Question

A magnetic 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

First, we should know that the relative permeability of a material is the measure of the degree of magnetization that a material attains when the magnetic field is applied to it.

Complete step by step solution:
We should know the relation of magnetic flux density and magnetic field with relative permeability and permeability of free space, $B = {\mu _0}{\mu _r}H$ where B is the magnetic flux density, H is the applied magnetic field, ${\mu _0}$ is the permeability of free space or vacuum and ${\mu _r}$ is the relative permeability of the material. The permeability of free space or vacuum is always constant. The relative permeability can also be defined as the ratio of magnetic permeability to the permeability of free space. The relative permeability defines the non-linear behavior of a material. In electromagnetism, the behavior of magnetic field H and magnetic flux density is represented by the B-H characteristics and the behavior of the material and the effect of H on B.

Given,
The magnetic field is $H = 2 \times {10^3}A/m$.
Magnetic flux density is, $B = 8\pi {\rm{ T}}$.
Permeability of free space, ${\mu _0} = 4\pi \times {10^{ - 7}}{\rm{H/m}}$
The equation of relative permeability with respect to magnetic field and magnetic flux density can be written as,
${\mu _r} = \dfrac{B}{{{\mu _0}H}}$ …… (1)
Substituting the values of $B,H$ and ${\mu _0}$ in equation (1).
$\begin{array}{c}{\mu _r} = \dfrac{{8\pi }}{{4\pi \times {{10}^{ - 7}} \times 2 \times {{10}^3}}}\\\ = {10^4}\\\ = {10^4}\end{array}$

Hence, the correct answer is (C).

Note: In the solution, the students can calculate magnetic permeability of the material by using the relation of magnetic flux density, B magnetic field, H and the permeability of free space. This relation is commonly represented in terms of the B-H characteristic graph.