JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 9)
An insulating thin rod of length $$l$$ has a linear charge density $$\rho \left( x \right)$$ = $${\rho _0}{x \over l}$$ 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 -
$${\pi \over 3}n\rho {l^3}$$
$${\pi \over 4}n\rho {l^3}$$
$$n\rho {l^3}$$
$$\pi n\rho {l^3}$$
Explanation
$$ \because $$ M = NIA
dq = $$\lambda $$dx & A = $$\pi $$x2
$$\int {dm} = \int {\left( x \right){{{\rho _0}x} \over \ell }} \,dx.\pi {x^2}$$
M = $${{n{\rho _0}\pi } \over \ell }.\int\limits_0^\ell {{x^3}.dx} = {{n{\rho _0}\pi } \over \ell }.\left[ {{{{L^4}} \over 4}} \right]$$
M = $${{n{\rho _0}\pi {\ell ^3}} \over 4}$$ or $${\pi \over 4}n\rho {\ell ^3}$$
dq = $$\lambda $$dx & A = $$\pi $$x2
$$\int {dm} = \int {\left( x \right){{{\rho _0}x} \over \ell }} \,dx.\pi {x^2}$$
M = $${{n{\rho _0}\pi } \over \ell }.\int\limits_0^\ell {{x^3}.dx} = {{n{\rho _0}\pi } \over \ell }.\left[ {{{{L^4}} \over 4}} \right]$$
M = $${{n{\rho _0}\pi {\ell ^3}} \over 4}$$ or $${\pi \over 4}n\rho {\ell ^3}$$
Comments (0)
