A conducting rod `AC` of length `4l` is rotate about a point `O` in a uniform magnetic field `vecB` directed into the paper. `AO=l` and `OC=3l`. Then

Options

A. `V_(A)-V_(O)=(B omegal^(2))/(2)`
B. `V_(O)-V_(C)=(9)/(2) B omegal^(2)`
C. `V_(A)-V_(C)=4Bomegal^(2)`
D. `V_(C)-V_(O)=(9)/(2)Bomegal^(2)`

Answer

C

Explanation

`V_(O)-V_(A)=(Bomegal)/(2)xxl=(B omegal^(2))/(2)`
`V_(O)-V_(C)=(Bomega3l)/(2)xx3l=(9Bomegal^(2))/(2)`
Hence, `V_(A)-V_(C)=4B omegal^(2)` or `V_(A) gt V_(C)`