# Positron Helix magnetic field, find pitch and radius

1. Oct 14, 2011

### JosephK

1. The problem statement, all variables and given/known data

A uniform magnetic field of magnitude 0.137 T is directed along the positive x axis. A positron moving at a speed of 5.40 106 m/s enters the field along a direction that makes an angle of θ = 85.0° with the x axis (see figure below). The motion of the particle is expected to be a helix.

(a) Calculate the pitch p of the trajectory as defined in figure.

(b) Calculate the radius r of the trajectory as defined in figure.

2. Relevant equations

vx = sqrt(vy^2 + vz^2)

R = (m (vx) )/ qB

T = (2 pi r ) / vx

3. The attempt at a solution

To find vx multiply v vector by sin(5 degrees). Plugging in values to radius. Answer is significantly wrong. To find pitch, p = (vx)T = 2 pi r.

2. Oct 15, 2011

### JosephK

I understand this problem now. Since the magnetic field is directed in the x direction, the magnetic force is not directed in the x direction. Thus, acceleration in the x direction is zero. And so, velocity in the x direction is constant. We obtain velocity in the x direction by multiplying velocity vector by cos 85 degrees. We now find the period. It follows that the period is the circumference of the circle divided by the velocity of the particle. Replacing the velocity by the equation v = qBr/m, the period is equal to ( 2 pi m ) / q B. Consequently, the pitch p is equal to the velocity in the x direction times T.

Now we solve part B. We assume that velocity in the y direction is equal to the velocity in the z direction. Then, velocity in the y direction is equal to the velocity vector times sin 85. Then, by equation, r = (v m) / qB, we find the radius.