Download Magnetic Field

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
1
‫‪ .2‬נחלק את התיל למספר קטעים ונרשום את הכוח לכל חלק‪.‬‬
‫עבור החלקים האופקיים נקבל‬
‫‬
‫ ‬
‫‪π‬‬
‫ˆ‪F = IL × B = ILB sin = I 6 RB = −0.075 Ry‬‬
‫‪2‬‬
‫את‬
‫הכיוון נקבל מכלל יד ימין‪.‬‬
‫עבור רבע המעגל נתבונן בכל רכיב של הכוח בנפרד‪:‬‬
‫‬
‫ ‬
‫‬
‫ ‬
‫‪π‬‬
‫‪F = IL × B ⇒ dF = IdL × B = IBdL sin‬‬
‫‪2‬‬
‫‪IB cos θ Rdθ = − IBR‬‬
‫‪2‬‬
‫‪3π‬‬
‫∫‬
‫‪π‬‬
‫‪IB sin θ Rdθ = − IBR‬‬
‫‪2‬‬
‫‪3π‬‬
‫∫‬
‫‪π‬‬
‫עבור הקטע האנכי נקבל‬
‫= ‪Fx = ∫ IBdL cos θ‬‬
‫= ‪Fy = ∫ IBdL sin θ‬‬
‫‬
‫ ‬
‫‪π‬‬
‫ˆ‪F = IL × B = ILB sin = I 2 RB = 2 IBRx‬‬
‫‪2‬‬
‫עבור החלק המשופע נמצא תחילה את זווית השיפוע‬
‫‪R‬‬
‫‪⇒ θ = 35.36‬‬
‫‪3R‬‬
‫= ‪sin θ‬‬
‫נמצא את רכיבי הכוח‪:‬‬
‫‪2‬‬
‫ˆ‪= − IBRx‬‬
‫‪3‬‬
‫‪Fx = IBL sin θ = IB 3R‬‬
‫‪1‬‬
‫ˆ‪= − 2 IBRy‬‬
‫‪3‬‬
‫‪Fy = IBL cos θ = IB 3R‬‬
‫בכיוון ‪ x‬נקבל שסכום הכוחות הוא ‪) 0‬זה הגיוני מפני שיש סימטריה בציר‬
‫התנועה האנכי של הזרם(‪ ,‬ובכיוון ‪ y‬נקבל שסכום הכוח הוא‬
‫‪( −7 − 2 ) IBR = −1.05R‬‬
Magnetic field
Submitted by: I.D. 200701894
The problem:
In the infinite wire flows a current I. Its mass for a length unit is µ. The magnetic field is
perpendicular to the plane of the page. Given that the wire levitates in this situation, find
1. What is the direction of the magnetic field?
2. What is the magnitude of the field?
3. The wire is rotated through the drawn line in angle α. What is the acceleration of the wire?
The solution:
1.
The force that acts on the wire is
~ ×B
~
F~ = I L
(1)
Since the wire levitates, the force due to the magnetic field should be into the upward direction.
Therefore, the magnetic field is into the ”‘inside of the page”’ direction.
2.
Writing the the second law for the forces per a unit length we get
µg = IB
µg
B =
I
(2)
(3)
3.
When the wire is rotated into the angle α (α is defined as the angle from the original direction of
the wire):
X
F = IB cos α − µg = µg cos α − µg = µg(cos α − 1)
(4)
a = g(cos α − 1)
(5)
1