Question

A particle having a mass of 0.5 g carries a charge of $2.5\times {10}^{-8}C$. The particle is given an initial horizontal velocity of $6\times {10}^{4}m{s}^{-1}$. To keep the particle moving in a horizontal direction

• A. the magnetic field may be perpendicular to the direction of the velocity
• B. the magnetic field should be along the direction of the velocity
• C. magnetic field should have a minimum value of 3.27 T
• D. no magnetic field is required

Answer 1

Answer: (A), (C)

The correct options are
A the magnetic field may be perpendicular to the direction of the velocity
C magnetic field should have a minimum value of 3.27 T

In the absence of a magnetic field, the particle will experience gravitational force mg. As a result, the particle will not continue moving in the horizontal direction but will described a parabolic path. So, a magnetic field must be present and its direction must be perpendicular to the direction of the velocity. The magnetic force experienced by the particle is given by $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. The magnitude of the force is $F=qvBsin\theta$. If the particle is to move in the horizontal direction, this force must balance the force of gravity, i.e., $mg=qvBsin\theta$
The minimum value of B corresponds to $sin\theta =1$ or $\theta ={90}^o$.
Thus, $mg=qvB$

or $B=\frac{mg}{qv}=\frac{0.5\times {10}^{-3}\times 9.8}{2.5\times {10}^{-8}\times 6\times {10}^4}=3.27T$.