Investigating the Lorentz Force Effect in Reducing Calcite ...
L6 Lorentz Force
Transcript of L6 Lorentz Force
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Ben Gurion University of the Negevwww.bgu.ac.il/atomchip
Week 6. The magnetic field and the Lorentz force Magnetic fields magnetic force on a moving charge Lorentz force charges a
currents in a uniform B field
torque on a current loopSource: Halliday, Resnick and Krane, 5 th Edition, Chap. 32.
Lecturer: Daniel Rohrlich
Teaching Assistants: Oren Rosenblatt, Shai Inbar
Physics 2B for Materials and Structural Engineering
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Magnetic fields
A combination of bar magnets and iron filings matches the
pretty diagram shown on the right
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Magnetic fields
A combination of bar magnets and iron filings matches the
pretty diagram shown on the right
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Magnetic fields
Combinations of bar magnets and iron filings match these
pretty diagrams
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Magnetic fields
Combinations of bar magnets and iron filings match these
pretty diagrams But these diagrams show to configurationsof the electric field E . Are there identical configurations of themagnetic field B?
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Magnetic fields
Yes, there are! Except for one big difference:
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Magnetic fields
Yes, there are! Except for one big difference: There are no
magnetic charges no magnetic monopoles only magneticdipoles.
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Magnetic fields
Yes, there are! Except for one big difference: There are no
magnetic charges no magnetic monopoles only magneticdipoles.
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 .
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Magnetic fields
Yes, there are! Except for one big difference: There are no
magnetic charges no magnetic monopoles only magneticdipoles.
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 andstrong nuclear forces predict magnetic monopoles. But recentsearches for these magnetic monopoles have not found any.
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The magnetic force on a moving charge
The relation between the electric field E and the electric force
F E on a point charge q is a simple one: F E = q E .
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The magnetic force on a moving charge
The relation between the magnetic field B and the magnetic
force F B on a point charge q is more complicated: F B = q v B ,where v is the velocity of the moving charge.
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The magnetic force on a moving charge
The relation between the magnetic field B and the magnetic
force F B on a point charge q is more complicated: F B = 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 :
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Right-hand rule (for a positive charge q):
F B
v B
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The magnetic force on a moving charge
The relation between the magnetic field B and the magnetic
force F B on a point charge q is more complicated: F B = 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 vB angle.
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The magnetic force on a moving charge
Example 1: How much work is done by the magnetic force F B
on the point charge q?
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The magnetic force on a moving charge
Example 2: What are the units of the magnetic field B?
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The magnetic force on a moving charge
Example 2: What are the units of the magnetic field B?
Answer: From the equation F B = q v B we infer that the unitsof B are (force) / (charge meters per second); this unit isknown as the tesla (T):
. mA
N
m/s
N/C T
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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 electronscollide 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.
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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 electronscollide 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 , anelectron must accelerate towards the center of the circle withacceleration a = v2 / r . This acceleration must equal the force F B on the electron divided by its mass m, hence a = F B / m = evB / m and we have v2 / r = evB / m, therefore e / m = v / Br .
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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 movein a helix.) What is the angular frequency of its motion?
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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 movein a helix.) What is the angular frequency of its motion?
Answer: The angular frequency is = v / r and we alreadyobtained v2 / r = qvB / m (for a particle of charge q and mass m);hence = qB / m. It is called the cyclotron frequency and it isindependent of r and v. A cyclotron exploits this independenceto accelerate many charged particles with different kineticenergies at the same time, via an electric field alternating withangular frequency . The kinetic energy of a particle increasesin the electric field, as do v and r , but is unchanged.
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The magnetic force on a moving charge
A cyclotron exploits this independence to accelerate many
charged particles with different kinetic energies at the sametime, via an electric field alternating with angular frequency .The kinetic energy of a particle increases in the electric field,as do v and r , but is unchanged.
AC current
Fast particles out
Slow particles in
bottom of magnet
D1 D
2
B
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The magnetic force on a moving charge
Example 5 : Confinement of a charged particle in a magnetic
bottle:
The magnetic field cannot change the speed of the charged
particle, but can change its direction.
Path of charged particle
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The Lorentz force
We can combine F B and F E into the Lorentz force F EM , which
is the total electromagnetic force on a point charge q:
F EM
= q ( E + v B ) .
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The Lorentz force
Example 1 (Hall effect): In 1879, E. Hall discovered that a
magnetic field B normal to this conducting bar induces apotential V H = vdrift Bd that is perpendicular to the current andto B. Why?
I
z
x
y
+
F B
F B
V H
B
B
t
I
d
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I
z
x
y
+
F B
F B
V H
B
B
t
I
d
The Lorentz force
The magnetic field deflects electrons up, where they collect
and produce an upward electric field E = E . The electrondensity there levels off when vdrift B = E . Since E = V H / d , theHall potential is V H = vdrift Bd . The direction of V H showsthat the charge carriers are indeed negatively charged.
z
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I
z
x
y
+
F B
F B
V H
B
B
t
I
d
The Lorentz force
From Slide 33 of Lecture 5 we have vdrift = I / neA. Here A = td ,
so vdrift = I / netd and V H = vdrift Bd = IB / net and n = IB / et V H .Thus we can infer n from measurements of I , B, e, t and V H.
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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 theelectric and magnetic forces on a point charge balance whenthe charge moves at the speed v such that E = vB. Only at thisspeed will charges move straight; at other speeds, the Lorentzforce will deflect them.
E
B
Source
Slit
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Halliday, Resnick and Krane, 5 th 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 itsmotion after its release?
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
B
E
(a)
(b)
(c)
(d)
(e)
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fieldBCharges and currents in a uniform
Current in a wire is the motion of charges. Thus the magnetic
force on a current-carrying wire is due to magnetic forces onall the charges in the wire.
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fieldBCharges and currents in a uniform
Current in a wire is the motion of charges. Thus the magnetic
force on a current-carrying wire is due to magnetic forces onall the charges in the wire.
Convention: B into page B out of page
I I
B BB
I = 0
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fieldBCharges and currents in a uniform
Let s consider an arbitrarily short element s of a current-carrying wire with cross-sectional area A. Its volume is
( A)(s), so it contains n ( A)(s) conduction electrons movingat average velocity vdrift . Hence the magnetic force d F B on the
wire element is d F B = evdrift nA (s) B, which we can write asd F B = I s B since I = evdrift nA.
The force F B on the whole wire is then an integral:
. BsFF d I d B B
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fieldBCharges and currents in a uniform
The formula for the force F B on the whole wire is especially
simple when B is uniform (constant over space) because thenwe can take B out of the integral:
We consider two cases:
. BsBsFF d I d I d B B
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BBd s
d s
L'
I
I
fieldBCharges and currents in a uniform
The formula for the force F B on the whole wire is especially
simple when B is uniform (constant over space) because thenwe can take B out of the integral:
We consider two cases:
. BsBsFF d I d I d B B
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fieldBCharges and currents in a uniform
If the integral is open, then it is just a vector sum over line
elements and equals the vector L' connecting the initial andfinal points. Then F B = I L' B.
BBd s
d s
L'
I
I
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fieldBCharges and currents in a uniform
If the integral is open, then it is just a vector sum over line
elements and equals the vector L' connecting the initial andfinal points. Then F B = I L' B.
If the integral is closed, then L' vanishes and so does F B !
BBd s
d s
L'
I
I
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I
BBd s
d s
L'
I
fieldBCharges and currents in a uniform
If the integral is open, then it is just a vector sum over line
elements and equals the vector L' connecting the initial andfinal points. Then F B = I L' B.
If the integral is closed, then L' vanishes and so does F B !
In a uniform magnetic
field, the net magnetic force on any closed current loop is zero.
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fieldBCharges and currents in a uniform
Example 1: Rank the magnitude of the magnetic force on these
four wires. ( B and I are identical.)
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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 wireand m is its mass, what should be the strength B of themagnetic field (if we neglect all other forces)?
I
B
B
F B
mg
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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 wireand m is its mass, what should be the strength B of themagnetic field (if we neglect all other forces)?
Answer: mg = ILB , so B = mg / IL.
I
B
B
F B
mg
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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 arectangular current looplying in the plane of B.Sides 1 and 3 do notcontribute; Sides 2 and4 each contribute ( IaB ) (b /2) to the torque , so = IBab = IBA, where
A = ab is the area of therectangle.
B
a
b
I
2
3
1
4
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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 side view of thecurrent loop, still lying inthe plane of B.
The torque is = IBabbut only for the instantthat the current loop is
parallel to B.
IBa
IBa
b /2
2 4
B
3
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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 side view of thecurrent loop, now makingan angle 90 with theplane of B. The torque nowis = IBa [b cos (90 )]= IBa [b sin ] = IBA sin .
2
4
B
IBa
IBa
3
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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 side view of thecurrent loop, now makingan angle 90 with theplane of B. The torque nowis = IBa [b cos (90 )]= IBa [b sin ] = IBA sin .
There are now also forceson and but they donot contribute to the torque.
1 3
2
4
B
IBa
IBa
3
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2
4
B
IBa
IBa
3
Torque on a current loop
Whenever moves to the right of , the torque switches
direction. The area vector A always tends to line up with B.
Defining the magneticdipole moment of thecurrent loop to be = I A ,we can write = B.
2 4
A
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Torque on a current loop
Whenever moves to the right of , the torque switches
direction. The area vector A always tends to line up with B.
If we integrate d ' startingfrom ' = 0, we get the work due to the magnetic torque:
so the potential energy U of a magneticdipole in a field B is U = B .
2
2
4
B
IBa
IBa
3
A
, cos
''sin
'
B
IBA
d IBA
d W B
4
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Halliday, Resnick and Krane, 5 th Edition, Chap. 32, Prob. 10:
In Bohrs model of the hydrogen atom, the electron moves in
circle of radius r around the proton. Suppose the atom, withthe proton at rest, is placed in a magnetic field B perpendicularto the plane of the electron motion. (a) If the electron movescounterclockwise around B (seen from above), will its angularfrequency increase or decrease? (b) What if the electronmoves clockwise?
B
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Halliday, Resnick and Krane, 5 th Edition, Chap. 32, Prob. 10:
Answer: (a) For a circular orbit, the centripetal force must
equal ma = m(v2
/ r ) = m2
r . The magnetic force on theelectron increases the centripetal force, so m 2r must increase.But the angular momentum mr 2 cannot change (because theforces are centripetal). Writing m 2r as
we see that must increase.
B
, 22 / 12 / 32 mr mr m
h
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Halliday, Resnick and Krane, 5 th Edition, Chap. 32, Prob. 10:
Answer: (b) For a circular orbit, the centripetal force must
equal ma = m(v2
/ r ) = m2
r . The magnetic force on theelectron decreases the centripetal force, so m 2r must decrease.But the angular momentum mr 2 cannot change (because theforces are centripetal). Writing m 2r as
we see that must decrease., 22 / 12 / 32 mr mr m
B
h
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Halliday, Resnick and Krane, 5 th 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 wireto jump to a height h. Given B = 0.12 T, m = 13 g, L = 20 cmand h = 3.1 m, calculate q.
B B
L
h
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Halliday, Resnick and Krane, 5 th Edition, Chap. 32, Prob. 15:
Answer: The magnetic force F B(t ) on the wire equals I (t ) LB.
The wire acquires momentum p = F B(t ) dt = I (t ) LB dt = qLB and kinetic energy p2 /2m = mgh ; solving for q, we obtain
B B
L
. C4.22 gh LB
mq