Download Magnetism Notes - Effingham County Schools

Magnetism & Electromagnetism
 Magnets form a magnetic
field around them, caused
by magnetic “poles.”
These are similar to
electric “poles” or
“charge.”
 Magnetic field lines leave
the magnet from the north
pole and reenter into the
south pole
 Magnetic field
lines are
continuous field
lines in loops with
no beginning or
end (not like
electric field lines
 The symbol for a
magnetic field is B
 Compasses are
 If they are allowed to select
their own orientation,
magnets align so that the
north pole points in the
direction of the magnetic
field
magnets that
can easily
rotate and align
themselves
A compass points to
the Earth’s North
magnetic pole (which
is near the north
geographic pole)
Is the North Magnetic
Pole the north pole of
the Earth’s Magnetic
field?
 Magnetic Monopoles DO NOT EXIST!!!
• Magnetic poles cannot be separated from each other in
the same way that electric poles (charges) can be
• Electric monopoles exist as either a negatively charged
object or a positively charged object
Units:
• Tesla (SI unit)
 N/(C m/s)
 N/(A m)
• Gauss
 1 Tesla = 104 gauss
 Magnetic fields cause the existence of
magnetic forces like electric fields cause
electric forces
 A magnetic force is exerted on a particle
within a magnetic field only if
• The particle has a charge
• The charged particle is moving with at least a portion
of its velocity PERPENDICULAR to the magnetic field
 Magnitude: F = qvBsinΘ
• q = charge in Coulombs (C)
• v = velocity in m/s
• B = magnetic field in Tesla
• Θ = angle between v and B
 Direction: Right hand rule if q is positive, left
hand rule if q is negative
 FB = q v x B (This is a “vector cross product” for
those of you who know your math)
Direction of the magnetic force?
Right Hand Rule To determine the DIRECTION of the
force on a POSITIVE charge we
use a special technique that
helps us understand the
3D/perpendicular nature of
magnetic fields.
Basically you hold your right
hand flat with your thumb
perpendicular to the rest of your
fingers
•The Fingers = Direction B-Field
•The Thumb = Direction of velocity
•The Palm = Direction of the Force
For NEGATIVE charges use left hand!
 Calculate the
magnitude of the force
on a 3.0 C charge
moving north at
300,000 m/s in a
magnetic field of 200
mT if the field is
directed
• North
• East
• South
• West
• Sample Problem:
– Calculate the magnitude of force exerted on a 3.0 μC
charge moving north at 300,000 m/s in a magnetic field
of 200 mT if the field is directed a)N, b)E, c)S, d)W
a) F = qvBsinθ
F = 3x10-6C ·3x105m/s ∙ 0.2T ∙ sin0o = 0 N
b) F = qvBsinθ
F = 3x10-6C ·3x105m/s ∙ 0.2T ∙ sin90o = 0.18 N
c) F = qvBsinθ
F = 3x10-6C ·3x105m/s ∙ 0.2T ∙ sin180o = 0 N
d) F = qvBsinθ
F = 3x10-6C ·3x105m/s ∙ 0.2T ∙ sin270o = -0.18 N
 Calculate the magnitude and direction of the
magnetic force
• Calculate the magnitude and direction of the
magnetic force.
v = 300,000 m/s
34o
B = 200 mT
q = 3.0μC
F = qvBsinθ
F = 3x10-6C ∙ 300000 m/s ∙ 0.2T ∙ sin(34o)
F =0.101N upward
 Magnetic forces are always orthogonal (at
right angles) to the plane established by the
velocity and magnetic field vectors
 Magnetic forces can accelerate charged
particles by changing their direction
 Magnetic forces can cause charged particles
to move in circular or helical paths
 Magnetic Forces CANNOT change the speed
or KE of charged particles
 Magnetic Forces CANNOT do work on
charged particles (F is perpendicular)
Magnetic forces ARE
centripetal
• Remember centripetal
acceleration is v2/r
• Centripetal force is mv2/r
SF = ma
FB = Fc
qvBsin = mv2/r
qB = mv/r
q/m = v/(rB)
V
V
F
F
F
F
V
V
B
 What is the orbital radius of a proton moving
at 20,000 m/s perpendicular to a 40 T
magnetic field?
Magnetic Forces on Charged Particles …
• …are centripetal.
• Remember centripetal force is mv2/r.
• For a charged particle moving perpendicular
to a magnetic field
– F = qvB = mv2/r
• Radius of curvature of the particle
– r = mv2/qvB = mv/qB
Sample Problem
What is the orbital radius of a proton moving at
20,000 m/s perpendicular to a 40 T magnetic field?
 What must be the speed of an electron if it is
to have the same orbital radius as the proton
in the magnetic field described in the previous
problem?
Sample Problem
What must be the speed of an electron if it is to
have the same orbital radius as the proton in
the magnetic field described in the previous
problem?
 An electric field of 2,000 N/C is directed to the
south. A proton is traveling at 300,000 m/s to
the west. What is the magnitude and
direction of the force on the proton? Describe
the path of the proton. Ignore gravitational
effects.
Sample Problem
An electric field of 2000 N/C is directed to the
south. A proton is traveling at 300,000 m/s to the
west. What is the magnitude and direction of the
force on the proton? Describe the path of the
proton? Ignore gravitational effects.
 A magnetic field of 2,000 mT is directed to the
south. A proton is traveling at 300,000 m/s to
the west. What is the magnitude and
direction of the force on the proton? Describe
the path of the proton. Ignore gravitational
effects.
Sample Problem
A magnetic field of 2000 mT is directed to the
south. A proton is traveling at 300,000 m/s to the
west. What is the magnitude and direction of the
force on the proton? Describe the path of the
proton? Ignore gravitational effects.
 Calculate the force and describe the path of
this electron if the electric field strength is
2000 N/C
Sample Problem

Calculate the force and describe the path
of this electron.
300,000 m/s
e
E = 2000 N/C
 How would you arrange a magnetic field and
an electric field so that a charged particle of
velocity v would pass straight through without
deflection?
Electric and Magnetic Fields
Together
B
e-
E
v = E/B
 It is found that protons when traveling at
20,000 m/s pass undeflected through the
velocity filter below. What is the magnetic
field between the plates?
e+
0.02 m
400 V
Sample Problem
It is found that protons traveling at 20,000 m/s pass
undeflected through the velocity filter below. What
is the magnitude and direction of the magnetic field
between the plates?
e
20,000 m/s
400 V
0.02 m
 F = I L B sin Θ
• I = current in Amps
• L = length in m
• B = magnetic field in Tesla
• Θ = angle between current and B field
 What is the force on a 100m long wire bearing
a 30A current flowing north if the wire is in a
downward-directed magnetic field of 400 mT?
Sample Problem
What is the force on a 100 m long wire bearing a 30
A current flowing north if the wire is in a
downward-directed magnetic field of 400 mT?
 What is the magnetic field strength if the
current in the wire is 15 A and the force is
downward with a magnitude of 40 N/m?
What is the direction of the current?
Sample Problem
What is the magnetic field strength if the current in the
wire is 15 A and the force is downward and has a
magnitude of 40 N/m? What is the direction of the
current?
 Magnetic Fields affect moving charge
• F = qvBsinΘ
• F = ILBsin Θ
 Magnetic fields are caused by moving charge
B = 0I/(2πr)
• 0 : 4 π x 10-7 T m/A
 Magnetic permeability of free space
• I: current (A)
• R: radial distance from center of wire (m)
1.
2.
3.
4.
Curve your fingers
Place your thumb in the
direction of the current
Curved fingers represent the
curve of the magnetic field
Field vector at any point is
tangent to the field line
 What is the magnitude and direction of the
magnetic field at P, which is 3.0 m away from a
wire bearing a 13.0 A current?
Sample Problem

What is the magnitude and direction of the
magnetic field at point P, which is 3.0 m away from
a wire bearing a 13.0 Amp current?
P
I = 13.0 A
3.0 m
 What is the magnitude and direction of the
force exerted on a 100 m long wire that passes
through point P which bears a current of 50
Amps in the same direction?
I2 = 50.0 A
I1 = 13.0 A
P
3.0 m
Sample Problem – not in packet

What is the magnitude and direction of the force
exerted on a 100 m long wire that passes through
point P which bears a current of 50 amps in the
same direction?
I2 = 50.0 A P
I1 = 13.0 A
3.0 m
 Remember this from electrostatics?
 When there are two or more currents forming
a magnetic field, calculate B due to each
current separately, and then add them
together using vector addition.
16. What is the magnitude and direction of the
electric field at point P if there are two wires
producing a magnetic field at this point?
I2 = 10.0 A
4.0 m
P
I1 = 13.0 A
3.0 m
Sample Problem

What is the magnitude and direction of the
electric field at point P if there are two wires
producing a magnetic field at this point?
I = 10.0 A
4.0 m
P
I = 13.0 A
3.0 m
 You learned that coils with current in them
make magnetic fields (electromagnets)
 The iron nail was not necessary to cause the
field, it only intensified it
 A solenoid is a coil of wire
 When current runs through a wire, it causes the coil
to become an “electromagnet”
 Air-core solenoids
have nothing inside
fo them
 Iron-core solenoids
are filled with iron
(a magnetic
material) to
intensify the
magnetic field
B
1.
2.
3.
4.
Curve your fingers
Place them along the wire loop
so that your fingers point in the
direction of the current
Your thumb gives the direction
of the magnetic field in the
center of the loop, where it is
straight
Field lines curve around and
make complete loops
 What is the direction of the magnetic field
produced by the current I at A? At B?
Sample Problem
What is the direction of the magnetic field
produced by the current I at A? At B?
I
B
A
 The product of magnetic field and area
 Can be thought of as total magnetic “effect”
on a coil of wire of a given area
 The area is aligned so that a perpendicular to
the area vector (orthogonal to area) points
parallel to the field
 The area is aligned so that a perpendicular to
the area vector points perpendicular to the
field
 The area is neither perpendicular nor parallel
 ΦB = B A cos Θ
• ΦB : magnetic flux in Webers (Tesla meters2)
• B: magnetic field in Tesla
• A: area in meters2
• Θ : the angle between the area vector and the
magnetic field
 ΦB = B A
 Calculate the magnetic flux through a
rectangular wire frame 3.0 m long and 2.0 m
wide if the magnetic field through the frame is
4.2 mT
• Assume that the magnetic field is perpendicular to the
area vector
• Assume that the magnetic field is parallel to the area
vector
• Assume that the angle between the magnetic field
and the area vector is 30 degrees
Sample Problem
Calculate the magnetic flux through a rectangular wire frame 3.0 m long
and 2.0 m wide if the magnetic field through the frame is 4.2 mT.
a)
Assume that the magnetic field is perpendicular to the area vector.
b)
Assume that the magnetic field is parallel to the area vector.
c)
Assume that the angle between the magnetic field and the area
vector is 30o.
 Assume the angle is 40 degrees, the magnetic
field is 50 mT, and the flux is 250 mWb. What
is the radius of the loop? (hint: A = πr2)
Sample Problem

Assume the angle is 40o, the magnetic field is 50 mT, and the
flux is 250 mWb. What is the radius of the loop?
A
B
 A system will respond so as to oppose
changes in magnetic flux
 A change in magnetic flux will be partially
offset by an induced magnetic field whenever
possible
 Changing the magnetic flux through a wire
loop causes current to flow in the loop
 This is because changing magnetic flux
induces an electric potential
 ε = -NΔΦB/Δt
• ε : induced potential (V)
• N : # loops
• ΦB : magnetic flux in Wb
• t : time (s)
If there is only ONE LOOP:
ε = -ΔΦB/Δt
ε = -Δ(B A cos Θ)/Δt
• To generate voltage:
 Change B
 Change A
 Change Θ
 A coil of radius 0.5 m consisting of 1000 loops
is placed in a 500 mT magnetic field such that
the flux is maximum. The field then drops to
zero in 10 ms. What is the induced potential in
the coil?
Sample Problem

A coil of radius 0.5 m consisting of 1000 loops is placed in
a 500 mT magnetic field such that the flux is maximum.
The field then drops to zero in 10 ms. What is the induced
potential in the coil?
 A single coil of radius 0.25 m is in a 100 mT
magnetic field such that the flux is maximum.
At time t = 1.0 s, the field increases at a
uniform rate so that at 11 s, it has a value of
600 mT. At time t = 11 s, the field stops
increasing. What is the induced potential:
a. At 0.5 s?
b. At 3.0 s?
c.
At 12 s?
Sample Problem
A single coil of radius 0.25 m is in a 100 mT magnetic field
such that the flux is maximum. At time t = 1.0 seconds,
field increases at a uniform rate so that at 11 seconds, it
has a value of 600 mT. At time t = 11 seconds, the field
stops increasing. What is the induced potential
 A) at t = 0.5 seconds?
 B) at t = 3.0 seconds?
 C) at t = 12 seconds?

 The current will flow in a direction so as to
oppose the change in flux
 Use in combination with the hand rule to
predict current direction
 The magnetic field is increasing at a rate of 4.0
mT/s. What is the direction fo the current in
the wire loop?
Sample Problem

The magnetic field is increasing at a rate of 4.0
mT/s. What is the direction of the current in
the wire loop?
 The magnetic field is increasing at a rate of 4.0
mT/s. What is the direction of the current in
the wire loop?
Sample Problem

The magnetic field is increasing at a rate of 4.0
mT/s. What is the direction of the current in
the wire loop?
 The magnetic field is decreasing at a rate of
4.0 mT/s. The radius of the loop is 3.0 m, and
the resistance is 4 ohms. What is the
magnitude and direction of the current?
Sample Problem

The magnetic field is decreasing at a rate of 4.0
mT/s. The radius of the loop is 3.0 m, and the
resistance is 4 W. What is the magnitude and
direction of the current?
 ε = BLv
• B: magnetic field
• L: length of bar moving through field
• v: speed of bar moving through field
 Bar must be “cutting through” the field lines.
It cannot be moving parallel to the field
 This formula is derivable from Faraday’s Law
of Induction
 How much current flows through the resistor?
How much power is dissipated by the resistor?
Sample Problem

How much current flows through the resistor?
How much power is dissipated by the resistor?





























3 W


50
 cm















B = 0.15

T



v = 2m/s









 In which direction is the induced current
through the resistor?
Sample Problem

In which direction is the induced current through
the resistor (up or down)?





























3 W


50
 cm















B = 0.15

T



v = 2m/s









 Assume the rod is being pulled so that it is
traveling at a constant 2 m/s. How much force
must be applied to keep it moving at this
constant speed?
Sample Problem

Assume the rod is being pulled so that it is traveling at a
constant 2 m/s. How much force must be applied to keep
it moving at this constant speed?





























3 W


50
 cm















B = 0.15

T



v = 2m/s








