NCERT Class XII Chapter
Magnetism and Matter
Questions With Solutions

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Question : 15
Total: 25
A short bar magnet of magnetic moment 5.25 × 102JT1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Solution:  

(a) At normal bisector
tan 45° =
BH
B
= 1 , BH = B =
µ0
4π
M
r3

0.42 × 104 =
107×5.25×102
r3

r3 = 12.5 × 105
r = 5 × 102 m = 5 cm
(b) At axis of magnet
tan 45° =
B
BH

B = BH =
µ02M
4πr3

0.42 × 104 =
107×2×5.25×102
r3

∴ r = 6.3 cm
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