Download 9. Charges in motion in a magnetic field

9. Charges in motion in a magnetic field
Exercise 9.1:
.1: Equilibrium of forces.
A beam of protons is accelerated to a speed of 5 x 106 m/s in a particle
accelerator and emergesr horizontally from the accelerator into a uniform
magnetic field. What B -field perpendicular to the velocity of the proton
would cancel the force of gravity and keep the beam moving exactly
horizontally?
Given:
v = 5 ∗ 10 6 m s
q = + e = + 1 . 6 ∗ 10
− 19
FB
C
+e
Find:
r
r
r
B = ? so that F B + m g = 0
or equivalently: F B = mg
B
v
mg
(it always points
downward !)
r
Because the force of gravity m g points downward,
r
the vector F B should point upward.
r
According to the right-hand rule the vector B should go into the page
SOLUTION:
⇒
F B = mg
⇒
evB sin 90° = mg
⇒
mg 1.67 ∗ 10 −27 kg 9.8 m s 2
B=
=
= 2.04 ∗ 10 −14 Τ
−
19
6
ev 1.6 ∗ 10 C 5 ∗ 10 m s
(
(
)(
)(
)
)
Exercise 9.2
9.2:
.2: Cross electric and magnetic fields
A charged particle travels undeflected through perpendicular electric and
magnetic fields whose magnitude are 3000 N/C and 30 mT, respectively.
Find the speed of the particle if it is (a) a proton or (b) an alpha particle. (An
alpha particle has a charge q = + 2e)
Given:
Find:
FE
E = 3000 N C
B = 30 m Τ = 0 . 03 Τ
v
FB
v=?
B
(out of page)
SOLUTION:
Take an arbitrary positive charge, q , moving on a straight line
⇒
equilibrium of forces
r
r
FE + FB = 0 ⇒ FE = FB
where: FE = qE and FB = qvB sin 90°
r
Using the right-hand rule, we find a magnetic field B pointing out of page.
⇒
qE = qvB
⇒
E 103 V m
v= =
= 105 m s
B
0.03Τ
The formula for v is charge independent. (same answer for (a) and (b)).
Exercise 9.3:
.3: Technique to identity a particle.
particle.
A single charged ion is moving on a straight path through perpendicular
magnetic and electric fields whose magnitude are 0.10 T and 1000 N/C,
respectively. Suddenly, the electric field is turned off and the ion is moving
on a circular path of radius 1.2 cm, due to the action of the magnetic field,
only. Find the mass of the ion.
This is a spectrometer device able to identify charged particles.
Given:
B = 0 . 10 T
E = 10 3 V m
R = 1 . 2 cm = 0 . 012 m
Find:
m=?
SOLUTION:
Use the result from the example 9.2:
r
r r
In cross fields, when a charged particle has v ⊥ E and B
E
simultaneously, the speed of the particle is always v = .
B
Remember:
In our case
E 103 V m
v= =
= 10 4 m s
B
0.1Τ
r
When the charged particle moves inside a magnetic field B , the path of the
charge is curved under the action of a magnetic force:
FB = qvB sin 90° = evB
(for a single charged ion q = e )
r
v2
FB creates a centripetal acceleration
.
R
v2
According to Newton’s 2 law: qvB = m
, from where we can solve for
R
qBR eBR
the mass of the particle: m =
=
= 1.9 ∗ 10 −26 kg .
v
v
nd