Question

A short bar magnet is placed with its axis in magnetic north south directions null pointer are found on axis of magnet 14 cm from the centre of the magnet b is equal to zero point 36 dozens and dip is equal to zero if the magnet is turned to 180 degree the distance between new null points

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A short bar magnet placed in a horizontal plane has its axis aligned along the magnet north - south direction . Null points are found on the axis of the of the magnet at 14cm from the center of the magnet . The earth magnetic field at the place is 0.36G and the cycle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null - point (14 cm) from the centre of the magnet?

(i)magnitude : 0.54 G (ii) direction : along the direction of earth's field(i)magnitude : 5.4 G (ii) direction: opposite to the direction of earth's field (i)magnitude : 0.54 G (ii) direction along the direction of earth's field (i)magnitude : 0.54 G (ii) direction along the direction of earth's field

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A)

Solution :

(At null points , field due to a magnet is equal and opposite to the horizontal component of earth magnetic field )

Given d=14cm

=0.14m

H=0.6G

B1=μ04π.2md3

But the magnetic field is equal to the horizontal component of earth's magnetic field .

(ie) B1=μ04π.2md3

on the equatorial line of magnet at same distance (d) magnetic field due to the magnet.

B2=μ04πmd3

=B12

=H2

The total magnetic field on equatorial line at this point is

B=B1+B2

=H+H2

=32H

∴B=32×0.36

=0.54G

Hence the magnetic of earth's magnetic field is 0.54G

Answer : (i)magnitude : 0.54 G (ii) direction : along the direction of earth's field