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The time period of oscillation of a free...

The time period of oscillation of a freely suspended bar magnet with usual notations is given by

A

`T=2pisqrt(I/(MB_(H)))`

B

`T=2pisqrt((MB_(H))/I)`

C

`T=sqrt(I/(MB_(H)))`

D

`T=2pisqrt(B_(H)/(MI))`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • The time period of oscillations of a freely suspended magnetic needle is given by

    A
    `T= 2pi sqrt((I)/(mB_H))`
    B
    `T= 2pi sqrt((mB_H)/(I))`
    C
    `T= sqrt((I)/(mB_H))`
    D
    `T= 2pi sqrt((B_H)/(mI))`

  • Time period of oscillation of a magnetic needle is

    A
    `T = sqrt ((T)/( MB))`
    B
    `T = 2pi sqrt((I)/( MB))`
    C
    `T = 2pi sqrt (( MB)/( I ))`
    D
    `T = pi sqrt (( MB)/( I ))`

  • If the period of oscillaion of freely suspended bar magnet in earth's horizontal field H is 4 sec. When another magnet is brought near it, the period of oscillation is reduced to 2s. The magnetic field of second bar magnet is

    A
    4H
    B
    3H
    C
    2H
    D
    `sqrt3 H`

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