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A metallic rod of 1 m length is rotated ...

A metallic rod of 1 m length is rotated with a frequency of 50 `rev//s`, with on end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular at to the plane of the ring. A constant uniform magnetic field of 1 T parallel to the axis is persent eveywhere. what is the e.m.f. between the centre and the metallic ring?

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To solve the problem of finding the e.m.f. generated between the center and the metallic ring when a metallic rod is rotated in a magnetic field, we can follow these steps: ### Step 1: Understand the Setup We have a metallic rod of length 1 m that is hinged at one end at the center of a circular metallic ring with a radius of 1 m. The rod is rotating about an axis perpendicular to the plane of the ring with a frequency of 50 revolutions per second. A uniform magnetic field of 1 T is present parallel to the axis of rotation. ### Step 2: Calculate Angular Velocity The angular velocity (ω) can be calculated using the formula: \[ ...
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Knowledge Check

  • Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is

    A
    `1//2MR^(2)`
    B
    `MR^(2)`
    C
    `1//4MR^(2)`
    D
    `3//4 MR^(2)`

  • Moment of inertia of a circular ring about an axis through its centre and perpendicular to its plane is

    A
    `I = (1)/(2) MR^2`
    B
    `I = MR^2`
    C
    `I = (3)/(2) MR^2`
    D
    `I = (5)/(2) MR^2`

  • The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is "4 kg m"^(2) . The diameter of the ring is

    A
    2 m
    B
    4 m
    C
    5 m
    D
    6 m

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