Question

A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field $\overrightarrow{B}$ directed into the paper. AO = l and OC = 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. $V_A-V_C=4B\omega l^2$
  
  
• D. $V_C-V_O=\frac{9}{2}B\omega l^2$
  

Answer 1

The correct option is C $V_A-V_C=4B\omega l^2$



By using $e=\frac{1}{2}B{l}^2\omega$
For part AO ; $e_{OA}=e_O-e_A=\frac{1}{2}B{l}^2\omega$
For part OC; $e_{OC}=e_O-e_C=\frac{1}{2}B(3l{)}^2\omega$
$\therefore e_A-e_C=4B{l}^2\omega$