Magnetic Force

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Magnetic Force
Force on Moving Charge
(−e)~
v ×~
B
~
v
−e
~
v ×~
B
~
B
The magnetic force on a moving charge is given
by
~ B = q~v × B
~.
F
~ can be determined using the
The direction of F
right hand rule.
Magnetic Force
Force on Moving Charge
(−e)~
v ×~
B
~
v
−e
~
v ×~
B
~
B
The magnetic force on a moving charge is given
by
~ B = q~v × B
~.
F
~ can be determined using the
The direction of F
right hand rule.
Magnetic forces are typically very small compared to electric force,
and are proportional to q/m.
Magnetic Force
Force on Moving Charge
(−e)~
v ×~
B
~
v
−e
~
v ×~
B
~
B
The magnetic force on a moving charge is given
by
~ B = q~v × B
~.
F
~ can be determined using the
The direction of F
right hand rule.
Magnetic forces are typically very small compared to electric force,
and are proportional to q/m.
~ B = q~v × B
~ is always perpendicular to displacement,
Note that F
so magnetic forces do no work.
Magnetic Force
Force on Moving Charge
Moving charged particles in uniform magnetic fields exhibit uniform
circular motion. Non-relativistically, their angular speed ωB (also
called the cyclotron frequency) is given by
ωB =
~
|q||B|
.
m
Magnetic Force
Force on Moving Charge
Moving charged particles in uniform magnetic fields exhibit uniform
circular motion. Non-relativistically, their angular speed ωB (also
called the cyclotron frequency) is given by
ωB =
~
|q||B|
.
m
Question: Why does a uniform magnetic field cause uniform
circular motion for a moving charged particle?
Magnetic Force
Force on Moving Charge
Moving charged particles in uniform magnetic fields exhibit uniform
circular motion. Non-relativistically, their angular speed ωB (also
called the cyclotron frequency) is given by
ωB =
~
|q||B|
.
m
Question: Why does a uniform magnetic field cause uniform
circular motion for a moving charged particle?
Exercise: Use the principles of uniform circular motion to find
expressions for both the period T and the momentum |~p | of a
moving chaged particle in a uniform magnetic field.
Magnetic Force
Force on Current-Carrying Wire
Using the principle of superposition, we can show that the
magnetic force on a short length ∆l of current-carrying wire is
given by
~ B = I ∆~l × B
~.
∆F
Magnetic Force
Force on Current-Carrying Wire
Using the principle of superposition, we can show that the
magnetic force on a short length ∆l of current-carrying wire is
given by
~ B = I ∆~l × B
~.
∆F
h
I
L
Exercise: Determine the direction and
magnitude of the magnetic force on the wire.
Magnetic Force
Force on Current-Carrying Wire
Using the principle of superposition, we can show that the
magnetic force on a short length ∆l of current-carrying wire is
given by
~ B = I ∆~l × B
~.
∆F
h
I
Exercise: Determine the direction and
magnitude of the magnetic force on the wire.
What happens to the force if the battery
terminals are reversed?
L
Magnetic Force
Parallel Wires
The magnetic field due to wire 1 at the
location of wire 2 is
L
~
F21
d
I1
I2
~
B
~ 1 = µ0 2I1
B
4π d
and into the page by the RHR.
Magnetic Force
Parallel Wires
The magnetic field due to wire 1 at the
location of wire 2 is
L
~
F21
d
I1
I2
~
B
~ 1 = µ0 2I1
B
4π d
and into the page by the RHR.
~ 21 , the force on wire 2 due to wire 1.
Exercise: Determine F
Magnetic Force
Parallel Wires
The magnetic field due to wire 1 at the
location of wire 2 is
L
~
F21
d
I1
I2
~
B
~ 1 = µ0 2I1
B
4π d
and into the page by the RHR.
~ 21 , the force on wire 2 due to wire 1.
Exercise: Determine F
~ 12 . You will need B
~ 2 at the location of
Exercise: Now determine F
wire 1.
Magnetic Force
Parallel Wires
The magnetic field due to wire 1 at the
location of wire 2 is
L
~
F21
d
I1
I2
~
B
~ 1 = µ0 2I1
B
4π d
and into the page by the RHR.
~ 21 , the force on wire 2 due to wire 1.
Exercise: Determine F
~ 12 . You will need B
~ 2 at the location of
Exercise: Now determine F
wire 1.
Question: What happens if the direction of one of the currents is
reversed?
Magnetic Force
~ and B
~ combined
E
A moving charged particle can experience electric and magnetic
forces simultaneously. Combining Fe and Fb gives us the Lorentz
Force Law:
~ = qE
~ + q~v × B
~.
F
Magnetic Force
~ and B
~ combined
E
A moving charged particle can experience electric and magnetic
forces simultaneously. Combining Fe and Fb gives us the Lorentz
Force Law:
~ = qE
~ + q~v × B
~.
F
Exercise: Determine the direction of the Lorentz force for each
situation below.
y
y
~
B
+
+
~
E
~
v
~
B
+
+
~
E
x
z
y
~
v
~
B
~
E
x
z
y
~
B
~
v
x
z
~
E
x
z
Magnetic Force
~ and B
~ combined
E
A moving charged particle can experience electric and magnetic
forces simultaneously. Combining Fe and Fb gives us the Lorentz
Force Law:
~ = qE
~ + q~v × B
~.
F
Exercise: Determine the direction of the Lorentz force for each
situation below.
y
y
~
B
+
+
~
E
~
v
~
B
+
+
~
E
x
z
y
~
v
~
B
~
E
x
z
y
~
B
~
v
x
z
~
E
x
z
Exercise: Repeat the exercise, but with a negative instead of a
positive charge.
Magnetic Force
~ and B
~ combined
E
~ and B
~ fields can be configured to select charged particles of a
E
particular velocity.
Magnetic Force
~ and B
~ combined
E
~ and B
~ fields can be configured to select charged particles of a
E
particular velocity.
y
~ and B
~
Exercise: Draw in a configuration of E
that could exert zero net foce on the given
particle.
+
~
v
x
z
Magnetic Force
~ and B
~ combined
E
~ and B
~ fields can be configured to select charged particles of a
E
particular velocity.
y
~
E
~ and B
~
Exercise: Draw in a configuration of E
that could exert zero net foce on the given
particle.
+
~
v
~
B
z
Question: If the magnetic field has magnitude B, what magnitude
of E will produce no net force?
x
Magnetic Force
~ and B
~ combined
E
~ and B
~ fields can be configured to select charged particles of a
E
particular velocity.
y
~
E
~ and B
~
Exercise: Draw in a configuration of E
that could exert zero net foce on the given
particle.
+
~
v
~
B
z
Question: If the magnetic field has magnitude B, what magnitude
of E will produce no net force?
Question: If the same particle comes in at a slower speed, why
won’t it move in a straight line?
x
Magnetic Force
~ and B
~ combined
E
Example: Given the details in the figure and
~ earth , what are the magnitude and
neglecting B
direction of I ?
2 turns, radius R
~
v
I
+Q
−e
3R
Magnetic Force
Motional EMF
+
L
~
v
~
E
−
~
B
A conducting bar of length L moving with
~
velocity ~v through a uniform magnetic field B
polarizes due to magnetic forces, inducing an
~.
electric field E
Magnetic Force
Motional EMF
+
L
~
v
~
E
−
~
B
A conducting bar of length L moving with
~
velocity ~v through a uniform magnetic field B
polarizes due to magnetic forces, inducing an
~.
electric field E
Question: If the bar is kept at a constant speed |~v |, at what value
~ | will the polarization process conclude?
of |E
Magnetic Force
Motional EMF
+
L
~
v
~
E
−
~
B
A conducting bar of length L moving with
~
velocity ~v through a uniform magnetic field B
polarizes due to magnetic forces, inducing an
~.
electric field E
Question: If the bar is kept at a constant speed |~v |, at what value
~ | will the polarization process conclude?
of |E
Question: In the steady state, what is ∆V between the top and
bottom of the bar? Is an external force needed to keep the bar
moving at |~v |? Why or why not?
Magnetic Force
Motional EMF
+
L
~
v
~
E
−
~
B
A conducting bar of length L moving with
~
velocity ~v through a uniform magnetic field B
polarizes due to magnetic forces, inducing an
~.
electric field E
Question: If the bar is kept at a constant speed |~v |, at what value
~ | will the polarization process conclude?
of |E
Question: In the steady state, what is ∆V between the top and
bottom of the bar? Is an external force needed to keep the bar
moving at |~v |? Why or why not?
Hard Question: If the bar is released (i.e., not maintained at a
constant speed) before the steady state is reached, what happens?
Magnetic Force
Motional EMF
+
R
~
ve
−
~
v
~
B
The conducting bar on rails acts like a
battery, where the source of emf is the
magnetic force on the charge carriers:
emf = vBL. This is known as motional
emf.
Magnetic Force
Motional EMF
+
R
~
ve
−
~
v
~
B
The conducting bar on rails acts like a
battery, where the source of emf is the
magnetic force on the charge carriers:
emf = vBL. This is known as motional
emf.
~ ext is required to
An external force F
maintain a constant bar speed |~v | because
the induced current I generates an
~ B = ILB.
opposing force F
+
I
~
Fext
R
~
v
−
L
~
B
Magnetic Force
Motional EMF
+
R
~
ve
−
~
v
~
B
The conducting bar on rails acts like a
battery, where the source of emf is the
magnetic force on the charge carriers:
emf = vBL. This is known as motional
emf.
~ ext is required to
An external force F
maintain a constant bar speed |~v | because
the induced current I generates an
~ B = ILB.
opposing force F
+
I
~
Fext
R
~
v
L
~
B
−
Exercise: Use the energy principle to show that the power
~ ext is equal to the power dissipated by the resistor.
generated by F
Magnetic Force
Motional EMF
w
~
B
−
+
~
v
− V +
Example: As shown, a bar on rails is moving
downward with speed |~v | in a uniform magnetic
~ What is the reading on the voltmeter?
field B.
The + and − refer to the positive and negative
terminals of the voltmeter, respectively.
Magnetic Force
~ and Inertial Reference Frames
B
An inertial reference frame is a coordinate system that is moving
with respect to other coordinate systems. For example: the inertial
reference frame inside an airplane is moving with respect to the
earth.
Magnetic Force
~ and Inertial Reference Frames
B
An inertial reference frame is a coordinate system that is moving
with respect to other coordinate systems. For example: the inertial
reference frame inside an airplane is moving with respect to the
earth.
A charged object observed from within its own inertial frame (i.e.,
at rest) will be seen to produce only an electric field. Seen from an
inertial frame moving relative to it, the charged object will produce
both electric and magnetic fields.
Magnetic Force
~ and Inertial Reference Frames
B
Lab Frame
+e1
~
B1
v
r̂
~
F21
B
v
+e2
~
E
~
F21
E
Given protons moving in parallel, the net force on e2
due to e1 (in the Lab frame) is
2
2
v
e
1
~ net =
1 − 2 (−ŷ ) ,
F
4π0 r 2
c
where µ0 0 = 1/c 2 and the ratio FB /FE = v 2 /c 2 .
Magnetic Force
~ and Inertial Reference Frames
B
Lab Frame
v
+e1
r̂
~
F21
~
B1
B
v
+e2
~
E
~
F21
E
Given protons moving in parallel, the net force on e2
due to e1 (in the Lab frame) is
2
2
v
e
1
~ net =
1 − 2 (−ŷ ) ,
F
4π0 r 2
c
where µ0 0 = 1/c 2 and the ratio FB /FE = v 2 /c 2 .
Particle Frame
+e1
r̂
+e2
~
E
~
F21
E
Question: If the protons are moving through a
chamber, in which inertial frame (particle or lab) do the
protons strike the chamber walls “first?” How is this
possible?
Magnetic Force
~ and Inertial Reference Frames
B
Lab Frame
v
+e1
r̂
~
F21
~
B1
B
v
+e2
~
E
~
F21
E
Given protons moving in parallel, the net force on e2
due to e1 (in the Lab frame) is
2
2
v
e
1
~ net =
1 − 2 (−ŷ ) ,
F
4π0 r 2
c
where µ0 0 = 1/c 2 and the ratio FB /FE = v 2 /c 2 .
Particle Frame
+e1
r̂
+e2
~
E
~
F21
E
Question: If the protons are moving through a
chamber, in which inertial frame (particle or lab) do the
protons strike the chamber walls “first?” How is this
possible?
Question: Why do we put “first” in quotes?
Magnetic Force
Magnetic Torque
Wire Loop Viewed Edge On
~
B
~
F =IwB I
axle
h
µ
~ =I ~
A
~
B
I
A wire loop of area A = wh and current I has
~ = Iwh(Â). In a
magnetic moment µ
~ = IA
~
uniform magnetic field B, the wire loop
experiences a torque
~.
~τ = µ
~ ×B
~
F =IwB
~
The torque will act to align µ
~ and B.
Magnetic Force
Magnetic Torque
Wire Loop Viewed Edge On
~
B
~
F =IwB I
axle
h
µ
~ =I ~
A
~
B
I
A wire loop of area A = wh and current I has
~ = Iwh(Â). In a
magnetic moment µ
~ = IA
~
uniform magnetic field B, the wire loop
experiences a torque
~.
~τ = µ
~ ×B
~
F =IwB
~
The torque will act to align µ
~ and B.
Exercise: Draw in ~τ and the direction of rotation on the figure.
Can you think of a RHR relating ~τ to the direction of rotation?
Magnetic Force
Magnetic Torque
Wire Loop Viewed Edge On
~
B
~
F =IwB I
axle
h
µ
~ =I ~
A
~
B
I
A wire loop of area A = wh and current I has
~ = Iwh(Â). In a
magnetic moment µ
~ = IA
~
uniform magnetic field B, the wire loop
experiences a torque
~.
~τ = µ
~ ×B
~
F =IwB
~
The torque will act to align µ
~ and B.
Exercise: Draw in ~τ and the direction of rotation on the figure.
Can you think of a RHR relating ~τ to the direction of rotation?
~ determined?
Question: How is the direction of A
Magnetic Force
Magnetic Torque
Wire Loop Viewed Edge On
~
B
~
F =IwB I
axle
h
µ
~ =I ~
A
~
B
I
A wire loop of area A = wh and current I has
~ = Iwh(Â). In a
magnetic moment µ
~ = IA
~
uniform magnetic field B, the wire loop
experiences a torque
~.
~τ = µ
~ ×B
~
F =IwB
~
The torque will act to align µ
~ and B.
Exercise: Draw in ~τ and the direction of rotation on the figure.
Can you think of a RHR relating ~τ to the direction of rotation?
~ determined?
Question: How is the direction of A
~ in which direction must µ
Question: Relative to B,
~ point for the
loop to be in unstable equilibrium?
Magnetic Force
Potential Energy of Magnetic Dipole
Since a magnetic dipole in a uniform magnetic field experiences a
torque, and that torque acts through some angle (i.e., work is
~ the dipole must have some magnetic potential
done) until µ
~ k B,
energy.
Magnetic Force
Potential Energy of Magnetic Dipole
Since a magnetic dipole in a uniform magnetic field experiences a
torque, and that torque acts through some angle (i.e., work is
~ the dipole must have some magnetic potential
done) until µ
~ k B,
energy.
Wire Loop Viewed Edge On
µ
~
~
B
~
Fext
I
θ
~
FB
I
h
~
FB
Exercise: Calculate the minimum work
required to rotate the wire loop through an
angle θ to verify that the potential energy of a
magnetic dipole in a magnetic field is given by
~.
Um = −~
µ•B
Magnetic Force
Potential Energy of Magnetic Dipole
Since a magnetic dipole in a uniform magnetic field experiences a
torque, and that torque acts through some angle (i.e., work is
~ the dipole must have some magnetic potential
done) until µ
~ k B,
energy.
Wire Loop Viewed Edge On
µ
~
~
B
~
Fext
I
θ
~
FB
I
h
~
FB
Exercise: Calculate the minimum work
required to rotate the wire loop through an
angle θ to verify that the potential energy of a
magnetic dipole in a magnetic field is given by
~.
Um = −~
µ•B
Question: What is the maximum potential energy of the loop for
~
a given value of B?
Magnetic Force
Force on a Magnetic Dipole
~
FB
~
B
θ
I
~
B
S
N
µ
~
R
~
FB
Exercise: Using Biot-Savart, integrate the
~ B about the loop to obtain the net
force F
force on the loop dipole due to the bar
~ is known.
magnet. Assume B
Magnetic Force
Force on a Magnetic Dipole
~
FB
~
B
θ
I
~
B
S
N
µ
~
R
Exercise: Using Biot-Savart, integrate the
~ B about the loop to obtain the net
force F
force on the loop dipole due to the bar
~ is known.
magnet. Assume B
~
FB
Using the energy principle, we can show that the force on a
magnetic dipole µ
~ is given by
~ B = −∇UB = ∇(~
~ .
F
µ • B)
Magnetic Force
Force on a Magnetic Dipole
~
FB
~
B
θ
I
~
B
S
N
µ
~
R
Exercise: Using Biot-Savart, integrate the
~ B about the loop to obtain the net
force F
force on the loop dipole due to the bar
~ is known.
magnet. Assume B
~
FB
Using the energy principle, we can show that the force on a
magnetic dipole µ
~ is given by
~ B = −∇UB = ∇(~
~ .
F
µ • B)
Exercise: Prove the statement above for
the case where the loop dipole is moved a
distance ∆x by an external force against
the force of the magnetic field
I
S
N
~
B1
µ
~
R
∆x
~
B2
Magnetic Force
Motors and Generators
A simple generator consists of a wire loop in a magnetic field.
~
B
~
B
θ
~
v = hω
2
Exercise: Show that a loop of area A = hw
turning at angular speed ω will generate an emf
around the loop equal to
ω
h
emf = ωBA sin(ωt) .
Magnetic Force
Motors and Generators
A simple generator consists of a wire loop in a magnetic field.
~
B
~
B
θ
~
v = hω
2
Exercise: Show that a loop of area A = hw
turning at angular speed ω will generate an emf
around the loop equal to
ω
emf = ωBA sin(ωt) .
h
Exercise: An external force must be exerted in
order to make the generator turn, resulting in an
input power equal to
P=
as required.
dW
= IBAω sin θ = IV ,
dt
~
B
~
FB
~
Fext
~
Fext
θ
ω
~
FB =I ωB
h
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