The relationship between the magnetic susceptibility $(\chi)$ and the magnetic permeability $(\mu)$ is given by (where $\mu_0$ is the permeability of free space and $\mu_{r}$ is relative permeability).

If $M$ is the magnetisation induced in the material,$H$ is the magnetic field intensity,and $B$ is the net magnetic field inside the material,then the correct relation between them is $(\mu_0 = \text{permeability of free space})$

When a piece of a ferromagnetic substance is placed in a uniform magnetic field,the flux density inside it is four times the flux density away from the piece. The magnetic permeability of the material is

When a piece of a magnetic substance is placed in a uniform magnetic field,the flux density inside it is four times the flux density away from the piece. The magnetic permeability of the material is

An iron bar having a cross-sectional area of $2 \times 10^{-5} \ m^2$ and a magnetising field of $2400 \ A/m$ produces a magnetic flux of $2.4 \pi \times 10^{-5} \ Wb$. What will be the value of permeability $(\mu)$ and susceptibility $(\chi)$ of the bar? (Given $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$)

$A$ domain in a ferromagnetic substance is in the form of a cube of side length $2 \mu m$. It contains $4 \times 10^{14}$ atoms and each atomic dipole has a moment of $16 \times 10^{-24} \text{ A m}^2$. The magnetization of the domain is . . . . . . .