A point particle ‘P’ of mass m and charge q and another point particle ‘Q’ of mass 2 m and charge q are attached at the ends of a massless rod of length ‘ $l$ ’. The system is now rotated with uniform angular velocity ‘w’ about an axis passing through the centre of mass & perpendicular to the length of the rod. The magnetic moment of the system is

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$\frac{4 q l^{2} w}{9}$

$\frac{5 q l^{2} w}{18}$

$\frac{8 q l^{2} w}{9}$

$\frac{q l^{2} w}{9}$

answer is A.

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Detailed Solution

$M = i A = q \frac{w}{2 π} π r^{2} = \frac{q ω r^{2}}{2} = \frac{q ω}{2} ({(\frac{l}{3})}^{2} + {(\frac{2 l}{3})}^{2}) = \frac{5 q l^{2} w}{18}$

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