Download Particle Accelerator

Particle Accelerator
Submitted by: I.D. 061016754
Let there be a ray of electrons with kinetic energy κ coming out of a particle accelerator. A distance
d away from the exit point is a wall which is at a right angle to the velocity of the electrons as they
come out.
1. Show that the electrons
q can be prevented from reaching the wall if you employ a magnetic
field of strength B ≥ 2mκ
, where m being the mass of an electron and e its charge.
e2 d2
2. What should be the direction of the magnetic field?
The solution:
Part 1:
1
κ = mv02
2
r
⇒ v0 =
2κ
m
(1)
~ is Lorentz force and (in the absence of an electrical field) is directed always perpendicular
F~ = q~v × B
to the velocity of the charged particle and also perpendicular to the magnetic field.
The Lorentz force would be strongest when the magnetic field is itself perpendicular to the velocity
of the particle and therefore that’s the direction we choose for the field.
Since we have chosen the magnetic field to be perpendicular to the initial velocity, it will remain
perpendicular to the velocity all through the motion of the particles and the particles will move in
a circle.
r
r
2κ
2κ
π
~ =e
F~B = q~v × B
B sin θ = e
B, where θ =
m
m
2
2
2
mv
mv0
m 2κ
2κ
FB = maR =
=
=
=
Rr
d
d m
d
r
2κ m
2κm
B =
=
ed 2κ
e2 d2
(2)
(3)
(4)
This kind of field would make sure that the electrons move in a circle which radius is d and, therefore,
the electrons won’t hit the wall. Any field in the same direction with greater strength would produce
a circle of smaller radius and would therefore be adequate as well.
1
r
B≥
2κm
e2 d2
(5)
Part 2:
As it was already mentioned in the previous section, the magnetic field would have to be perpendicular to the electrons’ initial velocity in order to make sure they don’t hit the wall (assuming
minimum field strength). In order to make the electrons move upwards, as shown in the drawing,
the field would have to be in the direction of the positive x-axis. This is due to the vector product
of the velocity and the magnetic field found in Lorentz force. Also keep in mind the negative charge
of the electrons.
2
‫א( הכוח המגנטי שמניע את הרכבת הוא הכוח שפועל על הציר שמחבר בין הגלגלים‪:‬‬
‫‬
‫ ‬
‫‪F = iL × B‬‬
‫אם הזרם זורם בציר בכיוון היוצא מהדף‪ ,‬על פי חוק יד ימין הכוח שמופעל על הציר הוא בכיוון ̂‪ . − y‬אם‬
‫לא אוהבים את החוק עם היד‪ ,‬ניתן לחשב את המכפלה הוקטורית‪:‬‬
‫‬
‫‬
‫‪F = iL( xˆ ) × Bzˆ = iLB (1,0,0) × (0,0,1) = iLB (− yˆ ) ⇒ F = iLB‬‬
‫מכאן ניתן לבודד את זרם הדרוש‪:‬‬
‫‪F‬‬
‫‪10.000 N‬‬
‫‪= 3.33 × 10 8 A‬‬
‫=‬
‫‪−6‬‬
‫‪LB 3m ⋅ 10 × 10 T‬‬
‫=‪i‬‬
‫ב(‬
‫)‬
‫‪2‬‬
‫(‬
‫‪P = i 2 R = 3.33 × 10 8 A × 1Ω = 1.11 × 1017 Watt‬‬
‫ג( ההספק שחישבנו בסעיף ב הוא האנרגיה שמתבזבזת על ההתנגדות של פסי הרכבת לכל ‪ 1Ω‬לכל‬
‫שנייה‪ .‬אם האנרגיה הזאת תהפוך לחום‪ ,‬אין חומר שיוכל להחזיק מעמד‪ .‬לכן נכון להיום הפרויקט הוא‬
‫לגמרי לא מציאותי‪ ,‬אבל מי יודע‪...‬‬
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