Question

A non conducting rod of length $l=20\text{cm}$ carries a charge $q=3\text{C}$ uniformly distributed on it. The rod is pivoted at one of its ends as shown in the figure and is rotated at an angular frequency $\omega =10\text{rad/s}$ about a fixed axis perpendicular to rod and passing through pivot. The magnetic moment of the rod system is:

• A. $500{\text{A cm}}^2$
• B. $1000{\text{A cm}}^2$
• C. $2000{\text{A cm}}^2$
• D. Zero

Answer 1

The correct option is C $2000{\text{A cm}}^2$

Magnetic moment of rotating rod pivoted at one of its ends is given by $$\mu =\frac{q\omega l^2}{6}$$From the data given in the question,

$\mu =\frac{3\times 10\times (0.2{)}^2}{6}$

$\Rightarrow \mu =\frac{30\times 0.04}{6}=0.2{\text{Am}}^2\text{or}2000{\text{A cm}}^2$

Hence, option (c) is the correct answer.