physics 2 for electrical engineering

Physics 2 for Electrical EngineeringPhysics 2 for Electrical Engineering

Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter

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 loopSources: Halliday, Resnick and Krane, 5th Edition, Chap. 32.

http://www.bgu.ac.il/atomchip

http://www.bgu.ac.il/nanocenter

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


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


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 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 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.


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

on the point charge q?


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


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):

. mA

N

m/s

N/C T


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.


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.


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?


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.


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.

AC current

Fast particles out

Slow particles in

bottom of magnet

“D1”

“D2”

B


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

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?

I

z

x

y

vdrift

vdrift +

– FB

FB

ΔVH

B

B

t

I

d

I

z

x

y

vdrift

vdrift +

– FB

FB

ΔVH

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 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

I

z

x

y

vdrift

vdrift +

– FB

FB

ΔVH

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 ΔVH = vdrift Bd = IB/net and n = IB/et ΔVH .

Thus we can infer n from measurements of I, B, e, t and ΔVH.

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.

E

B

Source

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?

. . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

B

E

– (a)

(b)

(c)

(d)

(e)

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.


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

I I

B BB

I = 0


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: . BsFF dId BB


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:

We consider two cases:

. BsBsFF dIdId BB

BBds

ds

L'

I

I


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:

We consider two cases:

. BsBsFF dIdId BB


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.

BBds

ds

L'

I

I



If the integral is closed, then L' vanishes and so does FB !

BBds

ds

L'

I

I

I

BBds

ds

L'

I



If the integral is closed, then L' vanishes and so does FB !

In a uniform magnetic field, the net magnetic force on any closed current loop is zero.


Example 1: Rank the magnitude of the magnetic force on these four wires. (B and I are identical.)


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)?

× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×

I

B

B

FB

mg


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.

× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×

I

B

B

FB

mg

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 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.

B

a

b

I

2

3

1

4



Here is a side view of thecurrent loop, still lying in the plane of B.

The torque is τ = IBabbut only for the instant that the current loop isparallel to B.

IBa

IBa

b/2

2 4

B

3



Here is a side view of thecurrent 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.

2

4

B

IBa

IBa

θ θ3



Here is a side view of thecurrent 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 and but they do not contribute to the torque.

1 3

2

4

B

IBa

IBa

θ θ3

2

4

B

IBa

IBa

θ θ3


Whenever moves to the right of , 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.

2 4

A


Whenever moves to the right of , 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:

so the potential energy U of a magnetic dipole μ in a field B is U = –μ · B.

2

2

4

B

IBa

IBa

θ θ3

A

, cos

''sin

'

Bμ

IBA

dIBA

dWB

4

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

–


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

we see that ω must increase.

B

–

, 22/12/32 mrmrm


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

we see that ω must decrease.

, 22/12/32 mrmrm

B

–


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.

× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×

B B

L


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

× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×× × × × × × × × × × × × × × × × × × × ×

B B

L

. C 4.2 2 ghLB

mq

physics 2 for electrical engineering

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