v20.2

A

50 V

B

25 V

C

100 V

D

200 V

Text Solution

Verified by Experts

The correct Answer is:
D
`e_(0_(1))=2pifNBA=100V`
`rArr2pifxx100xxBxxpixx(1)^(2)=100" "...(i)`
Total length = `100xx2pixx1=200pi`
Now, `r_(2)=2m`
no. of turn = `("Total length")/(2pir)=(100xx2pi)/(2pixx2)=50`
`e_(0_(2))=2pixxfxx50xxBxxpixx(2)^(2)`
`100/e_(0_(2))=2pixxfxx50xxBxxpixx(2)^(2)`
`100/(e_(0_(2)))=2/4rArre_(0_(2))=200V`.
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