Magnetization, Magnetic Induction Susceptibility Questions in English

(B) Given: Magnetic susceptibility, $\chi = 499$.
Permeability of vacuum, $\mu_0 = 4 \pi \times 10^{-7} \ H/m$.
Relative permeability of the rod is given by the relation: $\mu_r = 1 + \chi$.
Substituting the value of $\chi$: $\mu_r = 1 + 499 = 500$.
Absolute permeability $\mu$ is given by: $\mu = \mu_r \mu_0$.
Substituting the values: $\mu = 500 \times 4 \pi \times 10^{-7} \ H/m$.
$\mu = 2000 \pi \times 10^{-7} \ H/m$.
$\mu = 2 \pi \times 10^{-4} \ H/m$.

(C) Given:
Magnetic field intensity $H = 2000 \text{ A/m}$.
Magnetic flux $\phi = 6 \times 10^{-4} \text{ Wb}$.
Cross-sectional area $A = 3 \text{ cm}^2 = 3 \times 10^{-4} \text{ m}^2$.
We know that magnetic flux $\phi = B \cdot A$, where $B$ is the magnetic flux density.
$B = \frac{\phi}{A} = \frac{6 \times 10^{-4} \text{ Wb}}{3 \times 10^{-4} \text{ m}^2} = 2 \text{ T}$.
Also, $B = \mu H$, where $\mu$ is the magnetic permeability.
$\mu = \frac{B}{H} = \frac{2 \text{ T}}{2000 \text{ A/m}} = \frac{2}{2 \times 10^3} = 10^{-3} \text{ Wb/A} \cdot \text{m}$.
Therefore, the magnetic permeability is $10^{-3} \text{ Wb/A} \cdot \text{m}$.

(B) Given:
Magnetic field intensity, $H = 1000 \text{ A/m}$
Magnetic flux, $\phi = 3 \times 10^{-4} \text{ Wb}$
Cross-sectional area, $A = 1.5 \text{ cm}^2 = 1.5 \times 10^{-4} \text{ m}^2$
First, calculate the magnetic flux density $(B)$:
$B = \frac{\phi}{A} = \frac{3 \times 10^{-4} \text{ Wb}}{1.5 \times 10^{-4} \text{ m}^2} = 2 \text{ T (or Wb/m}^2)$
The relationship between magnetic flux density $(B)$, magnetic permeability $(\mu)$, and magnetic field intensity $(H)$ is given by:
$B = \mu H$
Therefore, the magnetic permeability $(\mu)$ is:
$\mu = \frac{B}{H} = \frac{2 \text{ Wb/m}^2}{1000 \text{ A/m}} = 2 \times 10^{-3} \text{ Wb/(A} \cdot \text{m)}$
Thus, the correct option is $B$.

(A) Magnetization $(M)$ is defined as the magnetic moment per unit volume of the sample.
Formula: $M = \frac{m}{V}$, where $m$ is the magnetic moment and $V$ is the volume.
Given: Magnetic moment $m = 8 \times 10^{-7} \text{ A} \cdot \text{m}^2$, mass $M_{mass} = 2 \text{ g}$, density $\rho = 4 \text{ g/cm}^3$.
First, calculate the volume $V = \frac{M_{mass}}{\rho} = \frac{2 \text{ g}}{4 \text{ g/cm}^3} = 0.5 \text{ cm}^3$.
Convert volume to $SI$ units: $V = 0.5 \times 10^{-6} \text{ m}^3$.
Now, calculate magnetization: $M = \frac{8 \times 10^{-7} \text{ A} \cdot \text{m}^2}{0.5 \times 10^{-6} \text{ m}^3} = 16 \times 10^{-1} \text{ A/m} = 1.6 \text{ A/m}$.
Note: Based on the provided options, there appears to be a discrepancy in the units or powers of the provided values. Recalculating with standard interpretation, the result is $1.6 \text{ A/m}$. Given the options provided, if we assume the magnetic moment was $8 \times 10^{-7} \text{ A} \cdot \text{m}^2$ and volume was $0.5 \text{ cm}^3$, the result is $1.6 \text{ A/m}$. If the question implies $8 \times 10^{-7} \text{ A} \cdot \text{m}^2$ and volume $0.5 \text{ m}^3$, the answer would be $1.6 \times 10^{-6} \text{ A/m}$.