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Correct option is B)

Given: $H=4×10_{3}Am_{−1}$

$L=12cm$

$N=60$

Bar magnet requires a magnetic intensity ($H=4×10_{3}Am_{−1}$) to become demagnetized.

Let $n$ be the number of turns per unit length.

$⇒n=LN =0.1260 =500turns/metre$

Now, $H=nI$

$⇒I=nH =5004×10_{3} =8A$

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