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the uniform magnetic field perpendicualr...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
(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`
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Knowledge Check

  • The magnetic field perpendicular to the plane of a conducting ring of radius r change ate the rate (dB)/(dt)

    A
    The emf induced in the ring is `pir^(2)(dB)/(dt)`
    B
    The emf induced in the rintg is `2pir(dB)/(dt)`
    C
    The potential difference between diametrically opposite points on the ring is half of the induced emf.
    D
    All points on the ring are at the same potential.

  • The magnetic field perpendicular to the plane of a conducting ring of radius r changes at the rate. (dB)/(dt) . Then

    A
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  • A conducting loop of radius R is present in a uniform magnetic field B perpendicular to the plane of the ring. If radius R varies as a function of time t, as R =R_(0) +t . The emf induced in the loop is

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