\(\mu_{ r }=1+\chi=1+1.2 \times 10^{-5}\)
Fractional Change
\(=\frac{\Delta B }{ B }=\frac{\mu_{0} \mu_{ I } ni -\mu_{0} ni }{\mu_{0} ni }=\left(\mu_{ r }-1\right)\)
\(=1.2 \times 10^{-5}\)
${\mu _o}$$=4$$\pi $$ \times 10^{-7}$ $\frac{{Tm}}{A}$ લો. પૃથ્વીનું ચુંબકીયક્ષેત્ર અવગણો.