The current required to be passed through a solenoid of 15 cm length and 60 turns in order to demagnetise a bar magnet of magnetic intensity 2.4 x 10 3 Am -1 is A
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(6)
The magnetic intensity is
For M = 0, it can be written
The data given is
H = 2.4 x 10 3 A m -1 
= 15 x 10
-2 m
N = 60
Using the relation, 
and H = ni, the value of the current is,
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