A charged particle having its charge to mass ratio as $β$ goes in a conical pendulum of length $L$ making an angle $θ$ with vertical and angular velocity $ω .$ If a magnetic field $B$ is directed vertically downwards (see figure) :The magnetic force on the charge particle is towards centre of its circular trajectory.
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Rate of change of angular momentum of the particle about the point of suspension is not a constant vector.

$B = \frac{1}{β} [ω - \frac{g}{ω L \cos θ}]$

Angular momentum of the particle about the point of suspension remains constant.

If the direction of B were reversed maintaining same $ω$ and $L,$ then $θ$ will remain unchanged.

answer is A, D.

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Detailed Solution

$T \sin θ + F = m ω^{2} (L \sin θ) T \sin θ = m g \frac{m g}{\cos θ} \sin θ + q (ω L \sin θ) B = m ω^{2} L \sin θ B = \frac{m ω^{2} L \sin θ}{(ω L \sin θ) q} - \frac{m g \sin θ}{\frac{\cos θ}{(ω L \sin θ) q}} = \frac{m ω}{q} - \frac{m g}{q ω L \cos θ} = \frac{1}{β} [ω - \frac{g}{ω L \cos θ}]$