# In principle

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```					                                  CH 19 - Magnetism
We most often associate magnetism with permanent magnets and with currents
(electromagnets). The space surrounding these magnets can be described in terms of a
magnetic field, somewhat like the electric field in the space surround electric charges.
All magnets have north and south poles. For an isolated magnet, such as the bar magnet
shown below, the field lines emerge from the north pole and return to the south pole. All
poles come in pairs. If you were to break the bar magnet into two pieces, you would
create new north and south poles.

S                  N

If you have two different magnets, their opposite poles attract and their like poles repel,
somewhat like charges.

There is a magnetic field associated with the earth. The magnetic north pole of the earth
is near the geographic south pole, and the magnetic south pole is near the geographic
north pole. The field lines of the earth point approximately ‘north’. So the north pole of
a compass needle, which is a bar magnet, points north since it is attracted to the earth’s
magnetic south pole (which is north). Actually, the earth’s field is generally not
horizontal and dips down as you move from the equator to the north pole.

Magnetic Field

Magnetic force on a charge

We can define the magnetic field, symbolized by B, by the force that it exerts on a charge.
Unlike the electric field E, a magnetic field will not exert a
force on an electric charge unless it is moving. Also, it is        F
only the component of field that is perpendicular to the
velocity that matters. The force is given by                                B

F  qvB sin                                                           
v

where  is the angle between the direction of the field and the
particle’s velocity. The direction of the force is perpendicular to both B and v, as shown.

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The direction of F can be determined by the ‘right hand rule’.
You point your fingers in the direction of v, then close you                        B
hand so that your fingers turn towards B, and your thumb is

then in the direction of F.                                                             v

For negative charges the force is in the opposite direction. Use
the right hand rule and then reverse directions (or use a ‘left                F
hand rule’).

The above equation for the force on a moving charge can be considered as a way of
defining the magnetic field. So we could write

F
B
qv sin 

The SI unit for B is tesla (T). Another unit that is used is gauss (G). 1 G = 10-4 T.

Example:

A proton travels at a speed of 5 x 105 m/s in a direction 30o east of north in a region
where the earth’s magnetic field is 3 x 10-5 T and points north. What is the magnitude and
direction of the force on the electron?

Solution:

From the right hand rule, the direction of F is up. The magnitude is

F  ( 1.6 x1019 C )( 5x105 m / s )( 3x105 T ) sin( 30o )  1.2 x1018 N

Although this may seem small, the resulting acceleration of the proton is huge. (Do the
calculation.)

Magnetic force on a current carrying wire

A wire carrying an electric current consists of moving charges. Therefore, a magnetic
field will exert a force on a current carrying wire if it has a component perpendicular to
the wire. The amount of moving charge in a section of a wire of length x is
q  It and the drift velocity is v  x / t . Thus, the force is

x
F  qvB sin   It      B sin   IxB sin 
t

Generalizing this to a length L of straight wire, we have

F  ILB sin

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where  is the angle between the field and current directions. Again, we use the right
hand rule to find the direction of F. It is perpendicular to both the wire and the field.

Torque on a current loop

A current loop in a magnetic field can experience forces on its sides which can generate a
torque. Consider a rectangular loop with the field in the plane of the loop.

B
I

b

a

The force on the left side is up and the force on the right side is down. There is no force
on the top and bottom sides since the field is parallel or antiparallel to the current. The
up and down forces on the left and right sides will create a torque. If there are no
restraining torques, then this torque would cause the loop to rotate about a vertical axis
a    a
   (left )   (right )  F  F  Fa
2    2
F  ILB  IbB , so
  IabB  IAB
F
where A = ab is the area of the loop.
I (out)
If the field is not in the plane of the loop, then
the torque is reduced by a factor sin. In
particular, if the field is perpendicular to the                a/2             
plane then the torque is zero. To the right is a                                    B
side view of the loop when the loop normal is
at angle  with respect to the field. The torque                             a/2      I (in)
is

  IABsin                                                               F

If the loop has N turns of wire, then the torque is

  NIABsin            (unit = Nm)

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This expression for the torque, although derived for a rectangular loop, applies to any
loop shape.

The term NIA is called the magnetic dipole moment (or magnetic moment) of the loop

  NIA         (unit = Am2)

So, we can write

   B sin 

Example:

A 20-turn circular loop of wire of radius 4 cm carrying a current of 5 A is placed in a
magnetic field of 0.1 T. The angle between the field direction and the plane of the loop is
30o. Find the torque on the loop.

Solution:

The angle between the normal to the plane and the field direction is 90o – 30o = 60o. So,

  NIABsin   (20)(5 A) (0.02m) 2 (0.1T ) sin 60o  0.011N  m

Electric motor:

The electric motor is a device which makes use of the torque exerted on a current in a
magnetic field to convert electrical energy into mechanical energy. An applet which
illustrates how the motor works can be found at

http://bama.ua.edu/~jharrell/Fendt_applets/ph14e/electricmotor.htm

Note how the commutator reverses the current direction each half rotation to keep the
motor continuously rotating in the same direction. Otherwise, it would just oscillate back
and forth.

Motion of a charged particle in a magnetic field

A charged particle moving perpendicular to a magnetic field will experience a force
given by F = qvB. Since this force is always perpendicular to v, then it will cause the
particle to move in a circle. Also, since the force is perpendicular to v it does no work on
the charge, and the kinetic energy and the speed of the particle don’t change. Thus, we
have uniform circular motion. So,

v2                    mv
F  qvB  m       ,    or       r
r                    qB

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v

B(in)             F

Example:

What is the radius of the circular motion of an electron moving at 106 m/s in a 0.05 T
magnetic field?

Solution:

mv (9.1x10 31 kg)(10 6 m / s )
r                                   1.1x10  4 m  0.11mm
qB           19
(1.6 x10 C )(0.05T )

A proton with the same speed in the same field would move in a circle with radius 21 cm.

The proton and electron would circulate in opposite directions in the field.

Magnetic Field due to a Current

Currents produce magnetic fields. In principle, the magnetic field due to any current
configuration can be calculated from the Biot-Savart law, which gives the field due to an
elemental current. We will not use the Biot-Savart law. Instead, we give the results for
three simple current configurations.

Magnetic field due to a long straight current-carrying wire

The field due to a long, straight wire circulates about the
wire and decreases with distance from the wire according
to the formula                                                                  I
B
0 I
B        ,
2 r

where 0 = 4 x 10-7 Tm/A and is called the permeability
of free space and r is the distance from the center of the wire.

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The direction of the magnetic field around the wire can be determined by another right
hand rule: Grab the wire with your hand with you thumb pointing in the direction of the
current. Then your fingers circulate around the wire in the direction of the magnetic field.

Example:

Two parallel wires 20 cm each carry a current of 5 A. Find the               I           I
magnitude and direction of the magnetic field at a point
between the wires 15 cm from one wire and 5 cm from the other.
r1    r2
Solution:

At the point in question, the field due to the left wire points into
the page and the field due to the right wire points out of the page. Letting out of the page
be the positive direction (this is arbitrary), we have

0 I   0 I     (4 x10 7 T  m / A)(5 A) (4 x10 7 T  m / A)(5 A)
B   B1  B2                                          
2  r1 2  r2          2  (0.15m)                2  (0.05m)
 6.67 x10  6 T  2 x10  5 T  1.33x10  5 T (out of page)

Force between two current carrying wires

Two parallel current-carrying wires will experience an
attractive or a repulsive force, depending on whether the
currents are in the same direction (attractive) or in
opposite directions (repulsive). Each wire sees the field
created by the current in the other wire. By combining
the formula for the field generated by one wire and the
force on the other wire due to this field, we get

 0 I1 I 2 L
F
2 d

where d is the separation of the wires and L is the length of the wires.

Example:

For the two wires in the previous example, the force per unit length is

F  0 I1I 2 (4 x10 7 T  m / A)(5 A) 2
                                      2.5 x10  5 N / m
L   2 d           2  (0.2m)

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Magnetic field due to a circular current loop

The field lines due to a current loop are shown qualitatively
to the right. The field passes through the loop in a direction
given by still another right hand rule. If your fingers
circulate around the loop in the direction of the current,
then the field passes through the loop in the direction of
your thumb. The field along the axis of the loop is given by

 0 IR 2
B
2 ( x 2  R 2 )3 / 2

where R is the radius of the loop and x is the distance from the center of the loop. In the
middle of the loop, this expression reduces to

0 I
B
2R

Magnetic field of a solenoid

A solenoid is a cylindrical coil of wire. The magnetic field due to
a solenoidal current is shown to the right. Outside the solenoid
the field is similar to that due to a bar magnet. If the solenoid is
long compared with its diameter, then the field is nearly uniform
inside, except near the ends, and is given by

B   0 nI

where n = N/L is the number of turns per unit length.

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