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.

Sol:

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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