At Calcutta, here the vertical component of the earth's field is 0.26(dyne per unit pole) it is found that a mass of 10mg placed 10cm from the axel of a dip-needle keeps it horizontal.Find the magnetic moment of the dip-needle( acceleration due to gravity is 1000cm/sec^2₎

Dear Sanket,

Ans:- Let the magnetic moment is M unit.

Then the moment about the axle of the dip needle is MV where V=0.26 dyne/pole

The moment given by the point mass=mgx where x=10 cm

Hence we get,

MV=mgx

M=mgx/V

  =(10*10-³ *10³ *10)/0.26=384.6 unit(ans)

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