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An insulting rod of length l carries a charge q distributed uniformly on it. The rod is pivoted at its mid point and is rotated at a frequency f about a fixed axis perpendicular to the rod passing through the pivot. The magnetic moment of the rod system is-

Option: 1

\frac{1}{12} \pi q f l^2


Option: 2

\pi q f 1^2


Option: 3

\frac{1}{6} \pi q f 1^2


Option: 4

\frac{1}{3} \pi q f 1^2


Answers (1)

best_answer

At a distance x consider a small element of width dx. 

Magnetic moment of the small elements is-

\\\quad d M=I A=(d p) f \pi x^2=\frac{q}{l} d \times f \pi x^2 \\ M=\frac{q f \pi}{l} \int_{1 / 2}^{1 / 2} x^2 d x=\frac{q \pi f l^2}{12}

Posted by

Nehul

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