The relation between total magnetic field (B), magnetic intensity $(\mathrm{H})$, permeability of free space…
The relation between total magnetic field (B), magnetic intensity $(\mathrm{H})$, permeability of free space $\left(\mu_0\right)$ and susceptibility $(\chi)$ is
$\frac{\mathrm{H}}{\mathrm{B}}=\mu_0(1+\chi)$
$\frac{B}{H}=\mu_0(1+\chi)$
$\frac{\mathrm{H}}{\mathrm{B}}=\mu_0(\chi-1)$
$\frac{\mathrm{B}}{\mathrm{H}}=\mu_0(1-\chi)$
Solution
The net magnetic field, B has two contributions:
- Contribution from external factors \(\left(\mathrm{B}_0\right)\). For example, the current-carrying wire will generate its own magnetic field. This is called Magnetic intensity (H).
- The contribution due to the specific nature of the material \(\left(B_m\right)\). For instance, some materials magnetize in the direction of the magnetic field and some against it. This is called magnetization (M).
- It is influenced by external factor \(\mathrm{H}\). This can be seen from the relation
\(\Rightarrow M=\chi H\)
Also, \(B_m=\mu_0 H\)
- The net field in the interior of the solenoid,
\(\begin{aligned}
& \Rightarrow B=B_0+B_m \\
& \Rightarrow B=\mu_0(H+M)
\end{aligned}\)
\(\Rightarrow \frac{B}{H} = \mu_O(1+\chi)\)