CP2 ELECTROMAGNETISM LECTURE 18: CHARGED PARTICLES …harnew/lectures/EM-lecture18-handout.pdf ·...
Transcript of CP2 ELECTROMAGNETISM LECTURE 18: CHARGED PARTICLES …harnew/lectures/EM-lecture18-handout.pdf ·...
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CP2 ELECTROMAGNETISMhttps://users.physics.ox.ac.uk/∼harnew/lectures/
LECTURE 18:
CHARGED PARTICLESIN E & B FIELDS
Neville Harnew1
University of Oxford
HT 2020
1With thanks to Prof Laura Herz1
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OUTLINE : 18. CHARGED PARTICLES IN E & B FIELDS
18.1 Motion of charged particles in E and B fields
18.2 Example : the mass spectrometer
18.3 Example : magnetic lenses
18.4 Electrodynamics “before Maxwell”
18.5 Revisit Ampere’s Law
18.6 Fixing Ampere’s Law : displacement current
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18.1 Motion of charged particles in E and B fields
I Force on a charged particle in an E and B field :
F = q(
E︸︷︷︸+ v ×B︸ ︷︷ ︸)along E ⊥ to both v and B
I Newton second law provides equation of motion :
F = m a = m r = q ( E + v ×B)
I Will demonstrate with 2 examples :
1. Mass spectrometer
2. Magnetic lens
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18.2 Example : the mass spectrometer
Used for detecting small charged particles (molecules, ions) bytheir mass m .
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Stage A : The velocity filter
I The particle will pass throughboth slits if it experiences no netforce inside the filter
I The region has both E and Bfields
F = q ( E + v ×B) = 0
→ need E = −v×B→ v = |E||B|
(E ⊥ v&B)
I Will filter particles with v = |E||B|
and the spread ±∆v is given bythe slit width
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Stage B : The mass filterI This region has only a B field
m r = q r×B
with B =
00B
and r =
xyz
→
xyz
= qm
y B−x B
0
→ z = 0 → vz = constant (= 0)
I r2 = x2 + y2 = q2
m2
(x2 + y2
)︸ ︷︷ ︸B2
v2
I Circular motion in x − y plane with : r = qm v B
For circular motion r = v2
R → R = mvqB
I Since q and v are constant, then R ∝ m6
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Mass spectrometer summary
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18.3 Example : magnetic lensesI Magnetic lenses are used for focusing and collimating
charged particle beams. Used in electron microscopes,particle accelerators etc.
I Quadrupole lens : four identical coils aligned in z-direction.I Sum of 4 dipole fields : for small values of x , y close to the
axis of symmetry, Bx ∝ y , By ∝ x
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Quadrupole lens
I Along x-axis : only By componentI Along y -axis : only Bx componentI No z-component (symmetry)I Inside the lens, close to the z-axis
B =
k yk x0
where k is a constant
I Equation of motion F = q v ×B
m
xyz
= q
∣∣∣∣∣∣i j kx y z
k y k x 0
∣∣∣∣∣∣ = q k
−x zy z
x x − y y
I Assume particle travels at a small angle wrt the z-axis :→ x , y ≈ 0 → z = 0 → z = v = constant → z = v t
I Equations of motion in the x − y plane :x = − q
m k v x and y = qm k v y
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Quadrupole lens continued
I Equ. of motion : x = −α2x & y = α2y , where α =√
q k vm
I Solutions : x(t) = A sinα t + B cosα t
Solutions : y(t) = C sinhα t + D coshα t
where cosh x , sinh x = (ex ± e−x )/2I Boundary conditions :
At t = 0 → z = 0, x = x0 and x = 0, y = y0 and y = 0I Solutions become x(t) = x0 cos α
v z : focusing
Solutions become y(t) = y0 cosh αv z : de-focusing
(using t = z/v ) → x = 0 for αv z = π
2 + n π
I Focal points in x direction at fn = π2
√m vq k + n π
√m vq k
I Use lens pair with 90◦ angle for collimating a chargedbeam
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Quadrupole lens continued
The lens pulls the beam on-axis in x and removes particlesdeviating in y
fn = π2
√m vq k + n π
√m vq k
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Magnetic lens summary
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18.4 Electrodynamics “before Maxwell”
1. Gauss Law :∮S E · da = Qencl./ε0 → ∇ ·E = ρ/ε0
2. No magnetic monopoles :∮S B · da = 0 → ∇ ·B = 0
3. Faraday’s Law :∮` E · d` = − ∂
∂t
∮S B · da → ∇×E = −∂B
∂t
4. Ampere’s Law :∮B · d` = µ0 Iencl. → ∇×B = µ0 J
Time-varying B-fields generate E-fields. However , time-varyingE-fields do not seem to create B-fields in this version.Is there something wrong ?
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18.5 Revisit Ampere’s Law
I Ampere’s Law : → ∇×B = µ0 J
Apply Div : → ∇ · (∇×B)︸ ︷︷ ︸always zero
= µ0∇ · J︸ ︷︷ ︸not always zero !!
I Recall the continuity equation :∫S J · da = − ∂
∂t
∫ν ρ(ν) dν → ∇ · J = − ∂
∂t (ρ)
[ Current leaving volume ] = [Change of charge inside volume]
I Therefore Ampere’s Law in its current form violates thecontinuity equation (and hence charge conservation) !
I But this is not surprising since we derived Ampere’s Lawassuming that ∂
∂t (ρ) = 0
→ We have to “fix” Ampere’s Law !
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18.6 Fixing Ampere’s Law : displacement current
I Add a term to Ampere’s Law to make it compatible with thecontinuity equation :
I ∇ · J = − ∂∂t (ρ)
Apply Gauss Law ∇ ·E = ρ/ε0
→ ∇ · J = − ∂∂t (ε0∇ ·E) = −∇ · (ε0 ∂E
∂t )
→ ∇ · (J + ε0∂E∂t ) = 0
I Implies we need to add(ε0
∂E∂t
)to J in Ampere’s law.
∇×B = µ0
(J+ ε0
∂E∂t
)(ε0
∂E∂t
)is called the displacement current JD (but is
actually a time-varying electric field)I Time-varying E fields now generate B fields and vice
versa. Also satisfies charge conservation.15