The coercivity of a magnet is 5 times 10^3 te | vedclass

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(A) The magnetic field intensity $H$ inside a long solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$

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$10$

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(A) The magnetic field intensity $H$ inside a long solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current.Given:Coercivity $H_c = 5 	imes 10^3 	ext{ A/m}$.Length of solenoid $L = 30 	ext{ cm} = 0.3 	ext{ m}$.Number of turns $N = 150$.Number of turns per unit length $n = \frac{N}{L} = \frac{150}{0.3} = 500 	ext{ turns/m}$.To demagnetize the magnet,the magnetic field intensity produced by the solenoid must be equal to the coercivity of the magnet:$H = H_c$$nI = 5 	imes 10^3$$500 	imes I = 5000$$I = \frac{5000}{500} = 10 	ext{ A}$.Thus,the required current is $10 	ext{ A}$.

The coercivity of a magnet is $5 \times 10^3 \text{ A/m}$. The amount of current required to be passed in a solenoid of length $30 \text{ cm}$ and the number of turns $150$,so that the magnet gets demagnetized when placed inside the solenoid is ............. $A$.

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