside the wire: (r < a)
0 I r
B=
2 a2
*
• Outside the wire: (r>a)
0 I
B=
2 r
* Just calculate the current enclosed vs. r. Similar to charge enclosed problems
with Gauss Law for uniformly charged shapes.
Clicker
Two cylindrical conductors each carry
current I into the screen as shown. The
conductor on the left is solid and has radius
R = 3a. The conductor on the right has a
hole in the middle and carries current only
between R = a and R = 3a.
3a
I
– What is the relation between the
magnetic fields at R = 6a for the
two cases (L=left, R=right)?
I
a
3a
2a
(a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a)
• Ampere’s Law with a circular loop at radius R = 6a
• The field in each case has cylindrical symmetry, being everywhere
tangent to the circle.
• So, field at R = 6a depends only on the total current enclosed
• In each case, a total current I is enclosed.
Clicker
Two cylindrical conductors each carry
current I into the screen as shown. The
conductor on the left is solid and has radius
R = 3a. The conductor on the right has a
hole in the middle and carries current only
3a
between R = a and R = 3a.
I
I
a
2a
3a
2a
What is the relation between the magnetic field at R = 2a for the two
cases (L=left, R=right)? (do the calculation)
(a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a)
• Field depends only on how much current is enclosed. Figure it out.
• For the LEFT conductor:
• For the RIGHT conductor:
 ( 2a ) 2
4

I
IL 
I
2
9
 (3a)




 ( 2a ) 2  a 2
3

I
IR 
I
2
2
8
 (3a)  a
( 0.444 I > 0.375 I )
B Field of  Current Sheet
Consider an  sheet of current described by n
wires/length each carrying current i into the
screen as shown.
•
•
Direction of the B field?
•
•
Symmetry  y direction!
Apply Ampere's law with square of side w:
•
 
 B d l  Bw  0  Bw  0  2 Bw
•
I  nwi

 
 B d l  μ 0 I

constant
B
μ 0 ni
2
y
x
x
x
x
x
x
x
x
x
x
x
x
x
w
constant
B Field of a Solenoid
• Purpose: makes a constant magnetic field
• A solenoid is defined by
•
a current I flowing through a wire which
is wrapped n turns per unit length on a
cylinder of radius a and length L.
L
a
•
If a << L, the B field is to first order contained within the
solenoid, in the axial direction, and of constant magnitude.
•
In this limit, we can calculate the field using Ampere's Law.
CMS
magnet at
the LHC
B Field of a  Solenoid
•
First, justify claim that the B field is 0 outside the solenoid.
•
View solenoid from the side as 2 
current sheets.
xxxxxxxxxxx
••••••••••••••
•
Same direction of fields between sheets
•
They cancel outside the sheets
•
Draw square path of side w:
 
 B d l  Bw
I  nwi
xxxxxxxx
• ••••••••••

B  μ 0 ni
CheckPoint
Use the right hand rule and curl your fingers along the direction of the current.
CheckPoint
Could you explain that last
problem with the cardboard
tube using Ampere's law. Is it
practical to do so because of
the way the current is
oriented? -James
Use the right hand rule and curl your fingers along the direction of the current.
•
Another important “real” magnet: Toroid•
•
•
Toroid defined by N total turns with
current i.
•
•
B = 0 outside toroid!
•
•
•
Consider integrating B on circle
outside toroid; net I enclosed = 0
x
x
x
x
xx
•
•
•
I  Ni
 
Apply Ampere’s Law:  B d l  μ 0 I
 B
•
xx x
For B inside, consider circle of radius r,
centered at the center of the toroid.
 
 B d l  B(2 πr)
•
μ 0 Ni
2 πr
x
•
x
x
r xx
xx
• B•
•
•
•
Example Problem
y
An infinitely long cylindrical shell with inner radius a and outer radius b carries a uniformly distributed current I out of the screen.
Sketch |B| as a function of r. a
I
x
b
Example Problem
y
What does |B| look like for r  a?
I
r
 
 B  d   o I enc
a

so B  0
0
A B C
b
x
Example Problem
y
What does |B| look like for r  b?
I
r
 
 B  d   o I enc
I
A B C
a
b
x
Example Problem
y
What does |B| look like for r  b?
 
LHS:  B  d    Bd  B  d  B  2r
RHS:
I enclosed  I
B
dl r
B
o I
2r
A B C
I
a
b
x
Example Problem
y
What is the current density j (Amp/m2) in the conductor?
I
a
b
I
I
I
j 2
j 2
j 2
A) B) C) 2
b
b  a
 b   a2
j = I / area
area   b   a
I
j 2
 b   a2
2
2
x
Example Problem
y
What does |B| look like for a  r  b ?
I
r
a
A B C
b
x
Example Problem
y
What does |B| look like for a  r  b ?
 
 B  d    oI enc
I
B  2r   o jAenc
I
2
2


(
r

a
)
B  2r   o
2
2
 (b  a )
r
o I ( r 2  a 2 )
B
 2
2r (b  a 2 )
A B C
a
b
x
Starts at 0 and increases almost linearly
Example Problem
y
An infinitely long cylindrical shell with inner radius a and outer radius b carries a uniformly distributed current I out of the screen. Sketch |B| as a function of r.
a
I
x
b
Circular Loop
0 i
R
Bz =
4  (z 2 +R2 ) 3/2
•
R
dl


r

z
R
dl = 2 R



Bz =
0
iR 2
•
Expressed in terms of the magnetic moment:

r
dB
2(z 2 +R2 ) 3/2
Note the form the field takes for z>>R:
=i
z
x
•
R2
dB
0 
Bz 
2  z3
Bz 
note the typical
dipole field
behavior!
0 iR 2
2z 3
Clicker
Equal currents I flow in identical circular
loops as shown in the diagram. The loop
on the right(left) carries current in the
ccw(cw) direction as seen looking along
the +z direction.
– What is the magnetic field Bz(A)
at point A, the midpoint
between the two loops?
(a) Bz(A) < 0
•
•
•
•
(b) Bz(A) = 0
I
o
I
x
B
A
x
(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!
o
z
Clicker
Equal currents I flow in identical circular
loops as shown in the diagram. The loop
on the right(left) carries current in the
ccw(cw) direction as seen looking along
the +z direction.
I
o
I
x
B
A
x
o
– 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
• 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!
z
Circular Loop
R
B
z

0
0
z
0 iR 2
Bz =
2(z 2 +R2 ) 3/2
1
z3