JEE PYQ: Magnetism And Matter Question 18
Question 18 - 2019 (10 Jan 2019 Shift 1)
An insulating thin rod of length $l$ has a linear charge density $\rho(x) = \rho_0 \frac{x}{l}$ 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:
(1) $\pi,n,\rho,l^3$
(2) $\frac{\pi}{3},n,\rho,l^3$
(3) $\frac{\pi}{4},n,\rho,l^3$
(4) $n,\rho,l^3$
Show Answer
Answer: (3) $\frac{\pi}{4},n,\rho,l^3$
Solution
$dM = \frac{\omega}{2\pi} \cdot \frac{\rho_0}{l} \cdot \pi x^2,dx$, integrating: $M = \frac{\pi}{4} n \rho l^3$
BETA
