Transcript P2magdefl

Deflections
We have already done programs on
Gravitational Deflection (trajectories PHYS 201, Vol. 1, #6) and Electric
Deflection (cathode ray tube, Vol. 3, #3).
Since we have introduced a third basic
force, magnetism, we will try to do a
Magnetic Deflection program (Vol. 4, #1).
Gravitational and Electric
In Gravitational Deflection, we had the force
of gravity (and the gravitational field)
directed down. It was a constant force (and
field) in a constant direction (down).
In Electric Deflection, we first accelerated a
charged particle, then we deflected it either
up or down with an electric field. This field
and force was also constant in magnitude
and constant in direction (either up or
down).
Magnetic Deflection
In the Magnetic Deflection, we will employ
again a field that is constant both in
magnitude and in direction. However,
because of the strange nature of the
magnetic force, the direction of the
magnetic force will not be constant (even
though the magnitude of the force does
remain constant)!
Magnetic Deflection
Our target is somewhere on the screen, with the
magnetic field either directed into  or out of 
the screen.
Our job, as in the gravitational deflection program, is
to indicate the magnitude and direction of the
velocity of our projectile so that it will hit the
target.
In the gravitational deflection program, we were
given the mass. In the magnetic deflection
program we are given not only the mass but also
the charge.
Particles
(These are given in the Introduction to the program.)
Electron: q = -e = -1.6 x 10-19 Coul; m = 9.1 x 10-31 kg
Proton: q = +e = 1.6 x 10-19 Coul; m = 1.67 x 10-27 kg
Neutron: q = 0 Coul;
m = 1.67 x 10-27 kg
Alpha: q = +2e = 3.2 x 10-19 Coul; m = 6.68 x 10-27 kg
Gamma: q = 0 Coul;
m(rest) = 0 kg
*** for gamma, can’t vary speed, it is v = c = 3 x 108 m/s
Basic Considerations
To start, we must consider the magnetic field
and how it exerts a magnetic force on our
particle: F = q v B sin(qvB) with the force
direction perpendicular to both v and B.
Any velocity component parallel to B will not
have a force associated with it. In our case,
this means that we do NOT want any
component of velocity to be into or out of
the screen (parallel or anti-parallel to B).
Basic Considerations
Since we must keep v in the plane of the
screen, we can see that the angle between v
and B will always be 90o, so sin(qvB) = 1.
Next, if we have no charge, there will be no
force! That means, if our projectile is
neutral, all we have to do is shoot it
directly at the target. The speed does not
matter. To determine this direction, we
simply use trig: tan(qparticle) = y/x, or
qtarget = tan-1 (y/x) .
Circular Motion
If our particle does have a charge, the
magnetic force will be perpendicular to the
magnetic field (and so be in the plane of the
screen) and the force will be perpendicular
to the velocity. This will cause the particle
to bend (go in a circle): F = qvB, F = ma,
where a = v2/r; putting these together give:
r = mv/qB .
Alternatives
As in the gravitational deflection problem,
there are many ways to hit the target. We
can choose different angles and then
determine the speed needed to hit the target
using that angle, or vice versa.
And as in the gravitational deflection
problem, there are limits to the angle we can
choose.
Limits
r = mv/qB
If we make the speed huge, r also becomes huge. A
huge r makes the circle so big that any small part
of that circle looks to be fairly straight. This
would make aiming the particle easy - just like it
was for no charge. However, we cannot
physically make the speed of a particle that huge.
The program takes this into account by limiting
the speed to about 1/10 the speed of light ( max
speed allowed is 3 x 107 m/s).
Limits
r = mv/qB
If we make the speed too small, the
radius will not be big enough so that
the particle will not go far enough to
reach the target.
A minimum size circle would have to
have a diameter equal to the distance
the target is from the firing point:
d = SQR[x2 + y2] = 2r .
But which way to head?
d
v
v
Direction for minimum circle
In circular motion, the force is perpendicular to the
velocity. The force is directed toward the center,
and the velocity is directed tangent to the circle.
In our case, to make the particle go in a circle of
minimum radius to hit the target, we must choose a
velocity, v, to fit: r = mv/qB = d/2.
We must choose a direction for the velocity that is
perpendicular to the direction to the target:
qv = qtarget +/- 90o where qtarget = tan-1 (y/x).
+ or - ?
Whether to choose the + sign or the - sign in
qv = qtarget +/- 90o where qtarget = tan-1 (y/x)
depends on which way the field is - which
affects which way the force is.
We need to use the right hand rule to
determine the direction of the force. We
must then direct the velocity so that the
force will bend the particle into the target
rather than away from it.
Example
An electron is to be fired in a magnetic field of
strength 6 gauss directed into the screen so that it
hits a target located at: x = +25 cm; y = + 42 cm.
The distance to the target is: d = SQR[x2 + y2] =
48.88 cm. If we choose the minimum circle
method, r = d/2 = 24.44 cm = .2444 m.
From r = mv/qB, v = qBr/m = (1.6 x 10-19 C) * (6
x 10-4 T) * (.2444 m) / (9.1 x 10-31 kg) =
2.58 x 107 m/s.
Example - direction
Since the field is directed into the screen, a velocity
directed up (+y) would experience a force to the
left (-x) if it were a positive charge. Since we
have an electron which has a negative charge, the
magnetic force would be to the right (+x).
The target is to the right and above. The direction
straight to the target is:
qtarget = tan-1 (y/x) = tan-1(42 cm / 25 cm) = 59.2o.
Because the force will push the electron to the right,
we must aim 90o to the left, which means we must
use the plus sign:
qv = qtarget +/- 90o = 59.2o + 90o = 149.2o .