Question

A plane metallic sheet is placed with its face parallel to lines of the magnetic induction $ B $ of a uniform field. A particle of mass $ m $ and charge $ q $ is projected with a velocity $ v $ from a distance $ d $ from the plane normal to the lines of induction. Then the maximum velocity of projection for which the particle does not hit the plane is,
(A) $ \dfrac{{2Bqd}}{m} $
(B) $ \dfrac{{Bqd}}{m} $
(C) $ \dfrac{{Bqd}}{{2m}} $
(D) $ \dfrac{{2Bqm}}{d} $

Answer 1

Hint: In the diagram, the bottom line represents the plane metallic sheet that is faced parallel to the magnetic field lines. If a charge is projected upwards, the magnetic force acts on it and deflects it in such a way that it follows a circular path, the radius of this path depends on the velocity of the charged particles, therefore the maximum radius of this circle can be $ d $ .

Complete Step by step solution:
It is given in the question that,
The magnetic field strength is, $ B $
The mass of the particle is, $ m $
The charge of the particle is, $ q $
Let the velocity of the particle be $ v $ , and the distance from where it is projected be $ d $ .
Then this charge will experience a force in a direction perpendicular to both the direction of the magnetic field and the velocity of charge, which will cause it to follow a circular path.
This force is given by,
 $ F = qvB\sin \theta $
Where $ \theta $ is the angle between the velocity of the particle and the magnetic field lines, since it moves perpendicular to the field, the angle $ \theta $ is,
 $ \theta = \dfrac{\pi }{2} $
We know that, $ \sin \dfrac{\pi }{2} = 1 $
Putting this value in the formula of force on the charged particle,
 $ F = qvB \times 1 $
 $ \Rightarrow F = qvB $
This force acts as a centripetal force causing the circular motion of the charged particle, as a reaction, a centrifugal force acts on the charge as well,
This centrifugal force is given by,
 $ {F_c} = \dfrac{{m{v^2}}}{r} $
Where $ r $ is the radius of the circle.
We know that to avoid the circle touching the plane, the maximum radius of the circle should be equal to $ d $ .
Therefore, equating both forces,
 $ F = {F_c} $
 $ \Rightarrow qvB = \dfrac{{m{v^2}}}{d} $
Rearranging this equation, we get-
 $ v = \dfrac{{Bdq}}{m} $
Thus, option (B) is the correct answer.

Note:
A charged particle does not experience a force as long as it is on rest, however when it is moved with a velocity, it experiences a magnetic force due to the field around it. A charged particle can move and still not experience magnetic force only when it moves parallel with the magnetic field lines and $ \sin \theta $ becomes zero.