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Ashok0210
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Asked: 2021-02-24T04:18:47+00:00 In:

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º 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.

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Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.10

NCERT Solutions for Class 12 Physics Chapter 5 Question 10

2020-2021cbse and up boardclass 12 chapter-5 physicsclass 12 physicsmagnetism and matterncert class 12
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    1 Answer

    1. Horizontal component of earth’s magnetic field,

      BH= 0.35 G

      Angle made by the needle with the horizontal plane = Angle of dip , δ = 22°

      Earth’s magnetic field strength = B We can relate B and BH as:

      BH = Bcos0

      Therefore  B = BH /cosδ

      0.35 /cos 22° =0.377 G

      Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

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