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

Vedclass · Accepted Answer

$2.51 	imes 10^{-2} \ T$

Vedclass · Answer

(A) The magnetic field $B$ inside a long solenoid is given by the formula: $B = \mu_0 n I$,where $n = \frac{N}{l}$ is the number of turns per unit length.Given:Length $l = 0.5 \ m$Number of turns $N = 1000$Current $I = 10 \ A$Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$Substituting the values:$n = \frac{1000}{0.5} = 2000 \ turns/m$$B = (4\pi 	imes 10^{-7}) 	imes 2000 	imes 10$$B = 8\pi 	imes 10^{-3} \ T$$B \approx 25.12 	imes 10^{-3} \ T = 2.51 	imes 10^{-2} \ T$.

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $1000$ turns. It carries a current of $10 \ A$. What is the magnitude of the magnetic field inside the solenoid?

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