UCSD Physics 2B
Summer Session
Unit 4 Lecture Notes
Chapter 29
Section 1
The Magnetic Field
Introduction
We begin by studying how magnetic fields affect charged particles and
afterwards how magnetic fields themselves are created. Natural magnets
called “Lodestones” were known centuries ago and, indeed, were used by
early sailors as an aid to navigation by aligning themselves with the earth’s
magnetic field. Not only did they attract certain metal objects but it was
found that iron articles could be “magnetized” and would either attract or
repel other such objects depending upon their orientation.
A few experimental facts:
1. Magnets have two “poles” – North and South.
2. Unlike poles attract each other, like poles repel.
3. If you break a magnet, each piece becomes a new magnet with its own
North and South. In other words, there are no magnetic monopoles.
Recall electric field lines begin on positive and end on negative
charges. But magnetic field lines never end – they loop right through.
4. The magnetic field lines of a bar magnet look similar to the electric
field lines of an electric dipole but, again, the lines loop in a
continuous direction.
5. Magnets attract ferromagnetic objects (iron and some other elements)
but other elements, such as aluminum, are not attracted.
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Section 2
Magnetic Force on a Moving Charge
Magnetic fields exert force on charged particles:
1. in proportion to the magnitude and sign of their electric charge q
2. in proportion to their velocity v – stationary charges feel no force
3. in a direction perpendicular to the velocity v and the field B
F = q v⊗B
Magnetic Force on moving charge
This is a vector cross-product which can be calculated in one of two ways:
Right Hand Rule
First use the right-hand rule (RHR) to determine the direction of the force.
Point the fingers in the direction of the velocity v and curl them towards the
magnetic field B. The thumb points in the direction of the force F.
The magnitude of the force is then found by the formula
F = q v B sin θ
Magnitude of Cross-product
where θ is the angle of rotation from v to B.
Matrix Method
The vector cross-product can be represented by the determinant of a matrix.
The first row is composed of the three cardinal unit vectors, the second row
by the component of the first vector and the third by the components of the
second vector. In particular, we have here
ˆi
ˆj
v ⊗ B = det vx
vy
vz = ( v y Bz − vz By ) ˆi − ( vx Bz − vz Bx ) ˆj + ( vx By − v y Bx ) kˆ
Bx
By
Bz
kˆ
CAVEAT: The vector cross-product is NOT commutative, in fact,
v ⊗ B = −B ⊗ v
2
EXAMPLE
Imagine that you are looking at an aerial map as shown below. The magnetic
field B points to the North and a particle with positive charge q is moving
with velocity v to the East. What is the magnitude and direction of the force?
N
j
B
i
W
v
E
k
S
Apply the RHR and the thumb points out of the page as shown. Since the
velocity and magnetic field vectors are perpendicular, the sine of 90º is 1.
Hence, the force is just F = qvB in the “up” direction (“out” of the page)
Cyclotron Motion
Consider a particle with electric charge q moving in a region of a uniform
magnetic field B and with velocity v perpendicular to B. Since the force is
always perpendicular to the velocity, the particle must move in a circular
orbit as it would under the influence of a “central force” which acts along
the radius of the orbit, continuously pulling the particle towards the center.
The radius of such an orbit can be calculated by equating the magnetic
(central) force with the centripetal force of the particle’s circular motion:
mv 2
FB = qvB
FC =
r
mv 2
FB = FC
⇒
qvB =
r
mv
rC =
Cyclotron Radius
qB
This is called the Cyclotron Radius. Note that, for a given charged particle
with fixed charge-to-mass ratio (a proton, for example), the radius is
proportional to the velocity and inversely proportional to the field strength.
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How long does it take for the particle to complete one orbit? The period
must be the distance traveled divided by the particle’s velocity:
C 2π r ⎛ 2π ⎞ ⎛ mv ⎞
=
=⎜
⎟
⎟⎜
v
v
⎝ v ⎠ ⎝ qB ⎠
2π m
T=
Cyclotron Period
qB
T=
This time is called the Cyclotron Period. Note that it is independent of the
particle’s velocity. For a given charged particle then, the period is
determined only by the magnetic field strength.
Typical variations on this theme are problems where the charged particle has
a velocity component parallel to the magnetic field. Since any vector can be
decomposed into components in different directions (see ‘Linear Algebra’)
an arbitrary velocity vector will have some component in the direction of B.
In these cases, we treat each component separately. In a magnetic field
region the perpendicular component “traps” the particle in a circular orbit,
while the parallel component is left undisturbed. The resulting path is a
helix, which looks like a long spring.
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Section 4
Magnetic Force on a Current
Suppose we have a wire which is electrically neutral but which carries an
electric current I. In other words, there are an equal number of positive and
negative charges in the wire, but some of them are moving. The wire
constrains the paths of the moving charges but, since the charges themselves
feel the magnetic force, there must be a force on the wire itself. Let’s do this
without a lot of hand-waving and, instead, just use dimensional analysis.
We begin by writing just the magnitude of the force equation without
worrying about the angle
F =q v B
But the velocity v is just some distance l divided by time t so that
F =q
l
B
t
If we move the time in the denominator under the charge instead, we get
F=
q
l B=I l B
t
since current is just charge divided by time. This is the force on a wire of
length l carrying current I in a perpendicular magnetic field B.
In order to generalize this relation, we remember that a wire can follow a
complex path through space and that the differential element dl may change
direction. Hence, for an arbitrary wire, we have
F = ∫ dF = ∫ I dl ⊗ B
Force on wire carrying a current
which must be integrated over the entire path of the wire.
Note that the current I must be constant everywhere along the wire by
conservation of charge. However, the magnetic field might vary at the
different points in space though which the wire passes.
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Section 5
Torque & Force on a Current Loop
Consider a square loop of side s in a region of uniform magnetic field B as
shown below. A current I flows clockwise around the loop.
B
j
i
s
k
I
Let’s look at the force on each side independently. Since the current through
the left side is parallel to the magnetic field B, and the right side anti-parallel
each force is zero. The left and right sides are perpendicular to B, so the
magnitude of the force F = IsB since sin(90º) = 1. Using the coordinate
system in the diagram, we apply the Right Hand Rule, and see that
F = IsB +kˆ out of the page
TOP
FBOT
( )
= IsB ( −kˆ )
into the page
Clearly, the net force on the loop is zero so that no translation takes place.
However, there is a torque and, if the loop is free, it will rotate. Recall the
definition of torque from mechanics
τ = r⊗F
force times the moment arm
If we draw a horizontal bisector through the loop (dotted line) r = s 2 and
since there are two applied torques – top and bottom – the total torque is
⎛s
⎞
τ = 2 ⎜ IsB ⎟ = Is 2 B
⎝2
⎠
If we define the magnetic dipole moment of the loop to be μ B = IA , then
τ = μB ⊗ B
Torque on a magnetic dipole
Note that the total force on a dipole in a uniform magnetic field is zero!
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Chapter 30
Section 1
Sources of the Magnetic Field
Biot-Savart Law
Mankind has been able to “trap” electric charge in Leyden jars for some time
but, until someone figured out how to maintain a constant current, the
intimate connection between electricity and magnetism was unknown.
The fact that one creates the other was formalized by two malodorous
French gymnasts, Biot and Savart. Their Law, like that of Coulomb (yet
another wacky Frenchman) defines the magnetic field created at some
distant point by a configuration of electric current
B=
μ0
4π
I dl ⊗ rˆ
∫ r2
Biot-Savart Law
This integral is performed over the entire region where current exists. Note
that it not only contains the unit vector r̂ but a nasty cross-product inside
the integral which must be performed for every differential line segment dl .
This integral is so nasty that it can readily be done only in a couple of
illustrative cases – the infinite wire and the circular current loop. Please
review these examples in your text as the pictures are better than I can draw.
Two important and useful results are –
B=
μ0 I
2π r
Field at a distance r from an infinite wire
The right thumb points in the direction of the current and the fingers
curl in the direction of the field which wraps around the straight wire
B=
μ0 I
2r
Field at center of circular loop of radius r
The right-hand fingers curl in the direction of the current and the
thumb points in the direction of the field at the center.
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Section 2
Force Between Two Current Carrying Wires
Wires were found to deflect magnetic compass needles when carrying
electric current. By moving the compass around the wire, the field thus
created was mapped out and found to encircle the wire. If the right hand’s
thumb points in the direction of the current, the fingers will curl around the
wire indicating the direction of the “curling” magnetic field.
Consider two infinite wires, each carrying current I and separated by
distance r as shown below.
I1
r
I2
The field on the bottom wire (2) created by the top wire (1) is
B1 =
μ0 I1
2π r
Since this is the field everywhere along the bottom wire, the magnitude of
the force per unit length on the bottom wire by the field of the top wire is
`
F μ0 I1 I 2
=
L
2π r
F2
⎛μ I ⎞
= I 2 B1 = I 2 ⎜ 0 1 ⎟
L
⎝ 2π r ⎠
Force between two wire currents
Note that the direction of the force depends on the directions of the
currents. The field on the bottom wire, according to the diagram and the
RHR, is into the page. Since I L ⊗ B is up or towards the top wire, we
conclude that
when the currents are in the same direction, the force is attractive
when the currents are opposite in direction, the force is repulsive
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Section 4
Ampere’s Law
Recall what a pain it was to calculate to Electric Field from a continuous
distribution of charge? But thanks to Gauss’s Law, we made quick work of
this chore in those cases where we had symmetry on our side. Gauss’s Law
said that the total flux passing out a closed surface is proportional to the total
charge enclosed.
The same can be done for the magnetic field and Ampere’s Law is the
magnetic analog of Gauss’s Law. Ampere’s Law states that the line integral
of the magnetic field around a closed loop is proportional to the flux of
current enclosed.
∫ B • dl
= μ0 I enclosed
Ampere’s Law
Consider a wire carrying current I coming out of the page as shown below.
To find the magnetic field at a distance r from the axis, we draw a circular
Amperean Loop around the wire which lies in the plane of the page.
r
I
We do this after using the RHR to determine the shape and direction of the
field due to the given current. We take the loop is counter-clockwise because
that’s the way our fingers curl when pointing the thumb in the current
direction – that is, out of the page. Since the loop is everywhere equidistant
from the current axis, it must be a constant along the loop and, further, it
must point in the direction of dl. Now we can do the same thing we did
before with Gauss’s Law, namely, collapse the dot product, take the constant
field outside, and get a simple line integral, which is just the loop length:
∫ B ( r ) • d l = ∫ B ( r ) d l = B ( r ) ∫ d l = B ( r ) 2π r = μ I
0 enclosed
⇒
B (r ) =
μ0 I
2π r
We obtain the same result we would have from Biot-Savart, only quicker.
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Section 5
Solenoids & Toroids
We can generalize this last equation (just like we did with Gauss’s Law)
turning an integral equation into an algebraic one – but must exercise care!
B=
μ0 I enclosed
Simplified Ampere’s Law
LLOOP
It’s very important to remember that we can use only that length of the loop
where the field actually exists!
Consider an infinite coil of wire on a common axis, called a solenoid, whose
cross-section is shown above. Each vertical pair represents one loop of the
coil and the pink dots show current out of the page, and blue crosses current
into the page. By using your right thumb to point in the current direction,
convince yourself that the red arrow represents the magnetic field direction.
There are several arguments which we’ll discuss in lecture to support the
claim that the field must be zero outside the solenoid. To find the field
inside, make a loop of length L as shown, and note that only the bottom leg
of the loop counts. The current enclosed is the wire current I times the
number of loops enclosed. Then you can apply Ampere’s Law:
B=
μ0 I enclosed
LLOOP
=
μ0 I wire N
B = μ0 In
L
= μ0 In
where n = turns/meter
Field Inside Solenoid
A toroid is a closed solenoid bent around on itself – it has no ends. You
should review the method of solution so you can see another example of
applying Ampere’s Law. Remember the keys to using this method:
1. Know what the field looks like before constructing the loop (RHR)
2. Only the current enclosed matters.
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Chapter 31
Section 1
Electromagnetic Induction
Induced Currents
We saw that an electric current produces a magnetic field. Then why
shouldn’t a magnetic field produce an electric current? Seems fair and
reasonable, doesn’t it? Well, it just doesn’t happen – hey, that’s nature.
But a changing magnetic field does drive a current. One way to see the
sense of this is to consider an old concept from Chemistry called Le
Chatelier’s Principle which gives Nature a mind of its own, and is based on
the idea that Nature likes balance, and abhors change. In other words, Nature
figures out a way to oppose the change.
Section 2
Faraday’s Law
We begin by defining Magnetic Flux as we did with Electric Flux
Φ B = ∫ B i dA
Magnetic Flux
Consider a conducting ring threaded by magnetic field lines. If the flux
inside the ring changes in time, Nature resists this change by trying to make
its own magnetic field in the opposite direction. It does this by creating a
circulating electric field or EMF (electromotive force) to drive free charges
into a current loop, thus making a magnet – in the opposite direction. This is
Lenz’ Law and accounts for the minus sign in the equation below
EMF =
∫ E • dl = −
dΦB
dt
Faraday’s Law
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There are three basic ways to change the magnetic flux through a loop:
1. Change the external magnetic field strength or move the loop itself to
a place where the field is different
2. Change the orientation (angle) of the loop and thus change the number
of field lines passing through – recall the dot-product depends on the
cosine of the angle θ
3. Change the size of the loop
Example
Suppose we have a square wire loop of side s = 10.0 cm and resistance
R = 456 Ω in a magnetic field of strength B = 25.0 × 10−2 T which makes
an angle θ = 20° with the loop normal as shown below.
Now suppose the magnetic field decays linearly to zero in a time t = 22 μ s .
Let’s calculate the emf. To do so, we need the change in flux, not the field:
−2
−2
ΔΦ Δ ( BA cos θ ) ( 25.0 × 10 T )(10 × 10 m ) ( cos 20° )
emf =
=
=
= 107 V
t
22.0 × 10 −6 s
Δt
2
The current in the loop is the same as if the loop were a resistor connected
to a battery whose voltage equals the emf, so apply Ohm’s Law:
I=
V 107V
=
= 234mA
R 456Ω
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