Question

A charged particle (q, m) enters a uniform magnetic field B (extends upto a length w) at right angles with speed v . The speed of the particle in magnetic field does not change. But it gets deviated in the magnetic field in a circular path and leaves the magnetic field with a angle theeta. Its lateral displacement is D.

Show that D= mv/qB(1-(1-(qbw/mv)^2)^1/2)  

Answer 1

Let us assume the cordinate axis for convinience as shown in figure

Figure below shows the path of the particle in the magnetic field.

To find the distance D

since radius is always perpendicular to the tangential velocity

now,   is a right angled triangle with

now,

So, from the figure

which is the required solution.

Note: here the radius of the path is greater than the width of the magnetic field.

If the radius was smaller than the width then the particle would have turned before reaching the end of the field and continue its circular path till it reaches the side from which it entered.

So, path in the field in this case is always semicircular in nature due to constant speed and force on the object.

Thank you.