Physics 212
Lecture 13
Forces and Torques on Currents
Physics 212 Lecture 13, Slide 1
Key Concepts:
•
•
05
Forces & Torques on loops of current due
to a magnetic field.
The magnetic dipole moment.
Physics 212 Lecture 13, Slide 2
Force on a wire carrying current I
Cross sectional area A, Length L
Number of charges/m3 = n
Each charge moving at Vavg
Force on each charge
Total force on wire
Fwire
B
L
06
I
Feach  q vavg  B
Fwire  N Feach  Nq vavg x B
Use
N  nAL
I  qnAvavg
Fwire  I L  B
Physics 212 Lecture 13, Slide 3
Current- carrying wire in uniform B field
B
I
L
Fwire  I L  B
Out of page
ACT
F  IL  B
A
B
C
08
Physics 212 Lecture 13, Slide 5
ACT
F  IL  B
A
B
C
10
Physics 212 Lecture 13, Slide 6
ACT
F  IL  B
What is the force on section d-a of the loop?
A) Zero
B) Out of the page
C) Into the page
12
Physics 212 Lecture 13, Slide 7
Checkpoint 1a
A
B
C
“The net force on any closed loop is zero.”
13
Physics 212 Lecture 13, Slide 8
Checkpoint 1b
y
x
In which direction will the loop rotate?
(assume the z axis is out of the page)
A)
B)
C)
D)
15
Around the x axis
Around the y axis
Around the z axis
It will not rotate
Physics 212 Lecture 13, Slide 9
Checkpoint 1c
y
R
F
  RF
A
B
C
D
E
17
Physics 212 Lecture 13, Slide 10
Magnetic Dipole Moment
Area vector
Magnitude = Area
Direction uses right
hand rule
Magnetic Dipole moment
  N turns I A
19
Physics 212 Lecture 13, Slide 11
  B
The torque always wants to line  up with B !
z
turns  toward B
  B

z
y
B

x
y
B
x
21
  B
turns  toward B
Physics 212 Lecture 13, Slide 12
Practice with  and 
  B
I
B

In this case  is out of the page (using right hand rule)
z
  B

x
22
is up (turns  toward B)
y
B
Physics 212 Lecture 13, Slide 13
Checkpoint 2a
Three different orientations of a magnetic dipole moment in a constant magnetic field are
shown below. Which orientation results in the largest magnetic torque on the dipole?
  B
Biggest when
24
B
Physics 212 Lecture 13, Slide 14
Potential energy of a magnetic dipole moment
B
1
Rotate  from 1 to 2
B
2


2
2
1
1
2
U 2  U1     B d    B sin d   B cos 

1
U 2  U1    B cos 2     B cos 1     2  B   1  B

U    B
27
Physics 212 Lecture 13, Slide 15
U=0
Lowest U
 parallel to B
Highest U
 antiparallel to B
1.5
U
1
0.5
0
0
-0.5
-1
-1.5
30
60
90
120

B
150
180
Checkpoint 2b
Three different orientations of a magnetic dipole moment in a constant magnetic field are
shown below. Which orientation has the most potential energy?

f
U = +Bcosf
U = 0
U = -Bcos
U    B
30
Physics 212 Lecture 13, Slide 17
ACT
Three different orientations of a magnetic dipole moment in a constant magnetic field are
shown below. We want to rotate the dipole in the CCW direction.
fa
a
c
B
First, consider rotating to position c. What are the signs of the work done by
you and the work done by the field?
A)
B)
C)
D)
30
Wyou
Wyou
Wyou
Wyou
>
>
<
<
0,
0,
0,
0,
Wfield
Wfield
Wfield
Wfield
>
<
>
<
0
0
0
0
W field  DU
• DU > 0, so Wfield < 0. Wyou must be opposite
Wfield
• Also, torque and displacement in opposite
directions  Wfield < 0 Physics 212 Lecture 13, Slide 18
ACT
Consider rotating the dipole to each of the three final orientations shown.
fa
a
c
B
Does the sign of the work done by you depend on which position (a, b, or c)
the dipole is rotated to?
A)Yes
B)No
U    B
The lowest potential energy state is with dipole parallel to B. The potential
energy will be higher at any of a, b, or c.
30
Physics 212 Lecture 13, Slide 19
Calculation
A square loop of side a lies in the x-z
plane with current I as shown. The loop
can rotate about x axis without friction. A
uniform field B points along the +z axis.
Assume a, I, and B are known.
How much does the potential energy of
the system change as the coil moves from
its initial position to its final position.
z
z
B
30˚
.
y
y
a
I
x
B
final
initial
• Conceptual Analysis
–
–
A current loop may experience a torque in a constant magnetic field
•
=XB
We can associate a potential energy with the orientation of loop
•
U=-∙B
• Strategic Analysis
– Find 
–
32
Calculate the change in potential energy from initial to final
Physics 212 Lecture 13, Slide 20
Calculation
A square loop of side a lies in the x-z
plane with current I as shown. The loop
can rotate about x axis without friction. A
uniform field B points along the +z axis.
Assume a, I, and B are known.
z
z
B
30˚
.
B
y
y
a
I
x
initial
final
• What is the direction of the magnetic moment of this current loop in
its initial position?
(A) +x
(B) -x
z
z
(C) +y
(D) -y
.
x
●
y

  IA
y
X
Right Hand Rule
34
Physics 212 Lecture 13, Slide 21
Calculation
A square loop of side a lies in the x-z
plane with current I as shown. The loop
can rotate about x axis without friction. A
uniform field B points along the +z axis.
Assume a, I, and B are known.
z
z
B
30˚
.
y
y
a
I
x
initial
B
final
• What is the direction of the torque on the current loop in the initial
position?
(A) +x
(B) -x
(C) +y
(D) -y
z
B
y
36
Physics 212 Lecture 13, Slide 22
Calculation
A square loop of side a lies in the x-z
plane with current I as shown. The loop
can rotate about x axis without friction. A
uniform field B points along the +z axis.
Assume a, I, and B are known.
U    B
z
z
B
30˚
.
y
y
a
I
x
B
final
initial
• What is the potential energy of the initial state?
(A) Uinitial < 0
(B) Uinitial = 0
(C) Uinitial > 0
z
  90
B

0
 
B 0

y
38
Physics 212 Lecture 13, Slide 23
Calculation
A square loop of side a lies in the x-z
plane with current I as shown. The loop
can rotate about x axis without friction. A
uniform field B points along the +z axis.
Assume a, I, and B are known.
z
z
B
30˚
.
y
y
a
I
U    B
x
B
final
initial
• What is the sign of the potential energy in the final state?
(A) Ufinal < 0
(B) Ufinal = 0
(C) Ufinal > 0
Check:  moves away from B
z
z
B
initial
B
  90o + 30o
  90o

final
y
y

40
  1200
 
B0
Energy must increase !
U    B  0
Physics 212 Lecture 13, Slide 24
Calculation
A square loop of side a lies in the x-z
plane with current I as shown. The loop
can rotate about x axis without friction. A
uniform field B points along the +z axis.
Assume a, I, and B are known.
z
z
B
30˚
.
B
y
a
I
U    B
x
final
initial
• What is the potential energy of the final state?
(A) U  Ia 2 B
z
B
1
cos(120o )  
2

120o
y

44
(B) U 
3 2
Ia B
2
(C) U 
1 2
Ia B
2
1
U     B    B cos(120 )   B
2
o
  Ia 2
1 2
U  Ia B
2
Physics 212 Lecture 13, Slide 25