Question

I have a Magnitisim question

If a proton is traveling in helical trajectory due to a magnetic field in +Z and a velocity in the x and y direction? I am trying to Find the initial magnetic force on the proton.

Answer 1

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Answer 2

According to the right hand rule,
* thumb points in the direction of a moving positive charge's velocity, v
* fingers point in the direction of magnetic field, B
* palm faces in the direction of the magnetic force, F

So, if the magnetic field is along +Z direction, and velocity in x direction,
the force will be downwards i.e along -Y direction.
And if the magnetic field is along +Z direction, and velocity in y direction,
the force will be downwards i.e along +X direction.

Since the velocities are both in X and Y directions, the force will be in the direction of the 'resultant' of the two direction (i.e -Y and +X )

Answer 3

So, this is basically a case where there is uniform magnetic field in space in the same direction Z. What happens when a proton enters this field? The important point is that when a moving charge enters a magnetic field, it suffers a force \[F = q(v \times B)\] That has the following characteristics:

1. There will be no force in the direction of field. So, there will no change in the component of velocity in the direction of field. That is, the z-component of initial velocity will remain same.

2. The other 2 components of velocity (vx and vy) are perpendicular to the field. Whatever be resulting velocity in x-y plane, there will be a force that is always perpendicular to it. Whenever, there is a force that is always perpendicular to instantaneous-velocity, this leads to uniform circular motion. So, the particle will go in circles along x-y plane. The important part is:

*The speed of circular motion (ie, the magnitude of resultant of vx and vy) remains same throughout
*The axial force (qvB) will also continue to be same as at initial time.

3. So it continues its movement along the z-direction, while it goes in circles in x-y pane. Together, this gives rise to the helical trajectory.

So, the initial force you need is F = q (magnitude of resultant of vx and vy) B. ie, \[F = qB \sqrt{v _{x0}^{2} + v _{y0}^{2}}\]
where initial velocity is: \[v _{x0} + v _{y0} + v _{z0}\]