$A$ magnetizing field of $100 \,A/m$ produces a magnetic flux of $2.4 \times 10^{-5} \,Wb$ in an iron bar of cross-sectional area $0.3 \,cm^2$. The magnetic permeability of the iron bar in the $SI$ unit is

If $\mu_0$ is the permeability of vacuum and $\chi_m$ is the magnetic susceptibility,then the permeability of the material $\mu$ is . . . . . . .

$A$ domain in ferromagnetic iron is in the form of a cube of side length $1\; \mu m$. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is $55\; g/mol$ and its density is $7.9\; g/cm^3$. Assume that each iron atom has a dipole moment of $9.27 \times 10^{-24}\; A m^2$.

Explain the magnetic susceptibility $(\chi)$ of a material. From it,explain the relative magnetic permeability and magnetic permeability of the material. Obtain the relation between them.

An iron rod is placed parallel to a magnetic field intensity of $1000 \text{ A/m}$. The magnetic flux through the rod is $3 \times 10^{-4} \text{ Wb}$ and its cross-sectional area is $1.5 \text{ cm}^2$. The magnetic permeability of the rod in $\text{Wb/(A} \cdot \text{m)}$ is:

$A$ magnetizing field of $5000 \ A/m$ produces a magnetic flux of $4 \times 10^{-5} \ Wb$ in an iron rod of cross-sectional area $0.4 \ cm^2$. The permeability of the rod in $Wb/(A \cdot m)$ is: