Download Problem 27.15 An electron at point A has a speed of 1.41 x 106 m/s

Problem 27.15
An electron at point A has a speed
of 1.41 x 106 m/s. Find
(a) the magnitude and direction of
the magnetic field that will cause the
electron to follow the semicircular path
from A to B, and
(b) the time required for the electron to
move from A to B.
Fig. 27.46
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Problem 27.31 Determining the mass of an isotope.
The electric field between the plates of the velocity selector in the
Bainbridge mass spectrometer (Fig. 27.24) is 1.12 x 105 V/m, and
the magnetic field is 0.540 T. A stream of singly charged selenium
ions moves in a circular path with a radius of 31.0 cm in the magnetic
field. Determine the mass of one selenium ion and the mass number
of the selenium isotope. (The mass number is equal to the mass of
the isotope in atomic mass units, rounded to the nearest integer. One
atomic mass unit = 1u = 1.66 x 10-27 kg).
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Fig. 27.24
2
The force exerted by a magnetic field.
x x x x x x x
The force on a single charge
is given by
→
→
x x x x x x x
→
•
F = qv xB
x x x x x x x
positive charge entering a magnetic field
When a wire carries a current, in a
magnetic field, there is a force on the
wire that is equal to the sum of the
magnetic forces on the charged particles
whose motion produces the current.
The number of charges in length l is
nAl , where n is the density and Al is the
volume.
→
→
→
∴ F = (q v d x B)nAl
But the current I = nqvd A
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Fig. 27.25
3
→
→
→
Hence the force can be written F = I l x B
(valid for both + and – charges)
→
Note: Current
I
is not a vector.
→
flows is given by
l
l
is. The direction in which current
I
→
, or
dl
.
D.C. motor
In everyday life, there are countless examples of motors. It is important to
understand how they work. First though, we have to learn about Forces and torques on a current loop.
Recall for a current carrying wire
in a magnetic field,
→
→
B
I
l
→
F = I l x B = IlB sin φ
F = IlB when φ = 90
I
o
B
The diagrams illustrate these two cases.
To understand this is important when
studying the case of torques on a current
carrying loop of wire in a B field.
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φ
l
4
This figure shows a current
carrying loop of wire in a
magnetic field.
A normal to the plane of the
loop makes an angle φ to
the magnetic field, B.
Consider the sides of length b.
The force on each side is
Fig.27.31
27.31
Fig.
F ' = IbB sin(90 − φ ) = IbB cos φ
∧
∧
They are designated F ' j and F ' (− j )
and are equal and opposite. There is no net force in the ±
y direction.
Consider the sides of length a. The force on each side is
F = IaB
∧
These forces are designated
∧
F i and F (− i )
and are equal and opposite. There is no net force in the ± x direction.
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But there is a torque acting on the sides of length a.
τ = 2 F ( b 2 ) sin φ = ( IaB)(b sin φ ) = IAB sin φ
o
o
Torque is a maximum when φ = 90 (Fig. 27.31b), and zero when φ = 0 , or 180o
(Fig. 27.31c).
We can write torque as
τ = µB sin φ where µ = IA (or NIA where N is the number of turns of wire)
is the magnetic dipole moment of the loop.
→
→
Since A is a vector, µ is a vector and the direction is given by the R.H. rule:
Place the fingers of the R.H. in the direction of I and the thumb points in
→
the direction of
→
→
µ (when it is perpendicular to the plane of the loop).
→
∴τ = µ x B
(recall the similarity to the torque exerted by an electric field on an electric dipole moment,
→
→
→
τ = p x E ).
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Using this symmetry and recalling that the potential energy of an electric
→
dipole is U
→
= − p• E ,
can make a reasonable guess that the potential
energy for a magnetic dipole moment is
→
→
U = − µ• B
U
→
Is greatest when
parallel.
U =0
µ
when
→
and
→
µ
B are antiparallel, and least when they are
→
and
B are perpendicular to each other.
Note that the rectangular current loop is not a special case. The relationships
above are valid for a plane loop of any shape. The
irregular shape to the right can be approximated by
an infinite number of rectangles. The currents in
adjacent sides cancel and that leaves just the
current in the outer boundary. The results above
are valid for each of the rectangles, and hence
valid for the irregular shape.
Fig. 27.33
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Question: What is the torque on a solenoid
situated in a magnetic field as shown?
For a single loop,
τ = IAB sin φ = µB sin φ
A solenoid is equivalent to N circular
loops, side by side. So in this case,
µ = NIA
The torque as usual is given by
τ = µB sin φ , so for the solenoid,
Fig. 27.34
τ = NIAB sin φ
Now we are ready to discuss the D-C motor.
The above ideas of current carrying loops and torques are applicable in this
case.
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Fig. 27.39
A current carrying loop lies in a magnetic field generated by a permanent
magnet. The battery is connected to the coil through commutators which
keep the current flowing in the same direction in the loop as it turns.
→
→
In (a), note the orientation of
the loop counter-clockwise.
µ
with respect to
B . The torque will turn
In (b), each brush which touches the commutators touches both
commutators simultaneously.
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→
Hence,
I =0
through the coil,
µ = 0 , and
→
τ =0.
Momentum keeps the loop turning.
In (c), the coil has turned
→
180o with respect to (a), but because of the
commutators, µ is in the same direction as it was in (a) and the torque
will continue to turn the loop counter-clockwise.
The coil will turn at top speed when the turning torque is balanced by a
resisting torque due to air resistance and friction.
For N turns of wire in the loop, the torque will be N times stronger.
Question:
What improvements could be made to this simple motor set-up to make
it more efficient?
Back emfs
When a wire moves in a magnetic field, a current is induced in it to
oppose the motion causing it. In the motor discussed above, the induced
current is in the direction opposite to the current shown. This induced emf
is called a back emf. We will study this in more detail later.
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Hall Effect
The Hall effect is the classic way of
determining the sign of the mobile
charge carriers in a material, and the
density of current carrying charges in
the material.
A current flows to the right in the +x
direction. For a metal conductor, the
drift velocity, vd, of the charge carriers
(electrons) is to the left (diagram (a)).
For a semiconductor, for example, the
drift velocity of the charge carriers (holes)
is to the right (diagram (b)).
In both cases, the magnetic field produces
a force on the moving charges in the +z
direction.
Fig. 27.41
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In (a), an excess of negative charges accumulates near the top edge of the
conductor. In (b), the charge accumulating along the top is positive. The net
result of this is to create an electric field to oppose the motion of the charge,
negative charge in (a), and positive charge in (b).
The transverse potential difference is called the Hall voltage.
Eventually, the electrostatic force balances the magnetic force.
∧
∧
, and
d
d
qE (− k ) + qv B(k ) = 0
E = −v B
E is positive, and vice versa).
From before, the current density J x = nqvd
vd
(when
Eliminate
nq =
Jx
is negative,
vd from these last two equation and obtain,
− J x By
and
(Hall effect)
E
By
are positive. If
E
is positive (diagram (a)), then nq is negative,
and vice versa (diagram (b)).
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Since the magnitude of q is known (equal to the electron charge), then n can
be calculated.
For a metal conductor,
n ~ 10 29 m −3
How magnets work
What will happen to a bar magnet when it is
placed in a magnetic field, and at some angle
to it?
B
It will rotate until it is aligned with the field. This
means that the bar magnet has a magnetic
dipole moment. The origin of this dipole moment
is the electron which spins and acts somewhat
like a current loop (look back on slide 5). An
electron has a magnetic moment. In a
permanent magnet, these dipoles are well
S
aligned. The magnetic dipole moment for a
permanent magnet points from S to N.
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bar magnet
B
→
µ
N
13
Fig. 27.37
In an unmagnetized object (e.g. iron in diagram (a)), there is no overall alignment
of the magnetic moments, and the vector sum is zero. In the presence of a
magnet, the magnetic moments align with the B field and the object acquires a
net magnetic moment (diagram (b)). This new magnetic moment of the object
(e.g. the magnet above, or an iron nail) is the reason why the object is attracted
by the B field of the magnet.
Attraction and repulsion
→
µ
Fig. 27.36
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→
µ
Fig. 27.37
14
In Fig. 27.36 (a), the current loop has a magnetic dipole moment which is
anti-aligned with the magnetic moment of the bar magnet. The force on a
section of the loop has a radial component, and a component to the right.
The radial components cancel and the net force on the loop is to the right.
So, when the magnetic moments are anti-aligned, the loop is repelled from
the bar magnet.
In (b), the current loop has a magnetic dipole moment which is aligned with
that of the bar magnet. The net force on the current loop is now to the left.
So, when the magnetic moments are aligned, the loop is attracted to the bar
magnet.
For a class of materials which are called ferromagnetic (iron, nickel, cobalt
and alloys of these) which are unmagnetized, the atomic magnetic moments
tend to align with the B field when in the presence of a permanent magnet.
The aligned dipole moments are represented by (b) in Fig. 27.36, and the
object is attracted to the magnet. Also (b) explains the attraction between
two bar magnets end to end, with the S pole of one next to the N pole of the
other.
Diagram (a) in Fig. 27.36 explains the repulsion between two bar magnets
when similar poles are put next to each other.
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Problem 27.51
The figure shows a portion of a
silver ribbon with z1=11.8 mm and
y1= 0.23 mm, carrying a current of
120 A in the +x direction. The ribbon
lies in a uniform magnetic field, in the
y direction, with magnitude 0.95 T.
If there are 5.85 x 1028 free electrons
per cubic metre, use the simplified model
of the Hall effect to find
(a) the magnitude of the drift velocity of the
electrons in the x direction,
(b) the magnitude and direction of the electric field in the z direction
due to the Hall effect,
(c) the Hall emf.
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