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

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.