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

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(D) The coercivity $H$ of a magnetic material is the magnetic field intensity required to demagnetize it.For a solenoid,the magnetic field intensity $

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$8$

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(D) The coercivity $H$ of a magnetic material is the magnetic field intensity required to demagnetize it.For a solenoid,the magnetic field intensity $H$ is given by the formula $H = nI$,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 $N = 60$ is the number of turns and $l = 12 \, cm = 0.12 \, m$ is the length of the solenoid.Substituting the values into the formula $H = \frac{N}{l} 	imes I$,we get $I = \frac{H 	imes l}{N}$.Given $H = 4 	imes 10^3 \, A/m$,$l = 0.12 \, m$,and $N = 60$.$I = \frac{4 	imes 10^3 	imes 0.12}{60} = \frac{4000 	imes 0.12}{60} = \frac{480}{60} = 8 \, A$.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 . . . . . . $A$.

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