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Chapter 19
Magnetic Force
on Charges
and CurrentCarrying Wires
DON’T FORGET:
Forces
have
directions!

 
 
F  qv  B  q v B sin 
WARNING: cross-product
(NOT simple multiplication!)
This force acts in the direction perpendicular
to the plane defined by the vectors v and B
as indicated by the right-hand rule!
Fingers point in the
direction of the magnetic
field.
Thumb points in the
direction of the current.
Palm faces the direction
of the magnetic force.

 
F  q v B sin 

B

B 

F

q v sin 

F
N
N



q v sin 
C m/ s A m
Tesla
Weber/m2
1 Tesla is a very strong magnetic field.
So, magnetic fields are often measured
in the cgs (i.e., centimeter, gram,
second) unit of Gauss.
1 Tesla = 104 Gauss
The Earth’s magnetic field is
about 0.5 Gauss near the
surface.
A proton moves at right angles to a
magnetic field of 0.1 T with a speed of
2.0 X 107 m/s. Find the magnitude of
the acceleration.

 
F  q v B sin 
 (16
.  1019 C)(2.0  107 m / s)(01
. T)(sin 90o )
 3.2  1013 N


F  ma

3.2  1013 N
 F
14
2
a 

 19
.  10 m / s
-27
m 1.67  10 kg
Remember when we talked about the motion
of charges in a wire….
Vd Dt
E
A
The charges move with an average velocity vd.
The magnetic force on a wire with N charge
carriers moving with velocity vd in a uniform
magnetic field B should just be the vector sum
of the force on each individual charge.
Since the average velocity is the same for all N
charge carriers, the magnetic force acts in the
same direction (on average) on all the charge
carriers. Therefore...

 
F  Nq vd B sin 
Where  is the angle between the long dimension
of the wire and the magnetic field. The force
acts in the direction perpendicular to both.

 
F  Nq vd B sin 
N  nAL
N = Number of charge carriers
n = number of charge carries per unit volume
A = cross-sectional area of wire
L = length of wire

 
F  nALq vd B sin 
But recall our definition of current in a wire...
I  nqvd A
Substituting, we get the simpler expression:


F  B I L sin 
So, the maximum value of the
magnetic force on a current
carrying wire occurs when the wire
is perpendicular to the magnetic
field and has the value...
Hey, Mr. Sluggo.
Magnetic Force on
a wire is named
after me!


Fmax  B I L
So what happens if we put a loop of wire
carrying current I in a magnetic field?
B
I


F  B I L sin 
The currents in the top and bottom of this loop
are anti-parallel and parallel to the magnetic field.
Therefore, sin  = 0. So the magnetic forces on
the top and bottom of the loop are 0!
B
I


F  B I L sin 
On the left side of the loop, we use the righthand rule to determine that the force is out
(toward us).
On the right side of the loop, we use the righthand rule to determine that the force is in
(away from us).
In each case,  = 90o, sin  = 1, so


F  B IL
So, the magnitude of the force on the left side
of the loop is the same as the magnitude on
the right side of the loop.
What, therefore, is going to happen to the
loop in this magnetic field?
The loop
rotates!
B
I
a/2

 left
b
a/2
The magnetic
force produces
a TORQUE on
the current loop,
causing it to
rotate. In this
case, the loop
rotates
counterclockwise
as viewed from
above...
F
I
G
J
HK



a
 F d  B ILd  B Ib
counter - clockwise
2
The loop
rotates!
B
I
a/2

 right
b
a/2
The magnetic
force produces
a TORQUE on
the current loop,
causing it to
rotate. In this
case, the loop
rotates
counterclockwise
as viewed from
above...
F
I
G
J
HK



a
 F d  B ILd  B Ib
counter - clockwise
2
The loop
rotates!
B
I
a/2



 total   left   right
b
a/2
The magnetic
force produces
a TORQUE on
the current loop,
causing it to
rotate. In this
case, the loop
rotates
counterclockwise
as viewed from
above...


 B Iab  B I (Area of loop)

  B I A sin

Where  is the angle between the normal
to the loop and the magnetic field.
normal

Top View:
B
I
x
The normal is the direction perpendicular
to the plane of the loop of wire.
If the loop has N turns, then the torque
becomes...

  N B I A sin

A small circular coil of 20 turns of wire
lies in a uniform magnetic of 0.5 T so
that the normal to the plane of the coil
makes an angle of 60o with the
direction of B. The radius of the coil is
4 cm, and it carries a current of 3 A.
What is the magnitude of the torque
on the coil?
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
side view
60o
top view
N = 20 turns
B = 0.5 T
A = p(.04)2
I=3A
 = 60o

  N B I A sin


  (20)(0.5 T)(3 A )(5  10 m ) sin60
3
 013
. Nm
2
o
x
x
x
x
x
x
I
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
I
60o
top view
side view
If the current flows clockwise
around the loop (viewed face on),
which way does the loop rotate?
The force on the left side of the loop is
toward the top of the top view diagram.
x
x
x
x
x
x
I
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
I
60o
top view
side view
The force on the right side of the loop is
toward the bottom of the top view diagram.
These two forces combine, both creating a
torque in the same direction, causing the loop
to rotate
CLOCKWISE!