value of magnetic field

Question

A particle of mass m and charge q is dropped  from a height h . In whole space there exists a constant magnetic field B inside the plane . prove that the value of B in tesla should be 2 such that the particle just grazes the ground . 

(h=20m , m=2Kg , q=1C and g= 10m/s²

Answer 1

Hi Samagra!

 

   I have not come up with a full solution to this yet, but I can tell you the most fruitful inroads I have made so far, and see if I can think of any more later, or if this spurs any ideas with you that may help you solve the problem.

 

   I found the path the worked the best for me was to look at the situation in terms of Newton's 2nd law of motion in the horizontal and vertical directions.  This becomes a differential equation which we can try to solve.

 

   In the x direction, the only force acting on the particle is whatever component of the magnetic force points in the x direction at that moment.  This yields:

 

m(dv_x/dt) = (F_B)_x = (qvB)_x

 

To define the x direction component, I chose to define an angle between the velocity vector at a given instant and the vertical axis.  I called this angle α.  One of the few things that is consistent in this problem is that the magnetic force must always be perpendicular to velocity, pointing in the direction that points away from the angle the velocity is making with the vertical (since it is what is causing the velocity vector to pull away from the vertical).  Because of this, the magnetic force vector must make an angle α with the horizontal.

 

Because of this, (F_B)_x = qvBcosα

 

Because q and B are two of the constants in the problem, I pull them to the front, rearranging this to:

 

qBvcosα 

 

And, because α is the angle between the velocity and the vertical, vcosα = v_y (at least, in magnitude)

 

Putting all this together, we have (F_B)_x = qBv_y , which leads to:

 

m(dv_x/dt) = qBv_y

 

By a similar argument, I get, from Newton's 2nd law in the y direction:

 

m(dv_y/dt) = qBv_x - mg

 

The way that I then tried to work with these was to differentiate one of these equations with respect to time, and then substitute.  For example:

 

m(d²v_x/dt²) = qB(dv_y/dt)

 

Then I would slightly algebraically rearrange the y direction equation and substitute in to get:

 

(d²v_x/dt²) = (qB/m)²v_x-(qB/m)g

 

Then, I would attempt to look up a solution to this ordinary differential equation, knowing I had to satisfy some conditions in the solution:

 

1.  That the initial x and y velocity at t = 0 had to be 0

2.  That, at some later point, the particle's velocity in the y direction would become 0, and it would have only x direction velocity for an instant, when it reached y = 0, after starting at a height y = h.  This corresponds to the particle just grazing the ground after being dropped from h.

 

To sketch out the following steps as I see them (even though I was not able to get them to work out yet):  once we have a formula for v_y(t), you could solve it for t when v_y=0, which should give a solution at the start and at time t_y later, when the particle is momentarily not moving vertically.  by integrating v_y(t) with respect to time, you would get a function for y(t), knowing that at t = 0, you have the particle at height h.  Then put in the later time when v_y(t) = 0 into the y(t) equation, and set y(t_y) = h (or -h, depending on your chosen origin), since the particle will be just at the ground at t = t_y..  Then solve for the value of B that makes y(t_y) = h. 

 

However, I may have been making a conceptual or mathematical error (or several), because I was not able to get a solution from this.  It could be as simple as a sign error.  Also, perhaps there is a more straightforward way that I am missing.  In any case, I will give it some more thought, as it is a very interesting and challenging problem.

 

Also, let me know if you have any questions about what I have posted so far.  Sorry it is not worked further toward a solution!