Download Lorenz Force

Lorenz Force
Submitted by: I.D. 303464531
The problem:
Proton with a velocity vx̂ comes in between two parallel metal plates positioned perpendicular to
the z axis. The distance between the plates is d. The upper plate is grounded and the lower one is
connected to a potential V .
~ that as a result of the presence of the two fields B
~ and E,
~
1. Can we find a magnetic field B,
~
the proton will continue its motion in a straight line? If so, find B.
2. Explain schematically what would happen if an electron with the same momentum would
enter between those conductor plates, while the field you have found is activated?
The solution:
1. The electric force is in the ẑ direction, so in order to cancel it, the magnetic field should be in
~ × B).
~ From the condition F~b = F~e ,we get
−ŷ direction (the Lorenz Force V
qV By = qEz
(1)
* We do not want forces in ŷ direction, so Bz has to be equal zero. Bx can get any value, since the
magnetic field in the direction of the motion makes no contribution to the force acting on q, hence
we can choose it zero.
Ez
~
B = 0,
,0
(2)
V
2. An electron’s mass is 1840 times smaller that one of a proton, so it’s velocity would be that much
bigger, so the magnetic force will be bigger and the electron will deviate in ẑ direction.
Because of the opposite charge of the electron, the electric field will act in −ẑ direction, so the
magnetic field was taken in −ŷ direction.
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Lorenz Force
Submitted by: I.D. 303464531
The problem:
~ The angle between
A particle with velocity ~v , mass m and charge q is moving in magnetic field B.
~ is θ, so the movement is circular. Find:
~v and B
1. the radius of the circle R.
2. the time period T
~ direction in time T .
3. the distance the particle moved in B
4. the equations of motion
The solution:
~ × B.
~ We’ll separate V
~ to it’s parallel and perpendicular to
1. The force on the particle is F~ = q V
~
B components
Vk = V cos θ
(1)
V⊥ = V sin θ
(2)
~ the movement is circular. The magnetic force FB balances the
In the plane perpendicular to B
centrifugal force
mV⊥2
= qV⊥ B
R
which leads to
mV sin θ
R=
qB
(3)
(4)
2.The time period is
T =
2πR
2πm
2π
=
=
ω
V⊥
qB
(5)
~ direction in time T is
3.the distance the particle moved in B
P = Vk T =
2πmV cos θ
qB
(6)
4.The equations of motion are
z(t) − z0 = ∆z = Vk t = V t cos θ
(7)
x(t) = ∆x = r cos(ωt)
(8)
y(t) = ∆z = r sin(ωt)
(9)
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Magnetic Field
Submitted by: I.D. 303464531
The problem:
Two ions of zinc isotopes 70 Zn and 68 Zn are accelerated with voltage V from point (0, 0, 0) in the x
direction. The field B is in the z direction and the ions have the same charge q. Find the distance
between points of impact of the isotopes on the y-axis.
The solution:
From the energy conservation we get that the potential energy the electron gets, equal to it’s kinetic
energy
mu2
2
= qV
r
u =
(1)
2qV
m
(2)
where u is the speed of the ion.
The motion of the particle in magnetic field is circular with radius R.
mu
R=
=
qB
2mV
qB 2
1
2
(3)
See the problem ”e 45 2 175” for the derivation.
The distance between the points of impact is
δ = D2 − D1 = 2 [R(m + ∆m) − R(m)]
(4)
where D is a diameter and m is the mass of the lighter ion (∆m m).
By the definition of a derivative we get
δ=2
∂R
R
∆m = ∆m
∂m
m
(5)
since
∆m
2
=
≈ 0.03
m
68
(6)
the distance will be
δ ≈ 0.03R
(7)
δ
≈ 0.03
R
(8)
or
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