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A thin circular ring of area A is held p...

A thin circular ring of area A is held perpendicular to a uniform field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is _____

A

`(BR)/A`

B

`(AB)/R`

C

ABR

D

`(B^2A)/R`

Text Solution

Verified by Experts

The correct Answer is:
B
When the area of the ring is A, the flux linked it is `phi_1`=AB, AB, when its area becomes zero the flux linked with it is zero.
`therefore` The change in flux associated with ring is `Deltaphi=phi_2-phi_1=-AB`
Now according to Faraday.s law `epsilon=-(Deltaphi)/(Deltat)`
`therefore IR=(AB)/(Deltat) " " [because epsilon=IR]`
`therefore Q/(Deltat) . R = (AB)/(Deltat) " " [because I=Q/(Deltat)]`
`therefore Q=(AB)/R`
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