Home
Class 12
PHYSICS
At 45^@ to the magnetic meridian, the a...

At `45^@` to the magnetic meridian, the apparent dip is `30^@`. Find the true dip.

Text Solution

Verified by Experts

At `45^@` to the magnetic meridian, the effective horizontal component of the earth's magnetic field is `B'_H = B_H cos 45^@ = 1/(sqrt(2)) B_H`. The apparent dip `delta` is given by
`(tandelta' = (B_v)/(b'_H) = (sqrt(2) B_v))/(B_H) = sqrt(2) tandelta`
or, `delta = tan^(-1) sqrt(1/6)`.
Promotional Banner

Topper's Solved these Questions

  • PERMANENT MAGNETS

    HC VERMA|Exercise Worked Out Examples|19 Videos
  • PERMANENT MAGNETS

    HC VERMA|Exercise Short Answer|11 Videos
  • OPTICAL INSTRUMENTS

    HC VERMA|Exercise Exercises|23 Videos
  • PHOTO ELECTRIC EFFECT AND WAVE PARTICLE DUALITY

    HC VERMA|Exercise Exercise|2 Videos

Similar Questions

Explore conceptually related problems

Define magnetic meridian.

A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. (a) Determine the horizontal component of the earth's magnetic field at the location. (b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30^@ with the magnetic meridian. The true value of the dip angle at the place is .....

A long straight horizontal cable carries a current of 2.5 A in the direction 10^@ south of west to 10^@ north of east. The magnetie meridian of the place happens to be 10^@ west of the geographie meridian. The earth's magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral potnts, magnetic field due to a current-carying cable is equal and opposite to the horizontal component of earth's magnetic field.)

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north up pointing down at 22^@ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field at the place.

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic Meld at the place is 0.36 G and the angle of dip 18 zero. What is the total magnetic feld on the normal bisector of the magnet at the same distance as the mull-point (i.e., 14 cm) from the centre of the magnet? At rull points, field due to a magnet 1s equal and opposite to the horizontal component of earth's magnetic field.)

In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is 0.26G and the dip angle is 60^(@) . What is the magnetic field of the earth at this location?

A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth's magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable ?

A magnetic needle is hung by an untwisted wire, so that it can rotate freely in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.1 g is kept or one end of the needle. If the magnetic pole strength of this needle is 10 Am, find the value of the vertical component of the earth's magnetic field. (g= 9.8 ms^(-2) )

HC VERMA-PERMANENT MAGNETS-Exercises
  1. At 45^@ to the magnetic meridian, the apparent dip is 30^@. Find the ...

    Text Solution

    |

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

    Text Solution

    |

  3. Two long bar magnets are placed with their axes coinciding in such a w...

    Text Solution

    |

  4. A uniform magnetic field of 0.20 xx 10^(-3) T exists in the space. Fin...

    Text Solution

    |

  5. Figure shows some of the equipotential surfaces of the magnetic scalar...

    Text Solution

    |

  6. The magnetic field at a point, 10cm away from a magnetic dipole, is fo...

    Text Solution

    |

  7. Show that the magnetic field at a point due to a magnetic dipole is pe...

    Text Solution

    |

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

    Text Solution

    |

  9. A magnetic dipole of magnetic moment 1.44 A m^2 is placed horizontally...

    Text Solution

    |

  10. A magnetic dipole of magnetic moment 0.72 Am^2 is planced horizontally...

    Text Solution

    |

  11. A magnetic dipole of magnetic moment 0.72(sqrt(2) Am^2) is placed hori...

    Text Solution

    |

  12. The magnetic moment of the assumed dipole at the earth's centre is 8.0...

    Text Solution

    |

  13. If the earth's magnetic field has a magnitude 3.4 xx 10^(-5) T at the ...

    Text Solution

    |

  14. The magnetic field due to the earth has a horizontal component of 26 m...

    Text Solution

    |

  15. A magnetic needle is free to rotate in a vertical plance which makes a...

    Text Solution

    |

  16. The needle of a dip circle shows an apparent dip of 45^@ in a particul...

    Text Solution

    |

  17. A tangent galvanometer shown a deflection of 45^@ when 10 mA of curren...

    Text Solution

    |

  18. A moving coil galvanometer has a 50-turn coil of size 2cm xx 2cm. It i...

    Text Solution

    |

  19. A short magnet produces a deflection of 37^@ in a deflection magnetome...

    Text Solution

    |

  20. The magnetometer of the previous problem is used with the same magent ...

    Text Solution

    |

  21. A deflection magnetometer is placed with its arms in north-south direc...

    Text Solution

    |