l6 lorentz force

7/31/2019 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
http://www.bgu.ac.il/atomchiphttp://www.bgu.ac.il/atomchip


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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



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



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 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 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 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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Example 1: How much work is done by the magnetic force F B

on the point charge q?


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Example 2: What are the units of the magnetic field B?


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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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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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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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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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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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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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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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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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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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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


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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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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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




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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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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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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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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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


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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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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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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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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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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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

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