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Magnetism And Matter

Question
CBSEENPH12037727

A short bar magnet of magnetic moment 5.25 x 10–2 JT–1 is placed with its axis perpendicular to earth's field direction. At what distance from the centre of the magnet, is the resultant field inclined at 45° with earth's field on (i) its normal bisector, (ii) its axis? Magnitude of earth's field at the place 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Solution
Here, a bar magnet is placed with it's axis perpendicular to the earth's field direction. 
Magnetic moment, M = 5.25 x 10
–2 JT–1
Earth's field  Be = 0.42 G = 0.42 × 10-4T 

(i) At a point P which is at a distance r from normal bisector [shown in fig. (a)], we will find the resultant magnetic field on it's normal bisector.


Therefore, 

Magnetic field B
2 due to magnet at equitorial line is, 
B2 = μ04πMr3  , along PA || NS . 

The resultant field R will be inclined at 45° to the earth's field along PQ’, only when 

                 B2  = Be 
       
           μ04π Mr3 = 0.42 × 10-4

10-7×5.25×10-2r3 = 0.42 × 10-4 

which gives, r = 0.05 m = 5 cm. 

(ii) When the point P lies on axis of the magnet such that OP = r, field due to magnet [Fig. (b),] is given by, 

B1 = μ04π 2Mr3, along PO. 

Earth's field Be is along PA. 

The resultant field R will be inclined at 45° to earth's field only when,

               B1 = Be 

     μ04π2Mr3 = 0.42 × 10-4 

which gives 

r = 6.3 × 10-2m = 6.3 cm.