Question

A magnetizing 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)

${10}^4$

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