Question

problem 20.61
a long thin solenoid has 420 loops of wire per meter, and a 410 - ma current flows through the wire.
part a
if the permeability of the iron is 3000μ0, what is the total field b inside the solenoid?
express your answer using two significant figures.
b = t

Explanation:

Step1: Recall the formula for the magnetic - field inside a solenoid with a magnetic core

The formula for the magnetic field $B$ inside a solenoid with a magnetic core is $B = \mu nI$, where $\mu$ is the permeability of the medium, $n$ is the number of turns per unit length, and $I$ is the current. Given that $\mu = 3000\mu_0$, $n = 420\ m^{-1}$, and $I=410\times10^{- 3}\ A$. Also, $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$.

Step2: Substitute the values into the formula

First, calculate $\mu = 3000\times4\pi\times10^{-7}\ T\cdot m/A$. Then, $B=\mu nI=(3000\times4\pi\times10^{-7}\ T\cdot m/A)\times420\ m^{-1}\times410\times10^{-3}\ A$.

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LATEXBLOCK0
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Answer:

$0.65$