Home
Class 12
PHYSICS
The needle of a dip circle shows an appa...

The needle of a dip circle shows an apparent dip of `45^@` in a particular position and `53^@` when the circle is rotated through `90^@`. Find the true dip.

Text Solution

Verified by Experts

If `delta_(1) and delta_(2)` be the apparent dips shown by the dip circle in the 2 perperdicular positions the true dip `delta` is diven by
`cot^(2) delta = cot^(2) delta_(1) + cot^(2)delta_(2)`
`rArr cot^(2) delta = cot^(2)45^(@) + cot^(2)53^(@)`
`cot^(2)delta = 1.56`
`cot^(2)delta = 1.56`
`delta = 38.6^(@) = 39^(@)
Promotional Banner

Topper's Solved these Questions

  • PERMANENT MAGNETS

    HC VERMA|Exercise Objective II|5 Videos
  • OPTICAL INSTRUMENTS

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

    HC VERMA|Exercise Exercise|35 Videos

Similar Questions

Explore conceptually related problems

A dip circle shows an apparent dip of 60^@ at a place where the true dip is 45^@ . If the dip circle is rotated through 90^@ , what apparent dip will it show?

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

At a certain location in Africa, a compass 12^@ west of the geographic north. The northe tip of the magnetic needle of a dip circle in the plane of magnetic meridian points 60^@ above the horizontal . The horizontal component of the earth's field measured to be 0.16 G. Specify the direction and magnitude of the earth's field at the location.

The lower end of a capillary tube is dipped into water and it is seen that the water rises through 7.5 cm in the capillary. Find the radius of the capilary. Surface tension of water =7.5xx10^-2Nm^-1 . Contact angle between water and glass =0^@. Take g=10ms^-2 .

A particle having a charge 20mu C and mass 20mug moves along a circle of radius 5.0 cm under the action of a magnetic field B = 1.0 T. When the particle is at a point P, a uniform electric field is switched on and it is found that the particle continues on the tangent through P with a uniform velocity. Find the electric field.

A dip circle is at right angles to the magnetic meridian. The apparent dip is

A compass needle oscillates 20 times per minute at a place where the dip is 45^@ and 30 times per minute where the dip is 30^@ . Compare the total magnetic field due to the earth at the two places.

The needle of a deflection magnetometer deflects through 45^@ from north to south when the instrument is used in Tan-A position with a magnet of length 10cm placed at a distance of 25cm. (a) Find the magnetic moment of the magnet if the earth's horizontal magnetic field is 20 muT . (b) If the magnetometer is used in Tan-B position with the same magnet at the same separation from the needle, what will be the deflection?

Figure shows a mass m placed on a frictionless horizontal table and attached toa string passing through a mall hole in the surface. Initially, themas movesin a circle of radius r_0 with a speed v_0 and the ree end of the string is held by a person. The person pulls on the string slowly to decrease the radius of th circle of r. a. Find the tension in the string when the mass moves in the circle of radius r. b. Calculate the chasnge in the kinetic energy of the mass

A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. a. What is the field at the centre of the coil ? b. What is the magnetic moment of this coil ? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A unifrom magnetic field of 2 T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of 90^@ under the influence of the magnetic field . c. What are the magnitudes of the torques on the coil in the initial and final position ? d. What is the angular speed acqired by the coil when it has rotated by 90^(@) ? The moment of inertia of the coil is 0.1 kg m^2

HC VERMA-PERMANENT MAGNETS-Exercise
  1. Show that the magnetic field at a point due to a magnetic dipole is pe...

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

  4. A magnetic dipole of magnetic moment 0.72 Am^2 is placed horizontally ...

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

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

    Text Solution

    |

  15. A deflection megnetometer is placed with its arms in north-south direc...

    Text Solution

    |

  16. A bar magnet takes pi/10 second to complete one oscillation in an osci...

    Text Solution

    |

  17. The combination of two bar magnets makes 10 oscillations per second in...

    Text Solution

    |

  18. A short magnet oscillates in a vibration magnetometer with a time per...

    Text Solution

    |

  19. A bar magnet makes 40 oscillations per minute in an oscillation magnet...

    Text Solution

    |

  20. A short magnet makes 40 oscillations per minute when used in an oscill...

    Text Solution

    |