MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR (22.5)
Important: basis of electric motors, microphones, speakers, transducers, etc.
 For wire carrying current I upward

q
v
o
d is upwards for each charge
o Force on EACH CHARGE points left
(perpendicular to wire and field)



FB  q v d  B
o Direction would be opposite for downward
current
 Next want to write force in terms of current
Bin
FB
I
MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR



F

q
v

B
 Force on each charge in wire is B
d
 How many charge carriers in a length l of wire?
Bin
A
FB
l
o For cross-section area A and n carriers per unit
volume, number of carriers in length l is n A l



 
 SO: force on wire segment of length l is FB  q vd  B n A l
 BUT: current in wire is I  n q v d A
 

 SO: force on length l of wire is FB  I l  B

o Vector l points in direction of current; magnitude is length of segment
I
Force on current-carrying wire of arbitrary shape:
 Divide wire into short,
 “straight” elements with
length/direction d s
o sum force on each element to get total
force on wire

 Force on segment d s is

 
dFB  I ds  B
 Force on length of wire from a to b is
b

 
FB  I  ds  B
a

d
s
may change with position along wire
o Direction of
ds

TORQUE ON A CURRENT LOOP IN A UNIFORM B FIELD
Look at rectangular loop, sides a and b, with current I flowing counterclockwise
I

1 case: B in plane of loop
 
 For segments 1 and 3, ds  B  0
o no force on these sides
st
1
4
2

 
 For segments 2 and 4, use dFB  I ds  B

o For 2, F2  IaB (up)

o For 4, F4  IaB (down)
(since area A = ab)
B
3
b
F
b/2
 Torque about axis through middle of 1 and 3 is
  2 b / 2  I a B  I A B
a
B
2
4
b/2
F
 This is maximum torque (force perp. to line from pivot to point of application)
o SO:  max  I A B
B
A

2 case: B not in plane of loop (general case)

 To specify direction of loop, define vector A

o Magnitude | A | is area
o Direction perp. to plane using Right Hand Rule

 Fingers → current; thumb → dir. A
nd
I
I
A
F2


 Forces on rotated loop:  is angle between A and B
and ALSO between force and line from pivot to wire
o Forces on 2 and 4, same as before (same

current, still perp. to B ) so:


F

IaB
(up)
F
and 4  IaB (down)
 2
b/2
  I A B sin 
B
2
o Lever arm is b / 2  sin  so torque is   2 I a B b / 2  sin 
 GENERAL:
θ
valid for loop of ANY shape
b/2
4
F4
TORQUE AND MAGNETIC MOMENT
τ
 
st


r
F
1 : Torque is a vector:

 Cross product
pivot

 r points from pivot to point where
force applied
 
   r  F points along axis


perpendicular to r AND F

 Direction of torque vector from Right-hand rule


o Fingers sweep from r → F
o Thumb shows direction of 
r
F
θ
2nd: Magnetic Dipole Moment
μ = IA
 Look at current I circulating around loop of area A

   IA is the magnetic dipole moment of the loop
o Magnitude:   IA

I
o Direction: 
 vector  perpendicular to the plane of the loop
 direction by Right-hand rule:
 fingers curl in direction of current

 thumb shows direction of 
 Units of magnetic dipole moment: A·m2
 For a coil of n loops,   nIA
A
3rd: Express torque as vector product of magnetic dipole moment and field
 Saw that magnitude of torque is   I A B sin 
μ = IA

o Same as magnitude of vector   B
F2


 Direction of   B is into screen/page

θ
B
2
o Same as direction for clockwise torque on loop
about line through midpoints of sides 1 and 3
τ
4
F4
 Conclude:

  B


 True for any current loop in a magnetic field!
EXAMPLE: based on problem 25, page 774
A rectangular coil is hinged so that it can turn
around one side that is parallel to the y-axis. The
sides parallel to the y-axis are 0.40 m in length
and the sides perpendicular to the y-axis are 0.30
m in length. The coil consists of 100 tightly
wrapped turns and the angle between the plane
of the coil and the x-axis is 30°. A current of 1.2
A circulates as shown. A magnetic field of 0.80 T
points in the x direction.
y
1.2 A
B = 0.8 T i
0.4 m
30°
0.3 m
z
What is the torque on the coil?
In what direction would it start to rotate if allowed to turn about the hinge?
x