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 Coul; m = 9.1 x kg Proton: q = +e = 1.6 x Coul; m = 1.67 x kg Neutron: q = 0 Coul; m = 1.67 x kg Alpha: q = +2e = 3.2 x Coul; m = 6.68 x kg Gamma: q = 0 Coul; m(rest) = 0 kg *** for gamma, can’t vary speed, it is v = c = 3 x 10 8 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(  vB ) 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 90 o, so sin(  vB ) = 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(  particle ) = y/x, or  target = 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 = v 2 /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 10 7 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[x 2 + y 2 ] = 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:  v =  target +/- 90 o where  target = tan -1 (y/x).

+ or - ? Whether to choose the + sign or the - sign in  v =  target +/- 90 o where  target = 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[x 2 + y 2 ] = cm. If we choose the minimum circle method, r = d/2 = cm =.2444 m. From r = mv/qB, v = qBr/m = (1.6 x C) * (6 x T) * (.2444 m) / (9.1 x kg) = 2.58 x 10 7 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:  target = tan -1 (y/x) = tan -1 (42 cm / 25 cm) = 59.2 o. Because the force will push the electron to the right, we must aim 90 o to the left, which means we must use the plus sign:  v =  target +/- 90 o = 59.2 o + 90 o = o.