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Shown in the figure is a circular loop o...

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction `B=B_(0)e^(-t)` is established inside the coil. If the key (K) is closed, the electric power developed right after closing the switch is equal to:-

A

`(B_(0)^(2)pir^(2))/(R )`

B

`(B_(0)pi r^(2))/(R )`

C

`(B_(0)^(2)pi r^(4)R)/(5)`

D

`(B_(0)^(2)pi^(2)r^(4))/(R )`

Text Solution

Verified by Experts

The correct Answer is:
D
Then induced emf E =`(dphi)/(dt)`
`=(d)/(dt) (B.A) = A (dB)/(dt)`
`= (pir^(2))B_(0) (d)/(dt) (e^(-t)) = -pir^(2) B_(0) e^(-1)`
`implies E_(0)= B_(0) pi r^(2) e^(-)I_(tau=0) = B_(0)pi r^(2)`
`:.` The electrical power developed in the resister just at the instant of closing the key = P
`(E_(0)^(2))/(R ) = (B_(0)^(2)pi^(2) r^(4))/(R ) `
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