Discussion
A point charge ‘q’ having mass ‘m’ follows a circular trajectory of radius ‘R’ under the action of magnetic field \(\vec{B}\). The magnitude of the angular momentum of the particle is :
Answer: D
The angular momentum is \(BqR^{2}\).
The magnetic force provides the centripetal force for circular motion:
\[ \begin{aligned} qvB &= \frac{mv^{2}}{R} \\ mv &= qBR \end{aligned} \]
The angular momentum about the centre of the circle is
\[ L = mvR = qBR \cdot R = BqR^{2} \]
It is not zero, because the particle moves in a circle about that centre. The options with \(q^{2}\) or with an extra m have the wrong dimensions; the mass cancels when mv is replaced by qBR.
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