Download Unit 16 - HKU Physics

Unit 16
Magnetism
16.1
Permanent magnets
16.2
The magnetic force on moving charge
16.3
The motion of charged particles in a magnetic field
16.4
The magnetic force exerted on a current-carrying wire
16.5
Current loops and magnetic torque
16.6
Biot and Savart’s law
16.1 Permanent magnets
A bar magnet can attract another magnet or repel it, depending on which ends of the
magnets are brought together. One end of a magnet is referred to as its north pole; the
other end is its south pole. The rule for whether two magnets attract or repel each
other: opposites attract; likes repel. Breaking a magnet in half results in the
appearance of two new poles on either side of the break. This behavior is
fundamentally different from that in electricity, where the two types of charge can
exist separately.
We saw a visual indication of the electric field E of a point charge using grass seed
suspended in oil. Similarly, the magnetic field B can be visualized using small iron
filings sprinkled onto a smooth surface. The filings are bunched together near the
poles of the magnets. This is where the magnetic field is most intense. The direction
of the magnetic field, B, at a given location is the direction in which the north pole of
a compass points when placed at that location. In general, magnetic field lines exit
from the north pole of a magnet and enter at the south pole.
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16.2 The magnetic force on moving charge
The magnetic force depends on several factors:
•
The charge of the particle, q;
•
The speed of the particle, v;
•
The magnitude of the magnetic field, B;
•
The angle between the velocity vector and the
magnetic field vector, θ.
The mathematical relation of them in vector form is

 
F = q (v × B ) .
One can rewrite it as a scalar expression, e.g. F = qvB sin θ . The maximum force is
obtained when θ = 90o. The force vanishes when θ = 0o. Now we define the magnetic
field B as B =
F
. The SI unit is 1 tesla = 1 T = 1 N/(A⋅m). The tesla is a fairly
qv sin θ
large unit of magnetic strength, especially when compared with the magnetic field at
the surface of the Earth, which is roughly 5.0 × 10 −5 T . Thus, another commonly used
unit of magnetism is the gauss (G), defined as follows: 1G = 10 −4 T . In terms of the
2
gauss, the Earth’s magnetic field on the surface of the Earth is approximately 0.5 G. A
bar magnet has a magnetic field of roughly 100 G.
Remark:
Magnetic field lines never cross one another. As the direction in which a compass
points at any given location is the direction of the magnetic field at that point. Since a
compass can point in one direction, there must be only one direction for the field B. If
field lines were to cross, however, there would be two directions for B at the crossing
point, and this is not allowed.
Example
Particle 1, with a charge q1 = 3.60 µC and a speed v1 = 862 m/s travels at right angles
to a uniform magnetic field. The magnetic force it experiences is 4.25 × 10−3 N .
Particle 2, with a charge q2 = 53.0 µC and a speed
=
v2 1.30 × 103 m / s moves at an angle of 55.0o
relative to the same magnetic field. Find (a) the
strength of the magnetic field and (b) the magnitude
of the magnetic force exerted on particle 2.
Answer
Apply the formula: F = qvB sin θ with θ = 90o. One obtains
B
=
4.25 × 10−3 N
F
=
= 1.37 T .
qv sin θ (3.60 × 10−6 C )(862 m / s ) sin 90o
Apply the formula: F = qvB sin θ with θ = 55.0o. One obtains
F=
(53.0 × 10−6 )(1.30 × 103 m / s ) (1.37 T ) sin 55.0o =
0.0773 N .
The direction of the magnetic force is given by the magnetic force right-hand rule
(RHR), which states as follows.
To find the direction of magnetic force on a positive charge, start by pointing the
fingers of your right hand in the direction of the velocity, v. Now curl your fingers
forward the direction of B. Your thumb points in the direction of F.
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If the charge is negative, the force points opposite to the
direction of your thumb. Note that the magnetic force F points
in a direction that is perpendicular to both B and the charge
velocity v.
As an example, the direction of a magnetic force F can be
indicated by the magnetic force RHR – extending our fingers to
the right (v) and then curling them into the page (B) – we see
that the magnetic force exerted on this particle is upward, as
indicated. If the charge is negative, the direction of F is
reversed.
Example
Three particles travel through a region of space where the magnetic field is out of the
page as shown in figure. For each of the three particles, state whether the particle’s
charge is positive, negative, or zero.
Answer
Use the RHR to obtain the following. Particle 1: negative; particle 2: zero and particle
3: positive.
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16.3 The motion of charged particles in a magnetic field
Figures (a) and (b) shows the motion of a positive charged particle moving under
electric field and magnetic field respectively.
Remarks:
1.
As the direction of velocity and the direction of magnetic
force are perpendicular to each other, the work done by the
magnetic force is always zero.
2.
A charged particle with a velocity v that is perpendicular to
the magnetic field moves in a circular path. The magnetic
force acts as the centripetal force, e.g. Fcp =
Fcp = qvB . Hence
path, r =
mv 2
and
r
mv 2
= qvB gives the radius of circular
r
mv 2
mv
. On the other hand, Fcp =
= mω 2 r = qvB ,
qB
r
we can write mω = qB , as v = rω . Plugging in the relation
3.
ω=
2π
2π
, we have m
= qB , and the period T is given by
T
T
T=
2πm
.
qB
Helical motion is a combination of linear motion and circular motion.
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Example
In a device called a velocity selector, charged particles move through a region of
space with both an electric and a magnetic field. If the speed of the particle has a
particular value, the net force acting on it is zero. Assume that a positively charged
particle moves in the positive x direction, and the electric field is in the positive y
direction. Should the magnetic field be in (a) the positive z direction, (b) the negative
y direction, or (c) the negative z direction in order to give zero net force?
Answer
The force exerted by the electric field is in the positive y
direction; hence, the magnetic force must be in the negative
y direction if it is to cancel the electric force. If we simply
try the three possible directions for B one at a time, applying
the magnetic force RHR in each case, we find that only a
magnetic field along the positive z axis gives rise to a force
in the negative y direction, as desired. The answer is (a).
Example
Two isotopes of uranium,
235
U ( 3.90 × 10−25 kg ) and
238
U ( 3.95 × 10−25 kg ), are sent
into a mass spectrometer with a speed of 1.05 × 105 m / s . Given that each isotope is
singly ionized, and that the strength of the magnetic field is 0.750 T, what is the
distance d between the two isotopes after they complete half a circular orbit?
Answer
The isotopes are singly ionized, which means that a single electron has been removed
from each atom. And the isotopes are now having charge of
=
e 1.60 × 10−19 C .
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From the relation
mv 2
rqB
mv
, and thus r =
.
= qvB , we obtain v =
m
r
qB
r=
235
mv (3.90 × 10−25 kg )(1.05 × 105 m / s )
=
= 34.1 cm .
qB
(1.60 × 10−19 C )(0.750 T )
r=
238
mv (3.95 × 10−25 kg )(1.05 × 105 m / s )
=
= 34.6 cm .
qB
(1.60 × 10−19 C )(0.750 T )
The separation between the isotopes: d = 2r238 − 2r235 = 2(34.6 cm − 34.1 cm) = 1cm .
16.4 The magnetic force exerted on a current-carrying wire
As a charged particle experiences a force when it moves across magnetic field lines.
The same thing happens when a wire carries current on it. Consider a straight segment
of length L of a wire with a current I flowing from left to right, presents in a magnetic
field B, as shown in figure. If the conducting charges move through the wire with an
average speed v, the time required for them to move
from one end of the wire segment to the other is
∆t = L / v . The amount of charge that flows through the
wire in this time is q = I ∆t = IL / v . Therefore, the
force exerted on the wire is
I L
F = qvB sin θ = 
vB sin θ .
 v 
Hence, we have F = ILB sin θ . Maximum force occurs
when the current is perpendicular to the magnetic field
(θ = 90o) and is zero if the current is in the same
direction as B (θ = 0o). The direction of the magnetic force is given by the RHR,
where the direction of charge velocity v is now the direction of current I.
Example
When the switch is closed in the circuit, the wire between the poles of the horseshoe
magnet deflects downward. Is the left end of the magnet (a) a north magnet pole of (b)
a south magnetic pole?
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Answer
Once the switch is closed, the current in the wire is into the page, as shown in the
right figure. Applying the magnetic force RHR, we see that the magnetic field must
point from left to right in order for the force to be downward. Since magnetic field
lines leave from north poles and enter at south poles, it follows that the left end of the
magnet must be a north magnetic pole. The answer is (a).
16.5 Current loops and magnetic torque
A torque is experienced by a current loop as shown in figure.
If we imagine an axis of rotation through the center of the
loop, at the point O, it is clear that the forces exert a torque
that tends to rotate the loop clockwise.
 w
2
 w
2
τ = ( IhB)  + ( IhB)  = IB(hw) = IAB ,
Plug in the area of the rectangular loop, A = hw. We have τ = IAB .
If the plane of loop makes an angle with the magnetic field, we have τ = IAB sin θ .
For the case that has n turns in a general loops, we have τ = nIAB sin θ .
Applications of the magnetic torque are electric motors and galvanometer.
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16.6 Biot and Savart’s law

B −field

E −Electric field
Charge
Current
Charge
Current
Recall that in the Coulomb’s law, we have, say, a point charge q

gives an electric field at P and E =
1
q
rˆ . For a charge
4πε 0 r 2

distribution, the electric field is at P due to dq is dE =
dq
dE

r
1
dq
rˆ .
4πε 0 r 2
Now, for a magnetic field, we have the law of Biot and Savart
dB =
µ 0 I ds sin θ
,
4π
r2
where µ0 =
4π × 10−7 T ⋅ m / A =
4π × 10−7 H / m
I
( 1 Henry
= 1H
= 1T ⋅ m / A ) and µ 0 is called the
2

ds
permeability
constant.
In
vector
form,
we
 µ 0 I ds × rˆ
have dB =
. Comparing with the Coulomb’s
4π r 2
Magnetic field
θ
r
r̂
I
1
law, we realize that both are 2 laws.
r
Example
Find the magnetic field at the center of a coil which carries a steady current I.
Answer
The magnetic field through the center of coil is the
superposition of the magnetic field due to current segment.
Hence, B = ∑
As
=
B
r
is
a
µ0 I ds sin θ
.
4π
r2
constant
and
θ = 90o, we can write
µ0 I
µ0 I
µ0 I
.
=
ds
=
(2
r
)
π
∑
2r
4π r 2
4π r 2
If the coil has N turns, the magnetic field is given by B =
µ0 NI
2r
.
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.×
P
Remark:
If the solenoid has N turns, length l and carries
a current I, the magnetic field B at a point O on
the axis near the center of the solenoid is found
to be
=
B
µ0 NI
= µ0 nI ,
l
where n is the number of turns per unit length of coil. If the coil is infinite long or
long enough that its length is ten times the diameter, the magnetic field at the end of
coil is half that at the center of coil, e.g. B =
µ0 nI
2
.
Example (Challenging)

Find the magnetic field B at a point P, a distance z from a long current wire, as
shown in the figure.
Answer
From Biot-Savart’s law, we have


 µ0 I dl × rˆ
, where dl =
B=
φ dl cos θ .
× rˆ dl sin
=
2
∫
4π
r
As l = z tan θ , we have dl = ( z / cos 2 θ ) dθ .
µI
We can write B = 0
4π
∫
π
2
−
π
2
dl cos θ
.
r2
P
z
1 cos θ
Now, z / r = cos θ ⇒ =
gives
r
z
µI
B= 0
4π
=
π
z cos 2 θ
2
∫−π2 cos2 θ z 2 cos θ dθ
I

r
θ
O
φ
l

dl
π
µ0 I π2
µ0 I
µ0 I
2
d
cos
sin
θ
θ
θ
=
=
π
π
∫
−
4π z − 2
4π z
2π z
2
Hence, the magnetic field at P due to a long wire of current is given by B =
µ0 I
.
2π z
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Remark 1:
The magnitude of the magnetic field 1 m from a long, straight wire carrying a current
of 1 A is B=
µ0 I (4π × 10−7 T ⋅ m / A)(1 A)
=
= 2 × 10−7 T .
2π z
2π (1 m)
This is a weak field, less than one hundredth the strength of the Earth’s magnetic field.
Remark 2
The magnetic field shown in the figure is due to the horizontal, current-carrying wire.
The current in the wire flows to the left. The following states the use of magnetic field
right-hand rule which points the direction of current:
If you point the thumb of your right hand along the wire to the left your fingers curl
into the page above the wire and out of the page below the wire. Thus, the current
flows to the left.
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