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(D) For a solenoid, the magnetic field is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length.Given: $I = 20 \ A$, $l = 2 \ m$,

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$150$

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(D) For a solenoid, the magnetic field is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length.Given: $I = 20 \ A$, $l = 2 \ m$, $B = 20 \ mT = 20 	imes 10^{-3} \ T$, and diameter $d = 3 \ cm = 3 	imes 10^{-2} \ m$.Radius $r = \frac{d}{2} = 1.5 	imes 10^{-2} \ m$.Calculating the number of turns per unit length $n$:$n = \frac{B}{\mu_0 I} = \frac{20 	imes 10^{-3}}{4 \pi 	imes 10^{-7} 	imes 20} = \frac{10^4}{4 \pi} \ m^{-1}$.Total number of turns $N = n 	imes l = \frac{10^4}{4 \pi} 	imes 2 = \frac{10^4}{2 \pi}$.The length of the wire is the total number of turns multiplied by the circumference of one turn:$L_{wire} = N 	imes (2 \pi r) = \left( \frac{10^4}{2 \pi} ight) 	imes (2 \pi 	imes 1.5 	imes 10^{-2}) = 10^4 	imes 1.5 	imes 10^{-2} = 1.5 	imes 10^2 \ m = 150 \ m$.

$A$ solenoid of length $2 \ m$ carries a current of $20 \ A$. The diameter of the solenoid is $3 \ cm$. If the magnetic field inside the solenoid is $20 \ mT$, then the length of wire forming the solenoid is (assume $\mu_0 = 4 \pi \times 10^{-7} \ H/m$) (in $m$)

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