Question Number 63124 by necx1 last updated on 29/Jun/19
$${A}\:{solenoid}\:{is}\:\mathrm{40}{cm}\:{long},{has}\:{a}\:{cross} \\ $$$${sectional}\:{area}\:{of}\:\mathrm{8}.\mathrm{0}{cm}^{\mathrm{2}} \:{and}\:{is}\:{wound} \\ $$$${with}\:\mathrm{309}\:{turns}\:{of}\:{wire}\:{that}\:{carries}\:{a} \\ $$$${current}\:{of}\:\mathrm{1}.\mathrm{2}{A}.{The}\:{relative} \\ $$$${permeability}\:{of}\:{the}\:{iron}\:{core}\:{is}\:\mathrm{600}. \\ $$$${Compute}\:{the}\:\boldsymbol{{B}}\:{for}\:{the}\:{interior}\:{point} \\ $$$${and}\:{the}\:{flux}\:{through}\:{the}\:{solenoid}. \\ $$
Commented by necx1 last updated on 29/Jun/19
$${please}\:{help} \\ $$
Commented by Hope last updated on 29/Jun/19
Commented by Hope last updated on 29/Jun/19
Commented by Hope last updated on 29/Jun/19
Commented by Hope last updated on 29/Jun/19
Answered by peter frank last updated on 30/Jun/19
$$\left.{flux}={A}\left({in}\:{meter}\right)\right)×{B}.......\left({i}\right) \\ $$$${B}=\mu_{{m}} {nI} \\ $$$${n}=\frac{{N}}{{L}\left({meter}\right)}........\left({ii}\right) \\ $$$$\mu_{{r}} \left({relative}\:{permeability}\right)=\frac{\mu_{{m}} }{\mu_{{o}} }.......\left({iii}\right) \\ $$$$\mu_{{m}} =\mu_{{r}} ×\mu_{{o}} \left({absolute}\:{p}\right) \\ $$$$ \\ $$