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9/21/2019
1
Electromagnetics:
Electromagnetic Field Theory
Lorentz Force Law
Outline
• Forces due to magnetic fields
• Example #1 – velocity filter
• Example #2 – deriving particle position
Slide 2
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Forces Due to Magnetic Fields
Slide 3
Mechanisms to Produce Force
• Force on a moving charged particle
• Force on a current element
• Force between two current elements
Slide 4
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Lorentz Force Law
Slide 5
F QE Qu B
• Force �⃗� in same direction as 𝐸.• Force is independent of velocity.• Can change velocity of the charge.• �⃗� is usually much greater than �⃗�except at very high velocity.
• Force �⃗� is perpendicular to 𝐵 and velocity 𝑢.• Force is only exerted on moving charges.• Cannot change kinetic energy of the charge (i.e. �⃗� · 𝑑ℓ 0).
• Magnetic field can only change the direction of velocity.
Electrostatic Force
eF QE Magnetostatic Force
mF Qu B
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6pop
dx t
dt
Recall 𝐹 𝑚𝑎
Slide 6
duF m QE Qu B
dt
This equation is used to calculate total force on a charge as well as its position, velocity, and acceleration.
5
5crackle
dx t
dt
4
4snap
dx t
dt
3
3jerk
dx t
dt
2
2acceleration
dx t
dt
velocitydx t
dt
positionx t
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Animation of Magnetic Force
Slide 7
Example #1: Velocity Filter
Slide 8
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Example #1 – Velocity Filter
Slide 9
Suppose we wish to create a “bandpass” velocity filter where the system only passes charges moving with velocity 4 m/s and deflects all others.
0F
Example #1 – Velocity Filter
Slide 10
Without any applied field, charges flow in a straight line and charges of all velocities are passed by the aperture.
0F
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Example #1 – Velocity Filter
Slide 11
Now apply an electric field in the vertical direction.
Let this be 𝐸 20 V/m 𝑎 .
F QE
E
x
y
z
Example #1 – Velocity Filter
Slide 12
All charges experience the same force in the downward direction.
Particles spread according to velocity due to inertia, not due to a velocity dependent force.
F QE
E
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Example #1 – Velocity Filter
Slide 13
It is desired to calculate the magnetic field who’s force will exactly oppose the force due to the electric field at the design velocity.
Let this be 𝐵 𝐵 𝑎 .
F Qu B
0B B a
x
y
z
Example #1 – Velocity Filter
Slide 14
With just the magnetic field applied, all charges experience a force in the upward direction, but that force is proportional to the velocity.
Deflection appears more uniform because the inertia and velocity dependent force counteract each other.
F Qu B
B
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Example #1 – Velocity Filter
Slide 15
The net force should be zero at the design velocity.
F QE Qu B
and BE
0 QE Qu B
E u B 0ˆ ˆ ˆ20 V m 4 m sx z ya a B a 0ˆ ˆ20 4x xa B a
20 5 Wb mB
Example #1 – Velocity Filter
Slide 16
The forces due to the fields counteract each other such that the charges moving at the design velocity experience zero net force and are not defected.
F QE Qu B
and BE
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Example #2:Deriving Particle
Position
Slide 17
Example #2 – Particle Position
Slide 18
A particle with charge 2.0 mC and mass 8 mg is moving at a velocity of 10 m/s in the
positive x direction in the presence of a static magnetic field of 𝐵 4𝑎 Wb/m . If the particle is at the origin at time t = 0, what is the particle’s position at t = 3 s?
Solution
duF m Q E u B
dt
Write the Lorentz force equation and include .du dt
Solve this for .du dt
du QE u B
dt m
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Example #2 – Particle Position
Slide 19
Expand the cross product.
du QE u B
dt m
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆx x y y z z x x y y z z y z z y x z x x z y x y x y z
d Qu a u a u a E a E a E a u B u B a u B u B a u B u B a
dt m
ˆ ˆ ˆ ˆ ˆ ˆyx zx y z x y z z y x y z x x z y z x y x y z
dudu du Q Q Qa a a E u B u B a E u B u B a E u B u B a
dt dt dt m m m
We can extract three scalar equations from this one vector equation.
Example #2 – Particle Position
Slide 20
Expand the cross product.
du QE u B
dt m
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆx x y y z z x x y y z z y z z y x z x x z y x y x y z
d Qu a u a u a E a E a E a u B u B a u B u B a u B u B a
dt m
ˆ ˆ ˆ ˆ ˆ ˆyx zx y z x y z z y x y z x x z y z x y x y z
dudu du Q Q Qa a a E u B u B a E u B u B a E u B u B a
dt dt dt m m m
We can extract three scalar equations from this one vector equation.
xx y z z y
du QE u B u B
dt m
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Example #2 – Particle Position
Slide 21
Expand the cross product.
du QE u B
dt m
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆx x y y z z x x y y z z y z z y x z x x z y x y x y z
d Qu a u a u a E a E a E a u B u B a u B u B a u B u B a
dt m
ˆ ˆ ˆ ˆ ˆ ˆyx zx y z x y z z y x y z x x z y z x y x y z
dudu du Q Q Qa a a E u B u B a E u B u B a E u B u B a
dt dt dt m m m
We can extract three scalar equations from this one vector equation.
xx y z z y
du QE u B u B
dt m
yy z x x z
du QE u B u B
dt m
Example #2 – Particle Position
Slide 22
Expand the cross product.
du QE u B
dt m
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆx x y y z z x x y y z z y z z y x z x x z y x y x y z
d Qu a u a u a E a E a E a u B u B a u B u B a u B u B a
dt m
ˆ ˆ ˆ ˆ ˆ ˆyx zx y z x y z z y x y z x x z y z x y x y z
dudu du Q Q Qa a a E u B u B a E u B u B a E u B u B a
dt dt dt m m m
We can extract three scalar equations from this one vector equation.
xx y z z y
du QE u B u B
dt m
zz x y x y
du QE u B u B
dt m
yy z x x z
du QE u B u B
dt m
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Example #2 – Particle Position
Slide 23
Substitute our known values into these equations to simplify them.
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2
3
2 10 C0 4 Wb m 0
8 10 kgy zu t u t
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3
2 10 C0 0 0
8 10 kgx yu t u t
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2
3
2 10 C0 0 4 Wb m
8 10 kgz xu t u t
xx y z z y
du t QE u t B u t B
dt m
zz x y x y
du t QE u t B u t B
dt m
yy z x x z
du t QE u t B u t B
dt m
yu t
0
xu t
Example #2 – Particle Position
Slide 24
Write the differential equation for the x component of velocity.
xy
du tu t
dt
yx du tdu td
dt dt dt
Differentiate with respect to time.
yx
du tu t
dt
We can replace duy/dt using the differential equation for the y component.
xx
du tdu t
dt dt
2
20x
x
d u tu t
dt
Put differential equation in standard form. Solve the differential equation.
cos sinxu t A t B t
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Example #2 – Particle Position
Slide 25
Substitute the solution back into the original differential equation for the x component to get the solution for uy(t).
xy
du tu t
dt
cos sin y
dA t B t u t
dt
sin cos yA t B t u t
Write the differential equation for the z component of velocity.
0zdu t
dt
zu t C
This is solved by integrating.
Example #2 – Particle Position
Slide 26
Apply the boundary conditions to calculate the constants A, B, and C.
zu t C
At time t = 0, . ˆ0 10 m s xu a
cos sinxu t A t B t 0 10 m s cos 0 sin 0xu A B A
10 m sA
0 0zu C 0C
Our equations for the components of velocity are now
0zu t
10cosxu t t
10sinyu t t
sin cosyu t A t B t 0 0 sin 0 cos0yu A B B
0B
ˆ ˆ10cos 10sinx yu t t a t a
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Example #2 – Particle Position
Slide 27
Position as a function of time is calculated by integrating the expression for velocity.
ˆ ˆ10cos 10sinx yu t t a t a
ˆ ˆ10cos 10sinx y
dr tu t t a t a
dt
ˆ ˆ10sin 10cosx yr t t A a t B a
The constants A and B are determined by applying the initial condition.
ˆ ˆ0 0 10sin 0 10cos 0x yr A a B a ˆ ˆ10x yAa B a
0A 10B The final expression for position is
ˆ ˆ10sin 10 cos 1x yr t t a t a
Example #2 – Particle Position
Slide 28
We can finally evaluate the position at t = 3 s.
ˆ ˆ10sin 10 cos 1x yr t t a t a
ˆ ˆ3 10sin 3 10 cos 3 1x yr a a
ˆ ˆ3 1.41 19.90x yr a a
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Example #2 – Particle Position
Slide 29
ˆ ˆ10sin 10 cos 1 0x yr t t a t a t
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