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A metallic rod of length I rotates at a...

A metallic rod of length I rotates at angular velocity `omega` about an axis passing through one end and perpendicular to the rod. If mass of electron is m and its charge is -e then the magnitude of potential difference between its two ends is

A

`(m omega^(2)l^(2))/((2e))`

B

`(m omega^(2)l^(2))/(e)`

C

`(m omega^(2)l)/(e)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A

When rod the centripetal acceleration of electron comes electric from electric field `E=(mromega^(2))/(e)`
Thus, `DeltaV=-int vec(E).dr=-underset(0)overset(l)(int)(mromega^(2))/(e)dr=(momega^(2)l^(2))/(2e)`
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