the uniform magnetic field perpendicualr to the plane of a conducting ring of radius a change at the rate of `alpha`, then

Options

A. (a) all the points on the ring are at the same potential
B. (b) the emf induced in the ring is `pia^(2) alpha`
C. ( c) electric field intensity `E` at any point on the ring is zero
D. (d) `E = (a alpha)//2`

Answer

A, B, D

Explanation

(a, b, d):
`phi =pia^(2)B`
`e = pia^(2)(dB)/(dt) = pia^(2)alpha`
`E2pia = e` `rarr` `E = (pia^(2)alpha)/(2pia) = (alpha a)/(2)`
Let `R` bethe resistance of the ring. Then current in the ring is `I = e//R`
Consider a small element `dl` on the ring,
emf induced in the element, `de = ((e)/(2pia))dl`
resistance of the element, `dR = ((R )/(2pia)dl`
:. Potentail difference across the element
`= de - idR`
`= ((e)/(2pia))dl - ((e)/(R ))((R )/(2pia))dl = 0`