$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then:
- A${V_A} - {V_O} = \frac{{B\omega {l^2}}}{2}$
- B${V_O} - {V_C} = \frac{9}{2}B\omega {l^2}$
- C${V_A} - {V_C} = 4B\omega {l^2}$
- D${V_C} - {V_O} = \frac{9}{2}B\omega {l^2}$
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