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(A) The magnetic field $B$ inside a long solenoid is given by the formula: $B = \mu_0 n I$, where $n = N/l$ is the number of turns per unit length.Giv

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$3.14 	imes 10^{-3} \ T$

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(A) The magnetic field $B$ inside a long solenoid is given by the formula: $B = \mu_0 n I$, where $n = N/l$ is the number of turns per unit length.Given:Length $l = 0.5 \ m$Number of turns $N = 250$Current $I = 5 \ A$Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$Substituting the values:$B = \frac{\mu_0 N I}{l}$$B = \frac{4 \pi 	imes 10^{-7} 	imes 250 	imes 5}{0.5}$$B = \frac{4 	imes 3.14159 	imes 10^{-7} 	imes 1250}{0.5}$$B = 8 \pi 	imes 10^{-7} 	imes 1250$$B = 10000 \pi 	imes 10^{-7} \ T$$B = \pi 	imes 10^{-3} \ T \approx 3.14 	imes 10^{-3} \ T$.

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $250$ turns. It carries a current of $5 \ A$. What is the magnitude of the magnetic field inside the solenoid?

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