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Transcript
```General Physics (PHY 2140)
Lecture 14
¾ Electricity and Magnetism
9Magnetism
9 Ampere’s law
9Applications of magnetic forces
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 19
10/6/2003
1
Lightning Review
Last lecture:
1. Magnetism
9 Galvanometer
9 Torque on a current loop
F = qvB sin θ
F = BIl sin θ
τ = NBIA sin θ
Review Problem: A rectangular loop is placed in a
uniform magnetic field with the plane of the loop
perpendicular to the direction of the field. If a current is
made to flow through the loop in the sense shown by
the arrows, the field exerts on the loop:
1. a net force.
2. a net torque.
3. a net force and a net torque.
4. neither a net force nor a net torque.
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2
19.7 Motion of Charged Particle in magnetic field
Consider positively charge
×
particle moving in a uniform
magnetic field.
Suppose the initial velocity of the
particle is perpendicular to the ×
direction of the field.
Then a magnetic force will be
exerted on the particle…
×
×
×
×
Bin
× ×
q
v
×
×F ×
×
×
×
×
×
×
r
×
×
×
×
×
×
×
×
×
×
×
Where is it directed?
… and make follow a circular
path.
Remember that
10/6/2003
v⊥F
3
The magnetic force produces a centripetal acceleration.
JG
G
F = ma c
mv 2
F = qvB =
2
2
v
ac =
r
sin θ = sin 90D = 1
The particle travels on a circular trajectory with a radius:
mv
r=
qB
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4
Example 1 : Proton moving in uniform magnetic field
A proton is moving in a circular orbit of radius 14 cm in a uniform
magnetic field of magnitude 0.35 T, directed perpendicular to the
velocity of the proton. Find the orbital speed of the proton.
Given:
r = 0.14 m
B = 0.35 T
m = 1.67x10-27 kg
q = 1.6 x 10-19 C
Find:
v=?
10/6/2003
mv
r=
qB
Recall that the proton’s radius would be
Thus
qBr
v=
m
1.6 ×10−19 C ( 0.35T ) 14 ×10−2 m
=
1.67 ×10−27 kg
(
(
)
(
)
)
= 4.7 ×106 m s
5
Example 2:
Consider the mass spectrometer. The electric field between the plates of the
velocity selector is 950 V/m, and the magnetic fields in both the velocity
selector and the deflection chamber have magnitudes of 0.930 T. Calculate
the radius of the path in the system for a singly charged ion with mass
m=2.18×10-26 kg.
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6
19.8 Magnetic Field of a long straight wire
Danish scientist Hans Oersted (1777-1851) discovered
(somewhat by accident) that an electric current in a wire
deflects a nearby compass needle.
In 1820, he performed a simple experiment with many
compasses that clearly showed the presence of a
magnetic field around a wire carrying a current.
I=0
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I
7
Magnetic Field due to Currents
The passage of a steady current in a wire produces a
magnetic field around the wire.


Field form concentric lines around the wire
Direction of the field given by the right hand rule.
If the wire is grasped in the right hand with the thumb in the
direction of the current, the fingers will curl in the direction of
the field.

Magnitude of the field
I
µo I
B=
2π r
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8
Magnitude of the field
µo I
B=
2π r
I
r
B
µo called the permeability of free space
µ0 = 4π ×10 T ⋅ m A
−7
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9
Ampere’s Law
Consider a circular path surrounding a current, divided
in segments ∆l, Ampere showed that the sum of the
products of the field by the length of the segment is
equal to µo times the current.
∑ B ∆l = µ I
&
Andre-Marie Ampere
I
o
r
∆l
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B
10
Consider a case where B is constant and uniform:
∑ B ∆l = B ∑ ∆l = B 2π r = µ I
&
&
&
o
Then one finds:
µo I
B& =
2π r
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11
19.9 Magnetic Force between two parallel conductors
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12
l
1
B2
2
F1
I1
d
I2
µo I 2
B2 =
2π d
µo I1I 2l
 µo I 2 
F1 = B2 I1l = 
I1l =

2π d
 2π d 
F1 µo I1I 2
=
l
2π d
10/6/2003
Force per
unit length
13
Definition of the SI unit Ampere
F1 µo I1I 2
=
2π d
l
Used to define the SI unit of current called
Ampere.
If two long, parallel wires 1 m apart carry the same current, and the
magnetic force per unit length on each wire is 2x10-7 N/m, then the
current is defined to be 1 A.
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14
Example: Levitating a wire
Two wires, each having a weight per units length of 1.0x10-4 N/m, are
strung parallel to one another above the surface of the Earth, one
directly above the other. The wires are aligned north-south. When their
distance of separation is 0.10 mm what must be the current in each in
order for the lower wire to levitate the upper wire. (Assume the two
wires carry the same current).
1
2
l
d
I1
I2
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15
Two wires, each having a weight per
units length of 1.0x10-4 N/m, are strung
parallel to one another above the surface
of the Earth, one directly above the
other. The wires are aligned north-south.
When their distance of separation is 0.10
mm what must be the current in each in
order for the lower wire to levitate the
upper wire. (Assume the two wires carry
the same current).
F1
1
I1
B2
mg/l
2
l
Weight of wire per unit
length:
mg/l = 1.0x10-4 N/m
Wire separation:
d=0.1 m
I1 = I2
10/6/2003
d
I2
F1 mg µo I 2
=
=
2π d
l
l
−4
1.0 ×10
(
N /m=
)( )
4π ×10−7 Tm A I 2
( 2π )( 0.10m)
I = 7.1 A
16
19.10 Magnetic Field of a current loop
Magnetic field produced by a wire can be enhanced
by having the wire in a loop.
∆x1
B
I
∆x2
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17
19.11 Magnetic Field of a solenoid
Solenoid magnet consists of a wire coil with multiple
loops.
It is often called an electromagnet.
10/6/2003
18
Solenoid Magnet
Field lines inside a solenoid magnet are parallel, uniformly spaced
and close together.
The field inside is uniform and strong.
The field outside is non uniform and much weaker.
One end of the solenoid acts as a north pole, the other as a south
pole.
For a long and tightly looped solenoid, the field inside has a value:
B = µo nI
10/6/2003
19
Solenoid Magnet
B = µo nI
n = N/l : number of (loop) turns per unit length.
I : current in the solenoid.
µo = 4π ×10 Tm / A
−7
10/6/2003
20
Example: Magnetic Field inside a Solenoid.
Consider a solenoid consisting of 100 turns of wire and
length of 10.0 cm. Find the magnetic field inside when it
carries a current of 0.500 A.
N = 100
l = 0.100 m
I = 0.500 A
µo = 4π ×10 Tm / A
−7
N 100turns
n= =
=1000turns/ m
l 0.10m
(
)
B= µonI = 4π ×10−7Tm/ A (1000turns/ m)( 0.500A)
B=6.28×10−4T
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21
Application: Cathode Ray Tube
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22
What is the direction of the magnetic field produced by these
solenoids?
(1) to the Left
(2) to the Right
What is the net force between the two solenoids?
(1) Attractive
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(2) Zero
(3) Repulsive
23
Comparison:
Electric Field vs. Magnetic Field
Source
Acts on
Force
Direction
Electric
Magnetic
Charges
Charges
F = Eq
Parallel E
Moving Charges
Moving Charges
F = q v B sin(θ)
Perpendicular to v,B
Field Lines
Opposites
10/6/2003
Charges Attract
Currents Repel
24
```