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The work done in rotating a bar magnet o...

The work done in rotating a bar magnet of magnetic moment M from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field B is

A

2MB

B

MB

C

`-MB`

D

`-2MB`

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • The work done in rotating a bar magnet f magnetic moment M from its unstable equilibrium positon to its stable equilibrium position in a uniform magnetic field B is

    A
    `2MB`
    B
    `MB`
    C
    `-MB`
    D
    `-2MB`

  • The work done in rotating a magnet of magnetic moment 2A-m^(2) in a magnetic field to opposite direction to the magnetic field, is

    A
    Zero
    B
    `2xx10^(-2)J`
    C
    `10^(-2)J`
    D
    `10J`

  • Stable equilibrium of a magnetic dipole in a uniform magnetic field will be at orientation when:

    A
    Dipole moment vector and magnetic field vector are in same direction
    B
    Dipole moment vector and magnetic field vector are in opposite direction
    C
    Dipole moment vector is perpendicular to magnetic field vector
    D
    Both a and c

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    Find the amount of work done in rotating an electric dipole of dipole moment 3.2 xx 10^(- 8) Cm from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field if intensity 10^(4) N// C .

    How does the energy of dipole change when it is rotated from unstable equilibrium to stable equilibrium in a uniform electric field.

    Calculate the amount of work done in turing an electric dipole of dipole moment 2 xx 10^(-8) C -m from its position of unstable eqquilibrium to its position of stable equilibrium, in a uniform electric field of 10^3 N C^(1) .

    What are the conditions for stable and unstable equilibrium of a magnet suspended in a uniform magnetic field?

    When will a magnet in an external magnetic field be in unstable equilibrium?

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