EMF 2005
Handout 7: The Magnetic Field
1
THE MAGNETIC FIELD
This handout covers:
• The magnetic force between two moving charges
• The magnetic field, B , and magnetic field lines
• Magnetic flux and Gauss’s Law for B
• Motion of a charged particle in E and B : the Lorentz force
Important special cases:
- Motion perpendicular to uniform B
- The velocity selector
- The Hall effect
Note:
1.
You do not need to remember the full vector treatment of the
magnetic force between moving charges.
2.
For exam purposes, the important relationship defining the magnetic
field, B is
FM = Q(v × B )
3.
However, it is important to understand the vector treatment so that
you can gain a good conceptual grasp of the subject.
4.
The key to understanding the magnetic field is to be able to use the
vector cross product ⇒ revise this if you are not sure of it.
EMF 2005
Handout 7: The Magnetic Field
2
The magnetic force
Up to now, we have considered the ELECTROSTATIC force, due to
charges at rest .
F2
Coulombs Law:
F2 =
Q1Q 2
4πε or
2
Q2
∧
r
∧
r 12
12
Q1
r
If the charges are BOTH moving, another force
exists between them: the MAGNETIC FORCE.
FM =
µ o Q1Q 2
4π r 2


 v 2 × ( v1 × r 12 )


∧
v2
Q2
∧
r 12
r
Q1
v1
Simpler case: v1 and v2 parallel
∧
v1 × r 12
Magnitude = v1r12 = v1
Direction = out of page
∧
v 2 × ( v1 × r 12 )
⇒
Q2
v2
FM
∧
Magnitude = v1v2
Direction = Towards Q1
∧
µ o Q1Q 2
FM = −
v1v 2 r 12
π r2
4π
Q1
r 12
v1
EMF 2005
Note:
Handout 7: The Magnetic Field
1.
FM ∝ Q1Q2
2.
FM ∝ v1v2
3.
FM exists in addition to the electric force
4.
The constant µo is called the
3
as for the electrostatic force
⇒
no force unless BOTH charges are
moving
PERMEABILITY CONSTANT or the
PERMEABILITY OF FREE SPACE
SI System:
µo = 4π x 10-7 N s2 C-2
Relative magnitudes of the electric and
magnetic forces
FM =
µ o Q1Q 2
v1 v 2
4π r 2
FE =
⇒
FM
v v
= µ o ε o v1 v 2 = 1 2 2
FE
c
 1 
where c = 

µoεo 
c = SPEED OF LIGHT
Q 1Q 2
4πε or 2
1/ 2
[To be explained later]
FM
v v
= 1 2 2 ⇒ FM << FE unless speeds are close to speed of light.
FE
c
BUT: The magnetic force can still dominate even at very low speed
because FE tends to be cancelled out due to the overall charge neutrality
of matter.
Example: Two wires carrying currents
FE = 0 (neither wire is charged)
FM ≠ 0 (and can be quite large)
FM
FM
I1
I2
EMF 2005
Handout 7: The Magnetic Field
4
The magnetic field
DEFINITION: The magnetic force experienced
by a charge Q moving with velocity v is
v
Q
B
F (inwards)
Fmag = Q(v × B )
This equation defines B , the MAGNETIC FIELD
v ⊥B
Special case:
⇒
F = QvB
Recall:
Q2
FM =
µ o Q1Q 2
4π r 2
∧


v
v
r
(
×
×
12 )
1
 2


v2
∧
Q1
But, by definition of B ,
r 12
r
v1
F2 = Q(v 2 × B1 )
where B1 = magnetic field at Q2 due to Q1.
So
B1 =
∧
µ o Q1 

v
r
×
1
12 
4πr 2 

Magnetic field of a
moving point charge
SI Units for Magnetic Field:
1 T (Tesla) = Field which exerts a force of 1 N on a 1-C charge moving
with velocity 1 m s-1 perpendicular to B .
B =
F
Qv
⇒
1 T ≡ 1 N C-1 m-1 s
EMF 2005
Handout 7: The Magnetic Field
5
Magnetic field lines
Recall: Electric field lines point in the direction of E
Similarly, magnetic field lines point along B
B1 =
Recall:
∧

µ o Q1 
v
r
= Magnetic field at P2 due to Q1
×
1

12 
4πr 2 

P2
Direction is outwards
by right hand rule
B1 (outwards)
∧
r 12
view along here
v1
Q1
Imagine we view ALONG THE DIRECTION OF MOTION, so that Q1
appears to be coming straight at us:
B
v1 is OUT OF PAGE
P2
If we consider points on a
circle (e.g., P2 - P5) then
the direction of the unit
∧
vector r 12 etc. depends
on which point we consider.
r 13
P3
B
∧
v1
∧
v1 is always out of the page.
Apply the right hand rule to
each point
∧
r 12
∧
r 15
r 14
P5
B
→
B is always tangential
⇒
Lines of B form CLOSED LOOPS
Convention: As before, line spacing is
used to indicate the magnitude of B
(closely spaced lines ⇒ strong field).
P4
B
Q moving
outwards
B
EMF 2005
Handout 7: The Magnetic Field
6
Magnetic flux, ψ
This is defined in exactly the same way as electric flux.
Consider a small flat area dA.
B
Let B be the magnetic field at its centre.
Assume that dA is so small that B can be
regarded as uniform over the whole of dA.
θ
dA
DEFINITION: The MAGNETIC FLUX, dψ
through the area dA is the product of dA
and the normal component of B .
or
dψ = (Bcosθ)dS
so
Normal
Normal
tovector
ds
Area
Areads
dA
dψ = B ⋅ dA
where dA is the NORMAL VECTOR of the area dA:
Magnitude of dA is:
Direction of dA is:
Note: 1.
dA
perpendicular to dA
Magnetic flux is a SCALAR.
2.
Young & Freedman uses the symbol ΦB for magnetic flux
3.
The SI unit of magnetic flux is the Weber (Wb):
1 Wb ≡ (1 Tesla)(1 m2) or 1 T ≡ 1 Wb m-2.
EMF 2005
Handout 7: The Magnetic Field
7
Magnetic flux for the case of a non-uniform
field passing through an arbitrary surface
Proceeding exactly as we did for the electric flux (see handout on
electric flux and Gauss’s Law), we can show that the total magnetic flux
crossing a surface S is
Ψ =
∫ B ⋅ dA
S
What is Ψ for a closed surface?
Recall: Gauss’s Law for E :
Φ =
∫
E ⋅ dA =
Q enclosed
εo
Lines of E begin and end on electric charges. But lines of B form closed
loops (there is no equivalent of charge - i.e., no “magnetic monopoles”).
⇒
as many magnetic field lines will leave a given volume as enter it
(no enclosed “magnetic charge”).
⇒
THE TOTAL MAGNETIC FLUX THROUGH A CLOSED SURFACE
IS ZERO
or
Ψ =
∫ B ⋅ dA = 0
THIS IS GAUSS’S LAW FOR THE MAGNETIC FIELD
MAXWELL’S 2nd EQUATION
This is sometimes written as div B = 0 or ∇ ⋅ B = 0.
EMF 2005
Handout 7: The Magnetic Field
8
Force on a charged particle moving
in a uniform magnetic field
Consider a uniform B coming out of the page (usually represented by
dots):
Let charge Q have velocity v
perpendicular to B
F = Q(v × B )
⇒
⇒
no component of force along
the direction of v
THE SPEED NEVER CHANGES
F = QvB = constant
Constant force
to velocity
⇒
Force required for circular motion is
⇒
mv 2
QvB =
r
⇒
r =
.
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F is always perpendicular to v and B
B. (out
. . of
. page)
. . .
.
.
v
Q
v
.
MOTION IN A CIRCLE
mv 2
F =
r
m = mass
r = radius
mv
QB
Speed = v, Circumference = 2πr
⇒
Frequency is
Note: 1.
2.
f =
v
QB
= CYCLOTRON FREQUENCY
=
2 πr
2 πm
f is independent of v
f depends only on B and fundamental constants
⇒ it can be used to find B.
EMF 2005
Handout 7: The Magnetic Field
9
What if v also has a component of motion along the direction of B ?
Call this component v parallel
B and v parallel are parallel,
so
Q
v parallel x B = 0
⇒
no force in this direction.
⇒
motion along B is not affected
B
vparallel
⇒
MOTION IN A SPIRAL
Motion of a charged particle in
combined electric and magnetic
fields: the Lorentz force
If a point charge Q moves with velocity v in an electric field E and a
magnetic field B , the resultant force on it is
F = Q[E + (v × B )]
THE LORENTZ FORCE
The velocity selector
Consider a charged particle moving with velocity v perpendicular to both
an electric field, E , and a magnetic field, B .
Q(v x B)
B (into the page)
x
x
x
Ex
x
x
x
x
Q
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x x
x x
xvx
x x
x
x
x
x
Q experiences
both electric
and magnetic
forces:
Screen with hole
Now v ⊥ B so
⇒
⇒
⇒
⇒
x
x
x
x
Q v × B = QvB
the two forces balance exactly if QvB = QE
the particle is not deflected if v = E/B
only particles for which v = E/B pass through the hole
the output beam is of uniform velocity (mono-energetic).
Q
QE
EMF 2005
Handout 7: The Magnetic Field
10
The Hall effect
• Rectangular sample of conductor
or semiconductor carrying a
current I
V1
d
b Current I
Top
• Magnetic field B applied
perpendicular to the
direction of the current
d
B
• Current could be carried by either
a
Bottom
B
v
electrons
V
2
-e
+e
or holes
v
• Voltmeter between top and bottom sides measures ∆V = V1 - V2
If I is carried by electrons:
If I is carried by holes:
V1
V1
vxB
v
∆V
-e
~ B (outwards)
+e
~
v
B (outwards)
F = -e(v x B)
F = e(v x B)
V2
V2
Build-up of negative charge
on the bottom side
V1 > V2
⇒
∆V
Build-up of positive charge
on the bottom side
∆V is positive
V2 > V1
∆V is negative
we can find out whether the current is carried by electrons
(n-type semiconductor) or holes (p-type semiconductor).
Charge build-up → electric field,
E , between top and bottom sides
b
E opposes further charge build-up
E
⇒ Equilibrium established when eE = evB
+ + + + + + + + + + +
⇒ e (∆V/b) = evB
⇒
∆V = vBb
eE
- - - - - - - - - - - - - -
e(v x B)
(b = distance between top and
bottom sides)
EMF 2005
Handout 7: The Magnetic Field
We can relate the speed, v, to
the current, I:
11
length vdt
Let n = no. of carriers per unit
volume (each with charge e)
b
a
Consider the amount of charge,
dQ, crossing area ab in time dt:
dQ = amount of charge in this volume
So
dQ = ne(ab)(vdt)
So
v =
I
neab
⇒
⇒
∆V =
IB
nea
I = dQ/dt = neabv
∆V is the HALL POTENTIAL
By measuring the sign and magnitude of ∆V we can :
- Find n if B is known - i.e., investigate the sign and number density of
the charge carriers in a sample of material
- Or find B if n is known - i.e., measure an unknown magnetic field with
a HALL PROBE