Question

A conducting rod $A C$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec{B}$ directed into the paper. $A O=I$ and $O C= 3l$. Then

• A. $V_{A}-V_{O}=\frac{B \omega l^{2}}{2}$
• B. $V_{O}-V_{C}=\frac{7}{2} B \omega l^{2}$
• C. $V_{A}-V_{C}=4 B \omega l^{2}$
• D. $V_{C}-V_{O}=\frac{9}{2} B \omega l^{2}$

Answer 1

Answer: (C)

By using $e=\frac{1}{2} Bl^{2} \omega$
For part $A O ; e _{ OA }= e _{0}- e _{ A }=\frac{1}{2} B l^{2} \omega$
For part $O C ; e _{ OC }= e _{ O }- e _{ C }=\frac{1}{2} B (3 l )^{2} \omega$
$\therefore e _{ A }- e _{ C }= 4 B l^{2} \omega$