Field inside a solenoid

Problem

A solenoid of length $\ell = 30$ cm, with $N = 500$ turns, carries $I = 4$ A. Calculate the internal magnetic field, assuming it is ideal.

Solution

For an ideal solenoid the internal field is uniform and equals $B=\mu_0\dfrac{N}{\ell}I$.

Number of turns per unit length: $\frac{N}{\ell}=\frac{500}{0{,}30}\approx1667\;\text{turns/m}$

Substituting the values: $B=\mu_0\frac{N}{\ell}I=4\pi\cdot10^{-7}\cdot\frac{500}{0{,}30}\cdot4=1{,}257\cdot10^{-6}\cdot1667\cdot4\approx8{,}4\cdot10^{-3}\;\text{T}$

$\ev{B \approx 8{,}4\;\text{mT}}$

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Topics: Magnetismo Concepts: Campo magnetico · Teorema di Ampère