Question

A conducting rod AC of length $ 4l $ is rotated about $ _{B}^{a}\to $ point $O$ in a uniform magnetic field $B$ directed into the paper. $ AO=l $ and $ OC=3l $ then :

• A. $ VA-Vo=\frac{B\omega l^{2}}{2} $
• B. $ Vo-Vc=\frac{9}{2}B\omega l^{2} $
• C. $ V_A-V_o= {{\beta \omega l^2}^{2}} $
• D. $ V_c-V_o=\frac{9}{2} \omega l^{2} $

Answer 1

Answer: (C)

$V_{O}-V_{A}=\frac{B \omega l}{2} \times l$
$=\frac{B \omega l^{2}}{2}$
$V_{O}-V_{C}=\frac{B \omega 3 l}{2} \times 3 l$
$=\frac{9 B \omega l^{2}}{2}$
Hence, $V_{A}-V_{C}=4 B \omega l^{2}$
or $V_{A} > V_{C}$