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)

${10}^4$

Given:

The magnetic field intensity is $2\times {10}^{3}A/m$.

The magnetic flux density in rod is $8\pi Tesla$.

The relative permeability of the rod will be:

${\mu }_r=\frac{B}{{\mu }_{0}H}$

${\mu }_r=\frac{8\pi }{4\pi \times {10}^{-7}\times 2\times {10}^{-3}}$

${\mu }_r={10}^4$