A bar magnet has a coercivity of 4 times 10^3 | vedclass

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(D) The coercivity of the bar magnet is $H = 4 	imes 10^3 \, A/m$. This is the magnetic intensity required to demagnetize the magnet.For a solenoid,t

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$8$

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(D) The coercivity of the bar magnet is $H = 4 	imes 10^3 \, A/m$. 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 is $n = \frac{N}{L}$,where $N = 60$ turns and $L = 12 \, cm = 0.12 \, m$.Thus,$n = \frac{60}{0.12} = 500 \, turns/m$.Using the formula $H = n \cdot i$,we get $i = \frac{H}{n}$.Substituting the values: $i = \frac{4 	imes 10^3}{500} = \frac{4000}{500} = 8 \, A$.Therefore,the required current is $8 \, A$.

$A$ bar magnet has a coercivity of $4 \times 10^3 \, A/m$. 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... (in $, A$)

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