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(D) The coercivity of the bar magnet is $H = 4 	imes 10^3 \, A \, m^{-1}$. This is the magnetic intensity required to demagnetize the magnet.For a so

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$8$

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(D) The coercivity of the bar magnet is $H = 4 	imes 10^3 \, A \, m^{-1}$. This is the magnetic intensity required to demagnetize the magnet.For a solenoid,the magnetic intensity $H$ is given by the formula $H = n \cdot i$,where $n$ is the number of turns per unit length and $i$ is the current.The number of turns per unit length $n$ is calculated as $n = \frac{N}{L} = \frac{60}{0.12 \, m} = 500 \, m^{-1}$.Substituting the values into the formula $H = n \cdot i$,we get $4 	imes 10^3 = 500 \cdot i$.Solving for $i$,we get $i = \frac{4000}{500} = 8.0 \, A$.

$A$ bar magnet has a coercivity of $4 \times 10^3 \, A \, m^{-1}$. It is desired to demagnetize it by inserting it inside a solenoid $12 \, cm$ long and having $60$ turns. The current that should be sent through the solenoid is . . . . . . $A$.

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