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

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

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$8$

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(D) The coercivity $H$ is the magnetic field intensity required to demagnetize the material. For a solenoid, the magnetic field intensity is given by $H = nI$, where $n$ is the number of turns per unit length and $I$ is the current.Given:Coercivity $H = 4 	imes 10^3 	ext{ A m}^{-1}$Length of solenoid $L = 12 	ext{ cm} = 0.12 	ext{ m}$Number of turns $N = 60$First, calculate the number of turns per unit length $n = \frac{N}{L} = \frac{60}{0.12} = 500 	ext{ turns m}^{-1}$.Now, use the formula $H = nI$ to find the current $I = \frac{H}{n}$.$I = \frac{4 	imes 10^3}{500} = \frac{4000}{500} = 8 	ext{ A}$.Thus, the current required is $8 	ext{ A}$.

$A$ bar magnet has a coercivity of $4 \times 10^3 \text{ A m}^{-1}$. It is placed inside a solenoid of $12 \text{ cm}$ length and $60$ turns. The current that should be passed through the solenoid to demagnetize the bar magnet is: (in $\text{ A}$)

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