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A thin non conducting ring of mass m, ra...

A thin non conducting ring of mass `m`, radius a carrying a charge `q` can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant `t=0`, a uniform magnetic field is switched on which is vertically downward and increases with time according to the law `B=B_0t`. Neglecting magnetism induced due to rotational motion of ring.
The magnitude of induced emf of the closed surface of ring will be

A

`pi a^(2) B_(0)`

B

`2a^(2)B_(0)`

C

zero

D

`(1)/(2) pi a^(2) B_(0)`

Text Solution

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The correct Answer is:
a
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Knowledge Check

  • A thin non conducting ring of mass m , radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant t=0 , a uniform magnetic field is switched on which is vertically downward and increases with time according to the law B=B_0t . Neglecting magnetism induced due to rotational motion of ring. Angular acceleration of ring is

    A
    `(qB_0)/(2m)`
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    `(qB_0)/(4m)`
    C
    `(qB_0)/m`
    D
    `(2qB_0)/m`

  • A thin non conducting ring of mass m , radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant t=0 , a uniform magnetic field is switched on which is vertically downward and increases with time according to the law B=B_0t . Neglecting magnetism induced due to rotational motion of ring. The magnitude of an electric field on the circumference of the ring is

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    zero

  • A thin non conducting ring of mass m , radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant t=0 , a uniform magnetic field is switched on which is vertically downward and increases with time according to the law B=B_0t . Neglecting magnetism induced due to rotational motion of ring. Find intantaneous power developed by electric force acting on the ring at t=1 s

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    B
    `(q^(2)B_(0)^(2)a^(2))/(8m)`
    C
    `(3q^(2)B_(0)^(2)a^(2))/(m)`
    D
    `(q^(2)B_(0)^(2)a^(2))/(4m)`

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