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Consider the situation shown in the figu...

Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA is made to rotate with a uniform angular speed `(omega)` as shown in the figure. Find the current in the rod when `-AOC = 90^(@)`

A

`5/3 (Ba^(2)omega)/(R)`

B

`8/3 (Ba^(2)omega)/(R)`

C

`(Ba^(2)omega)/(3R)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
The two resistance `r/4` and `(3r)/(4)` are in parallel.
Equivalent resistance `R'=((r)/(4)xx (3r)/(4))/(r ) = (3r)/(16)`
`e=Bvl=B xx(a)/(2)omega xx a =(Ba^(2)omega)/(2)`
`e/(R')=(Ba^(2)omega)/(2R') =(Ba^(2)omega)/(2 xx 3r//16) =(Ba^(2)omega 16)/(2 xx 3r)=(8 Ba^(2)omega)/(r)`.
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