Two Particles Each Of Mass M And Charge Q Are Attached To The Two Ends Of A Light Rigid Rod Of Length 2r. The Rod Is Rotated At Constant Angular Speed About A Perpendicular Axis Passing Through Its Centre. The Ratio Of The Magnitudes Of The Magnetic Moment Of The System And Its Angular Momentum About The Centre Of The Rod Is.


Let the angular speed be ω.

Thus the current due to the rotation of charges in circular path = \( 2 * \frac{q}{T}\)

= \( 2 * \frac{q\omega}{2\pi}\)

= \( \frac{q\omega}{2\pi}\)

Hence magnetic moment, M = iA =\( \frac{q\omega}{\pi} \pi R^{2}\)

= q ω R2

The angular momentum of the system of charges = L = 2 * m * ω R

Hence \( \frac{M}{L} = \frac{q}{2m}\)

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