MAGNETIC FIELDS
• THE MAGNETIC FIELD
• FORCES ON MOVING CHARGES
• THE MAGNETIC FIELD OF A
LONG, STRAIGHT WIRE
• THE MAGNETIC FIELD OF A
LOOP OR COIL
• THE MAGNETIC FIELD OF A
SOLENOID
Written by Dr. John K. Dayton
THE MAGNETIC FIELD:
The magnetic field is designated by the symbol B. The
source of all magnetic fields are moving electric
charges. Even the magnetic fields associated with
electromagnetic radiation have their source in moving
electric charge.
The SI unit for the magnetic field is the tesla, T. A
more common unit of magnetic field strength is the
gauss, G. Earth’s magnetic field at the surface is about
0.5 G.
104G = 1T
N
S
The magnetic field is characterized by what we call a north and a south pole.
Similar poles repel and opposite poles attract. Outside of magnetic materials
magnetic field lines extend away from north poles and toward south poles.
However, magnetic field lines, unlike electric field lines, are continuous. Each
magnetic field line forms a complete loop. The simplest of all magnetic fields is the
magnetic dipole, like a bar magnet.. Magnetic monopoles, such as an isolated north
pole, do not exist.
THE FORCE ON A MOVING CHARGE IN A MAGNETIC
FIELD:
F = Force on moving charge
F = qv  B
q = charge on particle / object
ˆ B sin  
F = nqv
v = velocity of moving charge
B = magnetic field
× = vector cross-product
v
F
B
I

 = angle between v and B
n = unit vector in correct direction
THE RIGHT-HAND RULE:
F
1.
2.
3.
4.
v
I
Use Right Hand
Point Fingers In Direction Of Current
Point Open Palm In direction Of Magnetic Field
Extended Thumb Is Direction Of Force On Charge
B
EXAMPLE: A proton traveling horizontally to the right with a
speed of 1.5x104 m/s encounters a uniform magnetic field of 0.55T
oriented 60o above the horizontal. What is the magnitude of the force
on the proton and what is its initial direction?
v
B
+e
F = qvB sin   = 1.6  1019 C 1.5  104 ms   0.55T  sin(60o )
F = 1.14  1015 N
According to the right-hand-rule this force will be
directed outward from the diagram.
MOTION OF A CHARGED PARTICLE WHOSE VELOCITY IS
PERPENDICULAR TO B:
F = qvB sin 90o = qvB
mv 2
F=
= qvB
r
mv = qrB
Using the right-hand-rule, can you
determine the sign of q in the diagram?
Click for the answer.
B
v
mv
r=
qB
v = r
mr
r=
qB
qB
=
m
q, m
F
r
According to the vectors shown in the diagram, and
the right-hand-rule, q must be a negative charge.
view from above
Is the charge q of the moving
particle positive or negative?
B
q
v
q is positive.
v
FB
B out of diagram
v(perpendicular)
v
B
B
q
q
EXAMPLE:
An alpha particle, q = 6.4x10-19C and
m = 6.68x10-27 kg, is following a circular path of radius 0.5m in a
uniform magnetic field of 1.2 T. (a) What is the angular speed of
the alpha particle? (b) What is the linear speed of the alpha
particle?
qB
=
=
m
19
6.4

10
C  1.2T 

27
6.68

10
kg 

 = 1.15 108 rads
v = r =  0.5m  1.15 108
v = 5.75 107
m
s
rad
s

FORCE ON A CURRENT-CARRYING WIRE:
B

wire
I
l
F = qvB sin  
q = neAl
v = v drift
F = neAv drift Bl sin  
I = neAv drift
n = density of mobile
electrons in wire.
F = IBl sin  
EXAMPLE: Calculate the net force on a wire of length 20m
and carrying a current of 12A through a magnetic field of 0.5
Gauss directed 70o to the wire.
B
12A
20m
F = IlB sin( )
F = 12 A 20m   0.5 104 T  sin(70o )
F = 1.13 102 N
According to the right-hand-rule, this force
will be directed outward from the diagram.
TORQUE ON A CURRENT-CARRYING LOOP IN A
MAGNETIC FIELD:
m
F
a


B
F = NIaB
 b
 = 2  F 2 sin  = Fbsin 


 = NIaBbsin = NIABsin 
NIA = m
 = mBsin 
b
I
 = I

F
 mB
= =
=  2
I
I
mB
=
I
EXAMPLE: A coil of wire composed of 100 turns of radius 6cm
carries a current of 5A. It is in a uniform magnetic field 0f 0.65T
oriented 75o to the plane of the coil. (a) What is the magnetic
moment of the coil? (b) What is the torque on the coil?
m = NIA = NI r 2
m = 100  5 A   .06m 
2
m = 5.65 A  m 2
 = mB sin 
 =  5.65 A  m
 = 0.95 N  m
2
  0.65T  sin 15 
o
 is the angle between B
and m.  = 90o – 75o
AMPERE’S LAW:
 Bl cos   =  I
o
I = the current that is encircled by a closed path.
l = small piece of the closed path.
B = local magnetic field at l
 = angle between B and direction of l
Sum is over all l comprising the closed path
MAGNETIC FIELD OF A LONG, STRAIGHT WIRE:
WIRE
B inward
B outward
I
o I
B=
2 r
Right-Hand Rule:
1-fingers in direction of I
2-Palm toward field point
3-Thumb points in
direction of B
 o = 4   10
 7 T m
A
SUPERPOSITION OF MAGNETIC FIELDS:




BP = BP ,1  BP ,2  BP ,3 ...
d = 5cm
I1 = 3A
Calculate the magnetic field mid
way between the two wires.
BP = BP ,2  BP ,1
BP ,1 =
P
2.5cm

B p ,1 is inward

B p,2 is outward
T m
A
  3 A
T m
A
   4 A
2   0.025m 
BP ,1 = 1.6  10 5 T
BP ,2 =
I2 = 4A
7
4


10

7
4


10

2   0.025m 
BP ,2 = 2.1  10 5 T
BP = 2.1  10 5 T  1.6  10 5 T
BP = 5.0  10 6 T
EXAMPLE: The central conducting wire of a coaxial cable carries
a current of 3A upward. The outer, cylindrical conducting wire of
radius 3mm carries a uniformly distributed current of 5A downward.
What are the magnetic field strengths at 2mm from the center and
5cm from the center of this wire?
0 I enclosed
B1 =
2 r1
4 10

=
B1 = 3.00 104 T
0 I enclosed
B2 =
=
2 r2
B2 = 8.00 106 T
7 Tm
A
  3 A
2 .002m 
7 Tm
4


10

A   2 A
2 .05m 
Only the net enclosed
current is used. Ienc=5A3A. The net current for B1
is upward and the net
current for B2 is
downward so these two
fields are oriented in
opposite directions.
MAGNETIC FORCE BETWEEN PARALLEL WIRES:
r
I1
I2
F2,1
l
 o I1
Bof 1 at 2 =
2r
F2,1 = I 2 Bof 1 at 2 l
 o I1 I 2 l
F2,1 =
2r
The force between parallel wires is
attractive if the currents are in the
same direction.
EXAMPLE: A long straight, vertical wire carries a current of 5A
upward. Near the wire is a rectangular loop of wire carrying a
current of 3A. The loop has dimensions 5cm by 10cm and is
positioned so that its long side is parallel to the long wire at a
distance of 5cm with its current also directed upward. What is the
net force on the loop?
The magnetic field produced by the wire in
the plane of the loop is directed inward, into
wire
the diagram. The strength of this field
decreases as you move further from the wire.
I2=3A
The force on side 1 is downward and the
3
5cm
force on side 3 is upward. Use the righthand-rule to confirm this. These forces are
I1=5A
10cm of equal magnitudes so cancel each other
4
2
F2
F4
out. The force on side 2 is toward the wire
and the force on side 4 is away from the
1
wire. These are also in opposite directions
loop
but the force on side 2 is stronger because it
5cm
B from the wire is inward is closer to the wire.
on this side of the wire
Solution continues on next slide.
continued from previous slide
o I1 I 2l o I1 I 2l
Fnet =F2  F4 =

2 r2
2 r4
Fnet
4 10

=
7 Tm
A
  5 A 3 A.1m  
2
Fnet = 3.00 106 N
Fnet is directed toward the wire.
1
1 



.0
5
m
.10
m


THE BIOT-SAVART LAW:
P
I
r

l
 0 Il sin 
B = 
n
2
4
r


B =  B
MAGNETIC FIELD OF A LOOP OR COIL:
I

r
B
a
x

P B
x
oa2 I
B=
3
2r
By
x-axis
a
B = B
B = Bx = B cos   = B
r
o I l sin   o I
B =
=
l
2
2
4r
4r
2

I
a

Ia

Ia

a
I
 o

o
o
o
B = 
l  =
l =
 2a  =
2
3
3 
3
4r
r
 4r
 r 4r
oa NI
B =
3
2r
2
If there are N turns of wire in the coil, then:
If x>>a, then:
oa2 NI
B =
3
2x
MAGNETIC FIELD AT THE CENTER OF A LOOP OR COIL:
I
a
x-axis
P
o a 2 NI
B=
2r 3
r=a
o NI
B=
2a
a = radius
I
B
Palm toward center
I
EXAMPLE: Calculate the magnetic field strength at the center
of a coil of radius 20cm, 200 turn of wire, and carrying a current
of 2.5A.
B=
o NI
2a
=
B = 0.0016T
7 Tm
4


10

A   200  2.5 A 
2 .2m 
MAGNETIC FIELD OF A SOLENOID:
I
o NI
B=
=  o nI
l
N = total number of turn
of wire
n = number of turns of
wire per unit length,
n = N/l
I
 o NI =  B cos l
1
I
 o NI =  B1 cos l   B2 cos  l
1
2
  B3 cos l   B4 cos  l
2
4
3
4
1 =  = 90o , 4 = 0o , B2  0
 o NI =  B4 cos  l = B  l = Bl
4
I
3
 o NI
=  o nI
B=
l
EXAMPLE: A solenoid is to be made so that a magnetic field of
0.5T is at its center. If the solenoid is to be 10cm in length and
wrapped with wire carrying a current of 3A, how many turns of
wire are required?
B=
o NI
l
0.5T  0.1m 

Bl
 N=
=
o I  4 107 TmA   3 A
N = 13263 turns
End Of Presentation