Home
Class 12
PHYSICS
A bar magnet of magnetic moment 1.5JT^-1...

A bar magnet of magnetic moment `1.5JT^-1` lies aligned with the direction of a uniform magnetic field of `0.22T`.
(a) What is the amount of work done to turn the magnet so as to align its mangetic moment
(i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?

Promotional Banner

Topper's Solved these Questions

  • MAGNETISM

    SL ARORA|Exercise Problems From Competitive Examinations|13 Videos

  • MAGNETISM

    SL ARORA|Exercise Based on Coulomb s law and dipole moment of a magnet|5 Videos

  • MAGNETIC EFFECTS OF CURRENT

    SL ARORA|Exercise TYPE C|4 Videos

  • OPTICAL INSTRUMENTS

    SL ARORA|Exercise Problems for self practice|35 Videos

Similar Questions

Explore conceptually related problems

A "bar" magnet of magnetic moment 1.5J//T is aligned with the direction of a uniform magnetic field of 0.22 T . The work done in turning the magnet so as to align its magnetic moment opposite to the field and the torque acting on it in this position are respectively.

A magnet of magnetic moment 2JT^(-1) is aligned in the direction of magnetic field of 0.1T . What is the net work done to bring the magnet normal to the magnrtic field?

A bar magnet of magnetic moment 6 J//T is aligned at 60^(@) with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. Calculate work done to align the magnetic moment (i) opposite to the field (ii) normal to field direction?

SL ARORA-MAGNETISM-Based on (i) Earth s Magnetism (ii) magnetic field of a bar magnet
  1. A bar magnet of magnetic moment 1.5JT^-1 lies aligned with the directi...

    Text Solution

    |

  2. A ship is sailing due west according in Mariner's compass . If the del...

    Text Solution

    |

  3. A ship is sailing due east according to Mariner's compass . If the dec...

    Text Solution

    |

  4. The horizontal component of earth's magnetic field is sqrt3 times the ...

    Text Solution

    |

  5. The ratio of the vertical component to the horizontal component of ear...

    Text Solution

    |

  6. The horizontal component of the earth's magnetic field is 0.22 Gauss a...

    Text Solution

    |

  7. Calculate earth's surface magnetic field at a place , where an angle o...

    Text Solution

    |

  8. Find the value of the vertical components of earth's magnetic at a pla...

    Text Solution

    |

  9. The vertical and horizontal components of earth's magnetic at a place ...

    Text Solution

    |

  10. A vertical wire in which current is flowing produces a neutral point w...

    Text Solution

    |

  11. A compass needle whose magnetic moment is 60Am^2 pointing geograhic no...

    Text Solution

    |

  12. A magnetic needle free to rotate about the vertical direction (compass...

    Text Solution

    |

  13. The true dip at a place is 30^(@) . What is the apparent dip when the ...

    Text Solution

    |

  14. The values of the apparent angles of dip in two planes at right angles...

    Text Solution

    |

  15. A dip circle lies initially in the magnetic meridian. If it is now rot...

    Text Solution

    |

  16. A magnetic needle of length 10cm, suspended at its middle point throug...

    Text Solution

    |

  17. A long bar magnet has a pole strength of 10 Am. Find the magnetic fiel...

    Text Solution

    |

  18. Calculate the magnetic field due to a bar magnet 2cm long and having p...

    Text Solution

    |

  19. A bar magnet of length 0.1 m has a pole strength of 50 Am . Calculate ...

    Text Solution

    |

  20. A bar magnet has a length of 8cm. The magnetic field at a point at a d...

    Text Solution

    |

  21. The magnetic moment of a current carrying loop is 2.1xx10^(-25) ampxxm...

    Text Solution

    |