Question

A magnetising field of $2\times 10^{3}A{m}^{-1}$ produces a magnetic flux density of $8\pi T$ in an iron rod. The relative permeability of the rod will be

• A. $10^2$
• B. $1$
• C. $10^4$
• D. $10^3$

Answer 1

Answer: (C)

Here, $H=2\times 10^{3}A{m}^{-1}$,
$B=8\pi T$, ${\mu }_0=4\pi \times 10^{-7}$
Since ${\mu }_r=\frac{\mu }{{\mu }_0}=\frac{\mu H}{{\mu }_{0}H}=\frac{B}{{\mu }_{0}H}$
$=\frac{8\pi }{4\pi \times 10^{-7}\times 2\times 10^3}=10^4$