Download Physics 101: Chapter 14 Electromagnetism

Physics 101: Chapter 14
Magnets and the Magnetic Force
z
Characteristics of magnets
Í they attracts metallic items made of IRON or STEEL or combination of iron, cobalt and
nickel
Í they do not attract items made of SILVER, Cooper, Aluminum, etc
Í every magnet has North (N) and South (S) pole,
Í they attract or repeal each other if they are properly aligned
» like poles repel and unlike poles attract.
Í The attractive or repelling force is similar to Coulomb’s electrostatic force
» the magnetic force decreases with the square of distance between the poles,
» the magnitude of the force is proportional to the strength of pole
Í Unlike electric charge (which can be isolated), a single magnetic pole cannot be isolated
Í Similarly with electric field lines, magnetic field lines ca be drawn around the magnet
Í Magnetic field lines are originated on N pole and they go toward S pole
Í Unlike electric field lines,magnetic field lines do not end
N
S
N
S
S
N
S
S
N
S
N S
N
N
Physics 101: Chapter 14, Pg 1
Magnetic Effect of Electric Current - 1
z
z
I=0
Consider a long conductor with no current in it.
Now let’s have a current going through the conductor
Í B - magnetic filed will be formed around the conductor
Í Direction of the magnetic filed lines are determined by the
right hand rule
z
I
Now let’s have two parallel current carrying conductors
I1
F
F
r
I2
Í In this case there will be an
B
attractive force exerted from one to
another conductor - Conductors will
be attracted to each other
Í This force is PERPENDICULAR to
the direction of current
Í The per-unit of length force (F/l) is
This force is used to define the unit
2 ⋅ kI 1 I 2
F
current Amp:
=
l
r
1 A is a current flowing in each of
two parallel wires 1 m apart, that
−7
2
will produce a force per unit of
k = 10
N / A
length equal to2 10 -7 N/n
Physics 101: Chapter 14, Pg 2
Magnetic Effect of Electric Current - 2
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Consider now electric charge particle in the magnetic field B:
B
q
+
v
F
Í If we solve for B:
Í B = F/(q v)
B
F
z
z
Í If the “q” is moving in the direction perpendicular to the
magnetic field lines, with velocity v the force F will be
exerted on the charge
Í This force is:
ÍF=qvB
v
Í The unit for B is:
Í Teslla (T) = N/A m
Exampel1: A particle with positive charge of 0.006 C is moving parallel to a magnetic field
strength of 0.3 T. The particle has a speed of 400 m/s. Calculate the magnitude of the
magnetic force exerted on the particle.
Í F = q v B = 0 N - because the particle is moving parallel to magnetic field lines
Example 2: A particle with positive charge of 0.006 C is moving perpendicular to a magnetic
field strength of 0.3 T. The particle has a speed of 400 m/s. Calculate the magnitude of the
magnetic force exerted on the particle.
Í F = q v B = 0.006C x 400 m/s x 0.3 T = 0.72 N
Physics 101: Chapter 14, Pg 3
Magnetic Effect of Electric Current - 3
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Consider now a current carrying conductor in the magnetic field B:
I
B
v
L
I
F
z
z
v
Í If the conductor is moving in the direction perpendicular to
the magnetic field lines, with velocity v the force F will be
exerted on the wire - conductor
Í This force is:
ÍF=ILB
» I - current through the conductor
» L - length of the conductor in magnetic field
» B - magnetic field strength
Í The direction of the force is determined by the right hand
rule
Exampel1: A straight segment of wire has a length of 30 cm and caries 4.0 A. It is oriented at
right angle to a magnetic field of 0.3 T. What is the magnitude of the magnetic force on this
segment of wire?
Í F = L I B = 0.2m x 5 A x 0.6T = 0.6 N
Example 2: Two parallel wires, each carrying a current of 10 A, lie a distance of 10 cm from
each other. What is the magnetic force per unit of length exerted by one wire on the other?
F 2 ⋅ k ' I1 I 2 2 ⋅ 10 −7 ⋅ 10 ⋅ 10
=
=
= 2 ⋅ 10 −4 N / m
l
r
0.1
Physics 101: Chapter 14, Pg 4
Magnetic Effect of Current Loops - 1
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Consider a current carrying conductor loop
ÍThe strength of the field is higher inside the
loop.
ÍFurther from the loop field is weaker.
i
z
Let’s consider now a PERPENDICULAR current carrying conductor loop in
external magnetic field
3
F3
Í Based on the rule of right hand, in each side
F2
4
F4
2
1
F1
of loop an electric force will be exerted
Í Forces F2 and F4 will be cancelled because
they are equal in magnitude and opposite
direction
Í Forces F1 and F3 produce a torque about the
axis of the loop
Í Since forces are proportional to the current,
the torque is also proportional to the current
Í This concept is used to build a single loop
electric meter.
Physics 101: Chapter 14, Pg 5
Magnetic Effect of Current Loops - 2
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Consider now a multi-loop coil - N loops with current
z
N
Coil produces electro magnetic
force proportional to the number
of turns N
E
Physics 101: Chapter 14, Pg 6
Faraday’s Law: Electromagnetic Induction - 1
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Faraday’s experiment
Iron frame
Iinduced
I
V1
N1
N2
galvanometer
Vinduced
Primary coil
Β
z
z
z
z
z
z
Secondary coil
At the moment when the switch is closed, the galvanometer will detect small current
When current I is steady no current will be detected by the galvanometer
Only when primary current is changing, there will be a current in secondary coil.
CONCLUSION
YOU CAN INDUCE ELECTRIC CURRENT BY MAGNETIC EFEFCTS THAT CHANGE IN
TIME
To determine induced voltage on the secondary side the concept of Magnetic Flux
need to be defined:
Physics 101: Chapter 14, Pg 7
Faraday’s Law: Electromagnetic Induction - 2
z
Magnetic flux is defined by the number of magnetic filed lines passing
through area bounded by a loop
z
B
Loop and magnetic filed lines are perpendicular:
ÍMagnetic flux is:
φ=BxA
loop
z
If loop and magnetic field lines are parallel:
ÍMagnetic flux is:
z
Faraday’s law is now:
ÍA voltage is induced in a circuit when there is a changing magnetic flux
passing through a circuit
Í
z
φ=BxA=0
ε = ∆φ /t = (φ2 − φ1 )/ t
Example: A coil of wire with 100 turns has a uniform magnetic field of 0.4 T passing through
the coil perpendicular to its plane. The coil encloses the area of 0.015 m2. If the flux through
the coil is reduced to zero by removing it from the field in a time of 0.2 second, what is the
induced voltage in the coil?
Í
Í
φ = N x B x A 100 x 0.4 x 0.015 = 0.6 T m2
ε = ∆φ /t = (φ2 − φ1 )/ t = ( 0.6 T m2 - 0)/(0.2 s) = 3.0 V
Physics 101: Chapter 14, Pg 8
Transformer
φ
I2
I1
N1
V1
N2
V2
z
Load
Neglecting the power loss
primary power is equal to
secondary power:
ÍV1 x I1 = V2 x I2
Íor
Primary coil
z
Β
Secondary coil
ÍV1 / N1 = V2 / N2
Example: A transformer has 20 turns of wire in its primary coil and 60 turns on its secondary
coil:
Í Is this a step-up or step-down transformer?
Í If an alternative voltage with effective voltage of 110 V is applied to the primary, what i
the effective voltage induced in the secondary?
Í
Í
Its step-up because N2 > N1
V1 = 110 V
V2 = V1 (N2 / N1) = 110 x (60/20) = 330 V
Physics 101: Chapter 14, Pg 9