Upload
ivana-hebert
View
63
Download
0
Embed Size (px)
DESCRIPTION
Applied Physics. Lecture 14. Electricity and Magnetism Magnetism Ampere’s law Applications of magnetic forces. Chapter 19. q. v. F. 19.7 Motion of Charged Particle in magnetic field. B in. Consider positively charge particle moving in a uniform magnetic field. - PowerPoint PPT Presentation
Citation preview
1104/20/2304/20/23
Applied Physics
Lecture 14Lecture 14 Electricity and Magnetism
Magnetism Ampere’s lawApplications of magnetic forces
Chapter 19
2204/20/2304/20/23
19.7 Motion of Charged Particle in magnetic field19.7 Motion of Charged Particle in magnetic field
Consider positively charge Consider positively charge particle moving in a uniform particle moving in a uniform magnetic field.magnetic field.
Suppose the initial velocity of Suppose the initial velocity of the particle is the particle is perpendicular to perpendicular to the direction of the fieldthe direction of the field..
Then a Then a magnetic forcemagnetic force will be will be exerted on the particle… exerted on the particle…
Fv
q
r
Bin
Where is it directed?Where is it directed?
… … and make it follow a and make it follow a circular circular pathpath..
Remember that v F������������� �
3304/20/2304/20/23
The magnetic force produces a centripetal acceleration.
The particle travels on a circular trajectory with a radius:
cF ma����������������������������
2
c
va
r
2
2
mvF qvB
mvr
qB
sin sin 90 1
4404/20/2304/20/23
Example 1 : Proton moving in uniform magnetic fieldExample 1 : Proton moving in uniform magnetic field
A proton is moving in a circular orbit of radius 14 cm in a uniform 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 magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton.velocity of the proton. Find the orbital speed of the proton.
Given:
r = 0.14 mB = 0.35 Tm = 1.67x10-27 kgq = 1.6 x 10-19 C
mvr
qB
qBrv
m
Find:
v = ?
Recall that the proton’s radius would be
19 2
27
6
1.6 10 0.35 14 10
1.67 10
4.7 10 ms
C T m
kg
Thus
5504/20/2304/20/23
19.8 Magnetic Field of a long straight wire19.8 Magnetic Field of a long straight wire
Danish scientist Hans Oersted (1777-1851) discovered Danish scientist Hans Oersted (1777-1851) discovered (somewhat by accident) that an electric current in a wire (somewhat by accident) that an electric current in a wire deflects a nearby compass needle.deflects a nearby compass needle.
In 1820, he performed a simple experiment with many In 1820, he performed a simple experiment with many compasses that clearly showed the presence of a compasses that clearly showed the presence of a magnetic field around a wire carrying a current.magnetic field around a wire carrying a current.
I=0 I
6604/20/2304/20/23
Magnetic Field due to CurrentsMagnetic Field due to Currents
The passage of a steady current in a wire produces a The passage of a steady current in a wire produces a magnetic field around the wire.magnetic field around the wire.
Field form concentric lines around the wireField form concentric lines around the wire Direction of the field given by the right hand rule.Direction of the field given by the right hand rule.
If the wire is grasped in the right hand with the thumb in the 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 direction of the current, the fingers will curl in the direction of the field (the field (second right-hand rulesecond right-hand rule).).
Magnitude of the field Magnitude of the field I
2oI
Br
7704/20/2304/20/23
Magnitude of the field I
r
B
o called the permeability of free space
2oI
Br
70 4 10 T m A
8804/20/2304/20/23
Andre-Marie Ampere
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 o times the current.
I
r
Bl
oB l I
9904/20/2304/20/23
2 oB l B l B r I
Consider a case where B is constant and uniform:
Then one finds:
2oI
Br
101004/20/2304/20/23
22 2
oIB
d
l
d
1
2 F1
B2 I1
I2
2 1 21 2 1 12 2
o oI I I lF B I l I l
d d
1 21
2oI IF
l d
Force per unit length
111104/20/2304/20/23
Definition of the SI unit AmpereDefinition of the SI unit Ampere
If two long, parallel wires 1 m apart carry the same current, and the If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10magnetic force per unit length on each wire is 2x10-7-7 N/m, then the N/m, then the current is defined to be 1 A.current is defined to be 1 A.
1 21
2oI IF
l d
Used to define the SI unit of current called Ampere.
121204/20/2304/20/23
Example 1: Levitating a wireExample 1: Levitating a wire
Two wires, each having a weight per units length of 1.0x10Two wires, each having a weight per units length of 1.0x10 -4-4 N/m, are N/m, are strung parallel to one another above the surface of the Earth, one strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their 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 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 order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current).wires carry the same current).
l
d
1
2I1
I2
131304/20/2304/20/23
l
d
1
2
F1
B2I1
I2
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).
mg/l
Weight of wire per unit length:
mg/l = 1.0x10-4 N/m
Wire separation: d=0.1 m
I1 = I2
21
7 2
4
2
4 101.0 10 /
2 0.10
oIF mg
l l d
Tm A IN m
m
7.1I A
141404/20/2304/20/23
19.10 Magnetic Field of a current loop19.10 Magnetic Field of a current loop
Magnetic field produced by a wire can be enhanced Magnetic field produced by a wire can be enhanced by having the wire in a loop.by having the wire in a loop.
x1
I
x2
B
151504/20/2304/20/23
19.11 Magnetic Field of a solenoid19.11 Magnetic Field of a solenoid
Solenoid magnet consists of a wire coil with multiple Solenoid magnet consists of a wire coil with multiple loops.loops.
It is often called an electromagnet.It is often called an electromagnet.
161604/20/2304/20/23
Solenoid MagnetSolenoid [email protected]@uettaxila.edu.pk
Field lines inside a solenoid magnet are parallel, uniformly spaced Field lines inside a solenoid magnet are parallel, uniformly spaced and close together.and close together.The field inside is uniform and strong.The field inside is uniform and strong.The field outside is non uniform and much weaker.The field outside is non uniform and much weaker.One end of the solenoid acts as a north pole, the other as a south One end of the solenoid acts as a north pole, the other as a south pole.pole.For a long and tightly looped solenoid, the field inside has a value:For a long and tightly looped solenoid, the field inside has a value:
oB nI
171704/20/2304/20/23
Solenoid MagnetSolenoid Magnet
n = N/n = N/ll : number of (loop) turns per unit length. : number of (loop) turns per unit length.
I : current in the solenoid.I : current in the solenoid.
oB nI
74 10 /o Tm A
181804/20/2304/20/23
Example: Magnetic Field inside a Solenoid.Example: Magnetic Field inside a Solenoid.
Consider a solenoid consisting of 100 turns of wire and Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it length of 10.0 cm. Find the magnetic field inside when it carries a current of 0.500 A.carries a current of 0.500 A.
N = 100l = 0.100 mI = 0.500 A 7
4
1001000 /
0.10
4 10 / 1000 / 0.500
6.28 10
o
N turnsn turns m
l m
B nI Tm A turns m A
B T
74 10 /o Tm A
191904/20/2304/20/23
Comparison:Electric Field vs. Magnetic Field
Electric MagneticSource Charges Moving ChargesActs on Charges Moving ChargesForce F = Eq F = q v B sin()Direction Parallel E Perpendicular to v,B
Field Lines
Opposites Charges Attract Currents Repel