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 103 Am^-1 is A

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Accepted Answer

Correct answer: 6. (6) The magnetic intensity is H= B _0-M For M = 0, it can be written H= ( B _0 )=ni The data given is H = 2.4 x 103 A m^-1 l = 15 x 10-2 m N = 60 Using the relation, n= Nl and H = ni, the value of the current is, i= Hn= 2.4 10^3 15 10^-260=6A

$The current required to be passed through a solenoid of 15 cm length and 60 turns in order to demagnetise a bar magnet of magnetic intensity 2.4 x 103 A m^{- 1} is A$

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