JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
An insulating thin rod of length \(l\) has a linear charge density \(\rho \left( x \right) = {\rho _0}\,\frac{x}{l}\) on it. The rod is rotated about an axis passing through the origin \((x = 0)\) and perpendicular to the rod. If the rod makes \(n\) rotations per second, then the time averaged magnetic moment of the rod is
- A \(\pi n\rho {l^3}\)
- B \(\frac{\pi }{3}n\rho {l^3}\)
- C \(\frac{\pi }{4}n\rho {l^3}\)
- D \( n\rho {l^3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi }{4}n\rho {l^3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{dM} =\mathrm{di} \mathrm{A} \) \(=\left(\frac{\mathrm{d} \mathrm{q} \omega}{2 \pi}\right) \pi \mathrm{x}^{2} \) \(=(\rho \mathrm{d} \mathrm{x}) \frac{\omega}{2 \pi} \pi \mathrm{x}^{2} \) \(\mathrm{M} =\int_{0}^{\mathrm{L}} \mathrm{dM}\)
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