Physics 2 for Electrical Engineering Ben Gurion University of the Negev , www.bgu.ac.il/atomchip

Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Grader: Gady Afek
Week 6. The magnetic field and the Lorentz force – Magnetic fields
• magnetic force on a moving charge • Lorentz force • charges and
currents in a uniform B field • torque on a current loop
Sources: Halliday, Resnick and Krane, 5th Edition, Chap. 32.
Magnetic fields
A combination of bar magnets and iron filings matches the
pretty diagram shown on the right…
Magnetic fields
A combination of bar magnets and iron filings matches the
pretty diagram shown on the right…
Magnetic fields
Combinations of bar magnets and iron filings match these
pretty diagrams…
Magnetic fields
Combinations of bar magnets and iron filings match these
pretty diagrams… But these diagrams show to configurations
of the electric field E. Are there identical configurations of the
magnetic field B?
Magnetic fields
Yes, there are! Except for one big difference:
Magnetic fields
Yes, there are! Except for one big difference: There are no
magnetic charges – no magnetic “monopoles” – only magnetic
dipoles.
Magnetic fields
Yes, there are! Except for one big difference: There are no
magnetic charges – no magnetic “monopoles” – only magnetic
dipoles.
This difference is remarkable and even surprising, for at least
two reasons:
1. Magnetic monopoles would make electromagnetism more
symmetric – electricity and magnetism would be dual.
Magnetic fields
Yes, there are! Except for one big difference: There are no
magnetic charges – no magnetic “monopoles” – only magnetic
dipoles.
This difference is remarkable and even surprising, for at least
two reasons:
1. Magnetic monopoles would make electromagnetism more
symmetric – electricity and magnetism would be dual.
2. Theories that “unify” electromagnetism with the weak and
strong nuclear forces predict magnetic monopoles. But recent
searches for these magnetic monopoles have not found any.
The magnetic force on a moving charge
The relation between the electric field E and the electric force
FE on a point charge q is a simple one: FE = q E.
The magnetic force on a moving charge
The relation between the magnetic field B and the magnetic
force FB on a point charge q is more complicated: FB = q v×B ,
where v is the velocity of the moving charge.
The magnetic force on a moving charge
The relation between the magnetic field B and the magnetic
force FB on a point charge q is more complicated: FB = q v×B ,
where v is the velocity of the moving charge.
• The force is proportional to q, including the sign of q.
• The force is proportional to B.
• The force is proportional to v.
• When B and v are parallel, the force vanishes.
• When B and v are not parallel, the force is perpendicular
to both of them, according to the right-hand rule:
Right-hand rule (for a positive charge q):
FB
v
B
The magnetic force on a moving charge
The relation between the magnetic field B and the magnetic
force FB on a point charge q is more complicated: FB = q v×B ,
where v is the velocity of the moving charge.
• The force is proportional to q, including the sign of q.
• The force is proportional to B.
• The force is proportional to v.
• When B and v are parallel, the force vanishes.
• When B and v are not parallel, the force is perpendicular
to both of them, according to the right-hand rule.
• The force is proportional to the sine of the v→B angle.
The magnetic force on a moving charge
Example 1: How much work is done by the magnetic force FB
on the point charge q?
The magnetic force on a moving charge
Example 2: What are the units of the magnetic field B?
The magnetic force on a moving charge
Example 2: What are the units of the magnetic field B?
Answer: From the equation FB = q v×B we infer that the units
of B are (force) / (charge · meters per second); this unit is
known as the tesla (T):
N/C
N
T

.
m/s A  m
The magnetic force on a moving charge
Example 3: The white arc in this picture is light showing a
beam of electrons in a uniform magnetic field. (The electrons
collide with a dilute gas of atoms, which then radiate light.)
Deduce the ratio e/m – where m is the mass of the electron –
from the radius r of the arc and the magnitudes B and v.
The magnetic force on a moving charge
Example 3: The white arc in this picture is light showing a
beam of electrons in a uniform magnetic field. (The electrons
collide with a dilute gas of atoms, which then radiate light.)
Deduce the ratio e/m – where m is the mass of the electron –
from the radius r of the arc and the magnitudes B and v.
Answer: To move at speed v in a circular orbit of radius r, an
electron must accelerate towards the center of the circle with
acceleration a = v2/r. This acceleration must equal the force FB
on the electron divided by its mass m, hence a = FB/m = evB/m
and we have v2/r = evB/m, therefore e/m = v/Br.
The magnetic force on a moving charge
Example 4: We have just seen that a point charge in a constant
magnetic field moves in a circle. (More generally, it can move
in a helix.) What is the angular frequency ω of its motion?
The magnetic force on a moving charge
Example 4: We have just seen that a point charge in a constant
magnetic field moves in a circle. (More generally, it can move
in a helix.) What is the angular frequency ω of its motion?
Answer: The angular frequency is ω = v/r and we already
obtained v2/r = qvB/m (for a particle of charge q and mass m);
hence ω = qB/m. It is called the cyclotron frequency and it is
independent of r and v. A cyclotron exploits this independence
to accelerate many charged particles with different kinetic
energies at the same time, via an electric field alternating with
angular frequency ω. The kinetic energy of a particle increases
in the electric field, as do v and r, but ω is unchanged.
The magnetic force on a moving charge
A cyclotron exploits this independence to accelerate many
charged particles with different kinetic energies at the same
time, via an electric field alternating with angular frequency ω.
The kinetic energy of a particle increases in the electric field,
as do v and r, but ω is unchanged.
Slow particles in
AC current
“D1”
“D2”
B
Fast particles out
bottom of magnet
The magnetic force on a moving charge
Example 5: Confinement of a charged particle in a “magnetic
bottle”:
Path of charged particle
The magnetic field cannot change the speed of the charged
particle, but can change its direction.
The Lorentz force
We can combine FB and FE into the Lorentz force FEM , which
is the total electromagnetic force on a point charge q:
FEM = q ( E + v × B )
.
The Lorentz force
Example 1 (Hall effect): In 1879, E. Hall discovered that a
magnetic field B normal to this conducting bar induces a
potential ΔVH = vdrift Bd that is perpendicular to the current and
to B. Why?
z
t
y
B
I
d
–
ΔVH
FB
FB
+
B
I
x
The Lorentz force
The magnetic field deflects electrons up, where they collect
and produce an upward electric field E = E ẑ . The electron
density there levels off when vdrift B = E. Since E = ΔVH/d, the
Hall potential is ΔVH = vdrift Bd. The direction of ΔVH shows
that the charge carriers are indeed negatively charged.
z
t
y
B
I
d
–
ΔVH
FB
FB
+
B
I
x
The Lorentz force
From Slide 33 of Lecture 5 we have vdrift = I/neA. Here A = td,
so vdrift = I/netd and ΔVH = vdrift Bd = IB/net and n = IB/et ΔVH .
Thus we can infer n from measurements of I, B, e, t and ΔVH.
z
t
y
B
I
d
–
ΔVH
FB
FB
+
B
I
x
The Lorentz force
Example 2: Crossed electric and magnetic fields can serve as a
velocity selector. We have just seen, in the Hall effect, that the
electric and magnetic forces on a point charge balance when
the charge moves at the speed v such that E = vB. Only at this
speed will charges move straight; at other speeds, the Lorentz
force will deflect them.
B
Source
E
Slit
Halliday, Resnick and Krane, 5th Edition, Chap. 32, MC 3:
An electron is released at rest in a region of crossed uniform
electric and magnetic fields. Which path best represents its
motion after its release?
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Charges and currents in a uniform B field
Current in a wire is the motion of charges. Thus the magnetic
force on a current-carrying wire is due to magnetic forces on
all the charges in the wire.
Charges and currents in a uniform B field
Current in a wire is the motion of charges. Thus the magnetic
force on a current-carrying wire is due to magnetic forces on
all the charges in the wire.
Convention: × B into page · B out of page
B
I=0
B
I
B
I
Charges and currents in a uniform B field
Let’s consider an arbitrarily short element Δs of a current-
carrying wire with cross-sectional area A. Its volume is
(A)(Δs), so it contains n (A)(Δs) conduction electrons moving
at average velocity vdrift. Hence the magnetic force dFB on the
wire element is dFB = evdrift nA (Δs) × B, which we can write as
dFB = I Δs × B since I = evdrift nA.
The force FB on the whole wire is then an integral:


FB  dFB  I ds  B .
Charges and currents in a uniform B field
The formula for the force FB on the whole wire is especially
simple when B is uniform (constant over space) because then
we can take B out of the integral:


 
FB  dFB  I ds  B  I ds  B .
We consider two cases:
Charges and currents in a uniform B field
The formula for the force FB on the whole wire is especially
simple when B is uniform (constant over space) because then
we can take B out of the integral:

 

FB  dFB  I ds  B  I ds  B .
We consider two cases:
ds
I
B
B
L'
I
ds
Charges and currents in a uniform B field
If the integral is open, then it is just a vector sum over line
elements and equals the vector L' connecting the initial and
final points. Then FB = I L' × B.
I
ds
B
B
L'
I
ds
Charges and currents in a uniform B field
If the integral is open, then it is just a vector sum over line
elements and equals the vector L' connecting the initial and
final points. Then FB = I L' × B.
If the integral is closed, then L' vanishes and so does FB !
I
ds
B
B
L'
I
ds
Charges and currents in a uniform B field
If the integral is open, then it is just a vector sum over line
elements and equals the vector L' connecting the initial and
final points. Then FB = I L' × B.
If the integral is closed, then L' vanishes and so does FB !
I
In a uniform magnetic
B
ds
field, the net magnetic
L'
force on any closed
current loop is zero.I
B
ds
Charges and currents in a uniform B field
Example 1: Rank the magnitude of the magnetic force on these
four wires. (B and I are identical.)
Charges and currents in a uniform B field
Example 2: A magnetic field can levitate a current-carrying
wire. If L is the length of the wire, I is the current in the wire
and m is its mass, what should be the strength B of the
magnetic field (if we neglect all other forces)?
FB
I
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Charges and currents in a uniform B field
Example 2: A magnetic field can levitate a current-carrying
wire. If L is the length of the wire, I is the current in the wire
and m is its mass, what should be the strength B of the
magnetic field (if we neglect all other forces)?
Answer: mg = ILB, so B = mg/IL.
FB
I
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Torque on a current loop
A uniform magnetic field does not exert a force on a closed
current loop, but it can exert a net torque!
Here is a top view of a
rectangular current loop
lying in the plane of B.
Sides 1 and 3 do not
contribute; Sides 2 and
4 each contribute (IaB)
× (b/2) to the torque τ, so
τ = IBab = IBA, where
A = ab is the area of the
rectangle.
I
B
1
2
4
3
b
a
Torque on a current loop
A uniform magnetic field does not exert a force on a closed
current loop, but it can exert a net torque!
Here is a side view of the
current loop, still lying in
the plane of B.
The torque is τ = IBab
but only for the instant
that the current loop is
parallel to B.
IBa
b/2
2
3
4
B
IBa
Torque on a current loop
A uniform magnetic field does not exert a force on a closed
current loop, but it can exert a net torque!
Here is a side view of the
current loop, now making
an angle 90° – θ with the
plane of B. The torque now
is τ = IBa[b cos (90° – θ)]
= IBa[b sin θ] = IBA sin θ.
IBa
2
θ
3
θ
B
4
IBa
Torque on a current loop
A uniform magnetic field does not exert a force on a closed
current loop, but it can exert a net torque!
Here is a side view of the
current loop, now making
an angle 90° – θ with the
plane of B. The torque now
is τ = IBa[b cos (90° – θ)]
= IBa[b sin θ] = IBA sin θ.
There are now also forces
on 1 and 3 but they do
not contribute to the torque.
IBa
2
θ
3
θ
B
4
IBa
Torque on a current loop
Whenever 2 moves to the right of 4 , the torque switches
direction. The area vector A always tends to line up with B.
Defining the magnetic
dipole moment of the
current loop to be μ = IA,
we can write τ = μ × B.
IBa
A
2
θ
3
θ
B
4
IBa
Torque on a current loop
Whenever 2 moves to the right of 4 , the torque switches
direction. The area vector A always tends to line up with B.
If we integrate τ dθ' starting
from θ' = 0, we get the work
due to the magnetic torque:

WB    d '
IBa
A
2

  IBA sin  ' d '
 IBA cos  μ  B ,
so the potential energy U of a magnetic
dipole μ in a field B is U = –μ · B.
θ
3
θ
B
4
IBa
Halliday, Resnick and Krane, 5th Edition, Chap. 32, Prob. 10:
In Bohr’s model of the hydrogen atom, the electron moves in a
circle of radius r around the proton. Suppose the atom, with
the proton at rest, is placed in a magnetic field B perpendicular
to the plane of the electron motion. (a) If the electron moves
counterclockwise around B (seen from above), will its angular
frequency ω increase or decrease? (b) What if the electron
moves clockwise?
B
–
Halliday, Resnick and Krane, 5th Edition, Chap. 32, Prob. 10:
Answer: (a) For a circular orbit, the centripetal force must
equal ma = m(v2/r) = mω2r. The magnetic force on the
electron increases the centripetal force, so mω2r must increase.
But the angular momentum mr2ω cannot change (because the
forces are centripetal). Writing mω2r as
m 2 r   3 / 2 m1/ 2 mr 2 ,
we see that ω must increase.
B
–
Halliday, Resnick and Krane, 5th Edition, Chap. 32, Prob. 10:
Answer: (b) For a circular orbit, the centripetal force must
equal ma = m(v2/r) = mω2r. The magnetic force on the
electron decreases the centripetal force, so mω2r must decrease.
But the angular momentum mr2ω cannot change (because the
forces are centripetal). Writing mω2r as
m 2 r   3 / 2 m1/ 2 mr 2 ,
we see that ω must decrease.
B
–
Halliday, Resnick and Krane, 5th Edition, Chap. 32, Prob. 15:
A staple-shaped wire of mass m and width L sits in a uniform
magnetic field B, with its two ends in two beakers of mercury.
A pulse of charge q passing through the wire causes the wire
to jump to a height h. Given B = 0.12 T, m = 13 g, L = 20 cm
and h = 3.1 m, calculate q.
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Halliday, Resnick and Krane, 5th Edition, Chap. 32, Prob. 15:
Answer: The magnetic force FB(t) on the wire equals I(t)LB.
The wire acquires momentum p = ∫FB(t) dt = ∫ I(t)LB dt = qLB
and kinetic energy p2/2m = mgh; solving for q, we obtain
m
q
2 gh  4.2 C .
LB
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