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Updated on : 2022-09-05

Solution

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Given:

where $H=3×10_{−3}Am_{−1}=$coercivity

$n=$number of turns per unit length $=0.150 =500m_{−1}$

To find: $I=$ current

Now

$H=nI$

$3×10_{−3}Am_{−1}=500m_{−1}×I$

$I=6×10_{−6}A=6μA$

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