p. 53
Force of a Magnetic Field on a Moving Charge
We defined the electric field in terms of the force per unit charge on a small positive test charge. We will again use a small positive test charge to define the magnetic field. Unfortunately, the definition of the magnetic field cannot be made as simply because the magnetic force does not act on the test charge unless it is moving . To complicate matters, the magnitude and direction of this force depend on the magnitude and direction of the velocity of the test charge. We find that the magnitude of the force is directly proportional to the component of the velocity that is perpendicular to the magnetic field.
Magnitude of the Magnetic Field q
0
F
 vsin

B
Let B be the magnetic field, v the velocity of the small positive test charge and F be the magnetic force on this charge. Here 0
  
180

. Since the magnitude of F is directly proportional to v sin

, we define the magnitude of the magnetic field to be v
B

F q v
0 sin

Units of the Magnetic Field
The units of the magnetic field are
C

N
 m C
N
 m s s

N

1 tesla (T) .
The tesla is a rather large unit. Small magnetic fields are measured in gauss , where 1 gauss = 10 -4 tesla.
Direction of the Magnetic Field
The direction of the magnetic field can be determined by replacing q
0
with a small compass needle. The north pole of the compass needle points in the direction of the magnetic field B .
Direction of the Magnetic Force
It is important to remember to think in three dimensions in this chapter. The vectors v and B form a plane in space. The magnetic force F on the test charge q
0 is perpendicular to this plane. The direction of the magnetic force F can be determined as follows. Imagine that the plane of B and v is a piece of wood. In the above figure replace the test charge q
0
with a right-hand screw embedded in this piece of wood. Next imagine the direction you would have to turn the velocity vector v to align it with the magnetic field vector B . (You want to turn v through the smallest angle possible.) To determine the direction of F imagine turning the screw in the same direction as you would turn v to align it with B . The screw advances in the direction of F .
Screw advances upward, in the direction of F . v
B
The right-hand screw may be replaced with a bottle top or any other convenient object with a right-hand thread. Note that the orientation of the object does not matter. For example, note that you obtain the same results with the screw regardless of whether the screw is embedded in the top of the wood or in the bottom of the wood, so long as you turn it in the direction that would align v with B through the smallest angle.

p. 54
Example
A magnetic field is directed parallel to the surface of Earth. It has a magnitude of 3.2 T and is directed at an angle of 30

north of east. A positive charge of 12

C moves with velocity 8.0 m/s directed west. Find the magnitude and direction of the magnetic force on the charge.
The magnitude of the magnetic force is
F
 qvB sin

        v
30

B
F
 
4
1.5 10 N
The head of a right-handed screw is shown in the diagram. The screw should be turned clockwise ; this causes the screw to move into the diagram. Hence the direction of the magnetic force is into the diagram.
F
 
4
1.5 10 N directed into the diagram.
Example
A uniform magnetic field in a laboratory has magnitude 18.0 T and is directed in the +y direction. An electron has velocity
 6
1.55 10 m/s the electron’s acceleration.
directed at an angle of 20

above the -x axis. Find the magnitude and direction of y
F
 qvB

 sin

 
19

 6
    v
20

B
F
  -12
4.19 10 N x
The acceleration of the electron is determined by Newton’s second law:
F

;

F m
; a

4.19 10

12
N
9.11 10

31
kg
; a
  2
The direction of the acceleration is in the direction of the force. If we imagine rotating v into B in the above diagram, a right-handed screw at the origin would advance into the diagram. BUT the electron has a negative charge. Remember that the magnetic field (as was the electric field) was defined for a positive test charge. If the charge is negative the force reverses direction . Hence the force on the electron is directed out of the diagram . And a
  18 2
4.60 10 m/s directed out of the diagram

p. 55
Motion of a Charge in a Uniform Magnetic Field
A uniform magnetic field can be produced between the poles of a permanent magnet that is in the shape of an broken ellipse. In the following diagrams the magnetic field lines will be directed either into or out of the diagram. A dot (

) will indicate a field line out of the diagram. A cross (

) will indicate a field line into the diagram. Consider a positive charge that enters a uniform magnetic field at right angles to the field.
What is the direction of the magnetic force? For the charge in the position shown in the figure on the left, the direction of the magnetic force is upward .
    
    
    
    
 
F
  
    
    
  q
 v
 
B
The acceleration of q is in the direction of the force F . We see that the acceleration and velocity of the charge make a right angle with one another. In fact, this is the case regardless of the direction of v , as shown on the right. It can be shown that if:







 v v
F








F
F q








N S
F v







 below v








B
 the velocity and acceleration of a particle always make a right angle;
 the magnitudes of the velocity and acceleration are constant;
 the motion of the particle is confined to a plane, then the particle moves in a circle with uniform circular motion.
If we know the speed v , the charge q and the magnetic field magnitude B , we can find the radius r of the circular path of the charge. We do this by equating the magnetic force to the centripetal force . qvB sin
  qvB
 mv
2 r mv
2
;
 
90 , sin90
 
1 r
; r
 mv
2 qvB
; r
 mv qB
Example
Find the radius of the circular path of an electron moving with speed
 8
1.20 10 m/s in a uniform magnetic field of 0.500 T. r
 mv

 qB
 
31

 8
9.11 10 kg 1.20 10 m/s

1.60 10

19
  


1.09 10

22
8.00 10

18
C


N
 
3
C

2 r
  
3 r

1.37 mm

p. 56
The Force on a Current in a Magnetic Field
We have seen that a force is exerted on a charge moving in a magnetic field. We know that a current-carrying conductor has charge moving through it. We might expect that a magnetic field can exert a force on a currentcarrying conductor.
Consider a thin conducting rod that carries a conventional current I . Suppose we place this in a uniform magnetic field B and that the direction of the conventional current makes an angle

with the magnetic field.
B Suppose that charge
 q is moving down the rod at some instant.
I

This charge experiences a magnetic force given by
F
  qvB sin

 q
F
 v
B
Suppose that the rod has length L . Let the charge
 q move through this distance in time
 t . The force can be expressed as
F

 q
 t sin

But

 q t
is the current and L
Hence
F

ILB sin

Force on a conductor carrying a current I
Example
One terminal of a 12.0 volt ideal battery is connected to 5.00

resistor. Two strands of frictionless conducting wire are then connected to the other terminal of the battery and the free terminal of the resistor. The strands are parallel. A short wire of mass 4.50 mg is placed across the frictionless wires as shown in the diagram. The entire circuit sits in a uniform magnetic field of 3.72 gauss directed out of the diagram. Find the magnitude and direction of the acceleration of the short wire.
The current in the short wire is directed downward. The magnitude of this current is
5.00

0.500 m
12.0 volts
I

12.0 V
5.00

; I

2.40 A
3.72 G
The force on the wire is given by
F

ILB
 the horizontal wires and 90 :
F
    

4
  
The acceleration of the short wire is given by a

F m
: a


6
4.50 10 kg
; a

99.2
s m
2
to the left
F


p. 57
The Torque on a Current-Carrying Coil in a Magnetic Field
Review of the torque due to a couple . Two forces of equal magnitude and opposite direction that act on an object are called a couple . The torque due to the couple is equal to the magnitude of either force multiplied by the distance separating their lines of action.
F -F
The torque due to a couple is

= xF and is measured in meter-newtons (m

N).
Torque measures the ability of a force to twist an object about a given axis. x
However, the torque due to a couple is independent of the axis about which it is calculated.
Consider a coil of one turn mounted on a vertical shaft that can turn. Suppose that the coil carries current I and is immersed in a uniform magnetic field B . (See the figure at the right.)
In the figure the current circulates clockwise around the coil. The magnetic field is directed from left to right. Side 1 of the coil has current directed upward in figure a and so is pushed backward by the magnetic force as shown in figure b .
Side 2 of the coil has current directed downward in figure a and so is pulled forward by the magnetic force as shown in figure b . The magnetic field is uniform so that the magnitudes of both the pushing and
Right-hand screw x pulling magnetic forces are equal. The lines of action of these forces are parallel so that these forces form a couple. From the geometry in figure b x = Wsin(

), where

is the angle between the coil’s normal line and the magnetic field. The magnitude of the force F is given by F

ILB sin
 
where L is the length of the coil’s side and

is the angle between the direction of the current and the magnetic field. Here

= 90

. The torque on the coil is therefore
 
Wsin
 
ILB
  or
 
IAB sin
 
where A = L W is the area of the coil. If the coil has
N turns,
 
NAIB sin
 
.
This formula turns out to be true for any planar coil, regardless of its shape, as long as A is the area enclosed by the coil. The quantity NAI is called the magnetic moment of the coil and is usually denoted by

:
 
NAI
The units of magnetic moment are amps
 m
2
(A
 m
2
). The torque on the coil can therefore be expressed as
  
B sin
 
.

Example
A circular coil of 25 turns and radius 15 cm carries a current of 2.0 A. The coil is immersed in a uniform magnetic field of magnitude 0.83 T that makes an angle of 60

with the plane of the coil. a.
What is the magnetic moment of the coil?
A
  r
2
; A
 
 
NIA ;
 
 
2
0.15 m ; A

0.071 m
   
2

2
   2 b.
What torque does the magnetic field exert on the coil?
Use
  
B
  
B sin sin
 
;
 
where

= 30

( Not 60

! Why?)
 

 2
 
0.83 N
A m
  
  p. 58

p. 59
Magnetic Fields Produced by Currents
For many centuries following the discoveries the ancient Greeks made with amber and magnetite the sciences of electricity and magnetism developed separately. Some scientists maintained that there was no connection between them. In 1820, however, Hans Christian Oersted found that an electric current in a wire can deflect a compass needle. (Ironically, Oersted discovered this while preparing a lecture demonstration for his physics students that was supposed to show that electricity and magnetism were unrelated!) From that point on the study of combined electrical and magnetic phenomena, called electromagnetism , was developed further by scientists in many countries. The most important contributions included those of the French physicists Biot and Savart,
Ampere, the English scientist Michael Faraday and the physicist James Clerk Maxwell.
An electric current in a wire produces a magnetic field. The precise formulation of this phenomenon was developed by Biot and Savart and later refined by Ampere and Maxwell. These details involve the use of vector calculus and will be omitted from this course. Instead we give several important results of these formulas.
In the following we assume that the wires are in air or empty space.
Long straight wire.
The magnetic field around a long, straight wire forms concentric circles. The intensity of the magnetic field a distance r from a long straight wire with current I is given by
B


0
I
2
 r where

0
is a constant called the permeability of free space:

6
1.26 10 T m/A .
Coil of wire.
The magnetic field along the axis of a coil of wire of radius R with N turns carrying current I is given by

2


0
NIR
R
2  z
2
2

3 / 2 where z is height of the point above the center of the loop. We can find the magnetic field at the center of the coil by setting z = 0:
B


0
NI
2 R where B = B (0) is the magnetic field at the center of the loop.
Solenoid.
A long, tightly wound helical coil of wire is called a solenoid . If the solenoid carries a current I it can be shown that inside a long solenoid the magnetic field is nearly uniform and has magnitude

p. 60
B
 
0
In where n is the number of turns of wire per unit length on the solenoid. The magnetic field outside a solenoid is usually negligible.
Toroid.
A toroid can be thought of as a solenoid bent into the shape of a hollow doughnut. It can be shown that inside the toroid the magnetic field is given by
B


0
IN
2

1 r where N is the number of turns in the toroid, I is the current in the toroid and r is the distance from the center of the toroid. Outside the toroid the magnetic field is zero.