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(D) The magnetic intensity $H$ inside a solenoid is given by the formula $H = ni$,where $n$ is the number of turns per unit length and $i$ is the curr

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$6$

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(D) The magnetic intensity $H$ inside a solenoid is given by the formula $H = ni$,where $n$ is the number of turns per unit length and $i$ is the current.Given:Length of solenoid $\ell = 15\,cm = 0.15\,m$Number of turns $N = 60$Magnetic intensity $H = 2.4 	imes 10^3\,A/m$First,calculate the number of turns per unit length $n$:$n = \frac{N}{\ell} = \frac{60}{0.15} = 400\,turns/m$Now,use the formula $H = ni$ to find the current $i$:$2.4 	imes 10^3 = 400 	imes i$$i = \frac{2400}{400} = 6\,A$Therefore,the required current is $6\,A$.

The current required to be passed through a solenoid of $15\,cm$ length and $60$ turns in order to demagnetize a bar magnet of magnetic intensity $2.4 \times 10^3\,A/m$ is $.........A$.

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