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An insulating rod of length $λ$ carries a charge q distributed uniformly on it. The rod is pivoted at its midpoint and is rotated at a frequency ’ f ‘ about a fixed axis perpendicular to rod and passing through the pivot. If the magnetic moment of the rod system is $\frac{{πqfλ}^{2}}{2 x}$ then the value of x is

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answer is 6.

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Detailed Solution

At a distance x consider small element of width dx.
Magnetic moment of the small element is : $dm = \frac{(\frac{q}{ℓ} dx) ω}{2 π} \cdot {πx}^{2}$

$M = \int_{- ℓ / 2}^{ℓ / 2} \frac{qω}{2 ℓ} x^{2} dx = - l / 2 M = \frac{{qωℓ}^{2}}{12} = \frac{{qπfℓ}^{2}}{6}$

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