Magnetization, Magnetic Induction Susceptibility Questions in English

(B) The Oersted $(Oe)$ is the $CGS$ unit of magnetic field strength $(H)$.
It is defined as the magnetic field strength in a vacuum at a distance of $1 \ cm$ from a unit magnetic pole.
In the $CGS$ system,the magnetic intensity is measured in Oersted.
Therefore,the correct option is $(B)$.

(C) The magnetic permeability $\mu$ is defined by the relation $B = \mu H$, where $B$ is the magnetic flux density and $H$ is the magnetic field intensity.
Since the unit of $B$ is Tesla $(T)$ or $Weber/m^2$ $(Wb/m^2)$ and the unit of $H$ is $Ampere/meter$ $(A/m)$,
the unit of $\mu$ is $(Wb/m^2) / (A/m) = Wb / (A m)$.
Since $1 \ Henry (H) = 1 \ Wb/A$, the unit of $\mu$ can be written as $H/m$ or $Henry m^{-1}$.
Therefore, the correct option is $C$.

(D) According to Curie's law,for a paramagnetic material,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$.
Mathematically,it is expressed as $\chi \propto \frac{1}{T}$ or $\chi = \frac{C}{T}$,where $C$ is the Curie constant.
Therefore,the magnetic susceptibility is proportional to $\frac{1}{T}$.

(A) Magnetic susceptibility $(\chi)$ is a dimensionless quantity that represents the degree to which a substance can be magnetized when placed in an external magnetic field.
It is defined as the ratio of the intensity of magnetization $(I)$ to the magnetic intensity $(H)$.
Mathematically,it is expressed as $\chi = \frac{I}{H}$.
Therefore,the correct option is $A$.

(D) The relationship between magnetic susceptibility $(\chi_m)$ and relative permeability $(\mu_r)$ is given by the formula:
$\chi_m = \mu_r - 1$
Given that the relative permeability of iron is $\mu_r = 5500$.
Substituting this value into the formula:
$\chi_m = 5500 - 1 = 5499$
Therefore,the magnetic susceptibility of iron is $5499$.

(D) The total magnetic induction $B$ in a magnetic material is the sum of the magnetic induction due to the external magnetic field $(B_0)$ and the magnetic induction due to the magnetization of the material $(B_m)$.
$B = B_0 + B_m$
Since $B_0 = \mu_0 H$ and $B_m = \mu_0 M$,we have:
$B = \mu_0 H + \mu_0 M$
$B = \mu_0(H + M)$
Therefore,the correct relation is $B = \mu_0(H + M)$.

(C) According to Curie's Law,the magnetic susceptibility $\chi$ of a paramagnetic material is inversely proportional to its absolute temperature $T$.
Mathematically,this is expressed as $\chi \propto \frac{1}{T}$ or $\chi = \frac{C}{T}$,where $C$ is the Curie constant.
Therefore,as the absolute temperature $T$ increases,the magnetic susceptibility $\chi$ decreases.

(D) The magnetic flux density $B$ inside a material is given by $B = \mu H$,where $\mu$ is the absolute permeability and $H$ is the magnetic field intensity.
The flux density away from the piece (in vacuum or air) is $B_0 = \mu_0 H$.
Given that the flux density inside the material is four times the flux density outside,we have $B = 4 B_0$.
Substituting the expressions,we get $\mu H = 4 \mu_0 H$.
Dividing both sides by $\mu_0 H$,we find the relative permeability $\mu_r = \frac{\mu}{\mu_0} = 4$.

(A) The magnetic susceptibility $(\chi)$ of a ferromagnetic substance is positive and very large, typically much greater than $1$.
Ferromagnetic materials, such as iron, nickel, and cobalt, exhibit high magnetic susceptibility, often exceeding values of $10,000$.

(A) The intensity of magnetization $I$ is defined as the net magnetic dipole moment per unit volume.
For maximum magnetization,all molecular dipoles must be aligned in the same direction.
Given:
Dipole moment of each molecule,$\mu = 1.5 \times 10^{-23} \, A \cdot m^2$
Number of molecules per unit volume,$n = \frac{N}{V} = 2 \times 10^{26} \, m^{-3}$
The formula for intensity of magnetization is $I = n \times \mu$.
Substituting the values:
$I = (2 \times 10^{26} \, m^{-3}) \times (1.5 \times 10^{-23} \, A \cdot m^2)$
$I = 3 \times 10^3 \, A/m$.
Thus,the maximum possible intensity of magnetization is $3 \times 10^3 \, A/m$.

(D) The magnetic intensity $H$ inside the solenoid is given by $H = ni$.
Given $n = 5 \, \text{turns/cm} = 500 \, \text{turns/m}$ and $i = 0.5 \, A$.
So, $H = 500 \times 0.5 = 250 \, A/m$.
The intensity of magnetization $I$ is given by $I = (\mu_r - 1)H$.
Given $\mu_r = 1000$, so $I = (1000 - 1) \times 250 = 999 \times 250 = 249750 \, A/m \approx 2.5 \times 10^5 \, A/m$.
The magnetic moment $M$ is given by $M = I \times V$, where $V = 10^{-4} \, m^3$.
$M = 2.5 \times 10^5 \times 10^{-4} = 25 \, A m^2$.

(A) Molar susceptibility is defined as the product of volume susceptibility and molar volume.
Mathematically,$\chi_m = \chi_v \times V_m$,where $\chi_v$ is the volume susceptibility (dimensionless) and $V_m$ is the molar volume.
The unit of molar volume $(V_m)$ is $m^3 \cdot mol^{-1}$.
However,in many physics contexts,molar susceptibility is defined as $\chi_m = \frac{\chi_v}{\rho} \times M$,where $\rho$ is density $(kg \cdot m^{-3})$ and $M$ is molar mass $(kg \cdot mol^{-1})$.
Thus,the unit is $\frac{1}{kg \cdot m^{-3}} \times kg \cdot mol^{-1} = m^3 \cdot mol^{-1}$.
Given the options provided,the standard $SI$ unit for molar susceptibility is $m^3 \cdot mol^{-1}$. Since the question asks for the unit and $m^3$ is the primary dimensional component listed in the options,option $A$ is the correct choice.

(B) The relationship between magnetic flux density $B$ and magnetic field intensity $H$ is given by $B = \mu_0 \mu_r H$.
From this,the relative permeability $\mu_r$ is proportional to the ratio $\frac{B}{H}$,which represents the slope of the $B-H$ curve at any given point.
$\mu_r \propto \frac{B}{H} = \text{slope of } B-H \text{ curve}$.
To find the highest relative permeability,we need to identify the point on the curve where the slope is maximum.
By observing the given $B-H$ graph,the curve is steepest in the region around point $Q$. Therefore,the slope is highest at point $Q$.

(A) According to Curie's law for paramagnetic materials,the magnetic susceptibility $\chi$ is given by $\chi = C \times (1/T)$,where $C$ is the Curie constant.
From the given graph,the slope of the line represents the Curie constant $C$.
Slope $C = \frac{\Delta \chi}{\Delta (1/T)} = \frac{0.4}{7 \times 10^{-3} \text{ K}^{-1}}$.
$C = \frac{0.4}{0.007} = \frac{400}{7} \approx 57.14 \text{ K}$.
Rounding to the nearest integer,the Curie constant is $57 \text{ K}$.

(B) The magnetic flux density $B$ is given by the ratio of magnetic flux $\phi$ to the cross-sectional area $A$:
$B = \frac{\phi}{A} = \frac{6.28 \times 10^{-4} \ Wb}{2 \times 10^{-5} \ m^2} = 31.4 \ T$
The relationship between magnetic flux density $B$,permeability $\mu$,and magnetizing field intensity $H$ is $B = \mu H$.
Therefore,the permeability $\mu$ is:
$\mu = \frac{B}{H} = \frac{31.4 \ T}{2000 \ A/m} = 0.0157 \ T \cdot m/A$
However,calculating the relative permeability $\mu_r$ using $\mu = \mu_0 \mu_r$:
$\mu_r = \frac{\mu}{\mu_0} = \frac{0.0157}{4\pi \times 10^{-7}} \approx 1.25 \times 10^4$
Thus,the magnetic permeability $\mu$ is approximately $1.25 \times 10^4 \ \mu_0$.

(B) Magnetization $I$ (or $M$) is defined as the magnetic dipole moment per unit volume: $I = \frac{M_{dipole}}{V}$.
Given:
Mass $m = 1 \, g = 10^{-3} \, kg$.
Density $\rho = 5 \, g/cm^3 = 5000 \, kg/m^3 = 5 \times 10^3 \, kg/m^3$.
Magnetic dipole moment $M_{dipole} = 6 \times 10^{-7} \, A \cdot m^2$.
First,calculate the volume $V$:
$V = \frac{m}{\rho} = \frac{1 \, g}{5 \, g/cm^3} = 0.2 \, cm^3 = 0.2 \times 10^{-6} \, m^3 = 2 \times 10^{-7} \, m^3$.
Now,calculate the magnetization $I$:
$I = \frac{6 \times 10^{-7} \, A \cdot m^2}{2 \times 10^{-7} \, m^3} = 3 \, A/m$.

(B) The magnetic flux $\phi$ is given by $\phi = B \cdot A$, where $B$ is the magnetic field induction and $A$ is the cross-sectional area.
$B = \frac{\phi}{A} = \frac{2.4 \times 10^{-5} \, Wb}{0.2 \times 10^{-4} \, m^2} = 1.2 \, T$.
The relation between magnetic field $B$, magnetic intensity $H$, and relative permeability $\mu_r$ is $B = \mu_0 \mu_r H$.
Thus, $\mu_r = \frac{B}{\mu_0 H} = \frac{1.2}{(4\pi \times 10^{-7}) \times 1600}$.
$\mu_r = \frac{1.2}{2.01 \times 10^{-3}} \approx 596.8$.
Since $\mu_r = 1 + \chi_m$, the magnetic susceptibility $\chi_m = \mu_r - 1$.
$\chi_m = 596.8 - 1 = 595.8 \approx 596$.

(C) The magnetization $M_{z}$ is defined as the magnetic dipole moment per unit volume: $M_{z} = \frac{m_{d}}{V}$.
Given mass $m = 0.075 \, kg$ and density $\rho = 7500 \, kg/m^3$,the volume $V$ is calculated as $V = \frac{m}{\rho} = \frac{0.075}{7500} = 10^{-5} \, m^3$.
The magnetic dipole moment $m_{d} = 8 \times 10^{-7} \, A \cdot m^2$.
Therefore,$M_{z} = \frac{8 \times 10^{-7}}{10^{-5}} = 8 \times 10^{-2} = 0.08 \, A/m$.

(B) According to Curie's Law for paramagnetic substances,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$,i.e.,$\chi \propto \frac{1}{T}$.
This implies $\chi_1 T_1 = \chi_2 T_2$.
Given: $\chi_1 = 1.2 \times 10^{-5}$,$T_1 = 300\, K$,and $\chi_2 = 1.8 \times 10^{-5}$.
Substituting the values: $(1.2 \times 10^{-5}) \times 300 = (1.8 \times 10^{-5}) \times T_2$.
$T_2 = \frac{1.2 \times 10^{-5} \times 300}{1.8 \times 10^{-5}} = \frac{1.2}{1.8} \times 300 = \frac{2}{3} \times 300 = 200\, K$.

(D) The relation between magnetic flux density $B$,relative permeability $\mu_r$,permeability of free space $\mu_0$,and magnetizing field $H$ is given by $B = \mu_r \mu_0 H$.
Given:
$H = 2 \times 10^3 \, A/m$
$B = 8 \pi \, T$
$\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$
Rearranging the formula for $\mu_r$:
$\mu_r = \frac{B}{\mu_0 H}$
Substituting the values:
$\mu_r = \frac{8 \pi}{(4 \pi \times 10^{-7}) \times (2 \times 10^3)}$
$\mu_r = \frac{8 \pi}{8 \pi \times 10^{-4}}$
$\mu_r = \frac{1}{10^{-4}} = 10^4 = 10000$

(B) The magnetic susceptibility $\chi$ is given as $7.54 \times 10^{-3} = 0.00754$.
Since $\chi$ is positive and small,the material is paramagnetic.
The relative permeability $\mu_r$ is related to susceptibility by the formula $\mu_r = 1 + \chi$.
Substituting the value: $\mu_r = 1 + 0.00754 = 1.00754$.
Therefore,the substance is paramagnetic and its relative permeability is $1.00754$.

(A) Given: Volume $V = 30\, cm^3 = 30 \times 10^{-6} \, m^3$,Magnetising field $H = \frac{1250}{\pi} \, A/m$,Magnetic moment $M = 6 \, A\cdot m^2$.
First,calculate the intensity of magnetisation $I$:
$I = \frac{M}{V} = \frac{6}{30 \times 10^{-6}} = 0.2 \times 10^6 = 2 \times 10^5 \, A/m$.
Next,calculate the magnetic susceptibility $\chi$:
$\chi = \frac{I}{H} = \frac{2 \times 10^5}{1250/\pi} = \frac{2 \times 10^5 \times \pi}{1250} = 160\pi$.
Calculate relative permeability $\mu_r$:
$\mu_r = 1 + \chi = 1 + 160\pi \approx 1 + 502.65 = 503.65$.
Finally,calculate magnetic induction $B$:
$B = \mu_0 \mu_r H = (4\pi \times 10^{-7}) \times (1 + 160\pi) \times \frac{1250}{\pi}$.
$B = 4 \times 10^{-7} \times (1 + 160\pi) \times 1250 = 5000 \times 10^{-7} \times (1 + 160\pi)$.
$B = 5 \times 10^{-4} \times (1 + 502.65) = 5 \times 10^{-4} \times 503.65 \approx 0.2518 \, T$.
Rounding to the nearest option,the value is $0.25 \, T$.

(A) Curie's Law states that for a paramagnetic material,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$.
The mathematical expression is given by:
$\chi = \frac{C}{T}$
where:
$\chi$ is the magnetic susceptibility.
$C$ is the Curie constant,which is specific to the material.
$T$ is the absolute temperature measured in kelvins $(K)$.
This law indicates that as the temperature increases,the magnetic susceptibility of a paramagnetic material decreases,meaning it becomes less magnetic in the presence of an external field.

(D) The magnetic induction $B$ is given by the formula $B = \mu_0(M_v + H)$,where $M_v$ is the intensity of magnetization and $H$ is the magnetic field strength.
First,calculate the volume $V$ of the rod: $V = 10 \, cm \times 0.5 \, cm \times 0.2 \, cm = 10^{-1} \, m \times 0.5 \times 10^{-2} \, m \times 0.2 \times 10^{-2} \, m = 10^{-6} \, m^3$.
The intensity of magnetization $M_v$ is $M_v = \frac{M}{V} = \frac{5 \, A \cdot m^2}{10^{-6} \, m^3} = 5 \times 10^6 \, A/m$.
The magnetic field strength $H = 0.5 \times 10^4 \, A/m = 5000 \, A/m$.
Now,$B = 4\pi \times 10^{-7} \times (5 \times 10^6 + 5000) = 4 \times 3.14 \times 10^{-7} \times (5,005,000) \approx 12.56 \times 10^{-7} \times 5 \times 10^6 = 6.28 \, T$.

(D) The volume of the cube is $V = (1\, cm)^3 = (10^{-2}\, m)^3 = 10^{-6}\, m^3$.
The intensity of magnetization $I$ is defined as the magnetic dipole moment per unit volume:
$I = \frac{M}{V} = \frac{20 \times 10^{-6}\, J/T}{10^{-6}\, m^3} = 20\, A/m$.
The magnetic susceptibility $\chi$ is given by the ratio of the intensity of magnetization $I$ to the magnetic intensity $H$:
$\chi = \frac{I}{H}$.
Given $H = 60 \times 10^3\, A/m$,we have:
$\chi = \frac{20}{60 \times 10^3} = \frac{1}{3} \times 10^{-3} = 0.333 \times 10^{-3} = 3.33 \times 10^{-4}$.

(B) Volume of the rod $V = 10 \times 0.5 \times 0.2 \, cm^3 = 1 \, cm^3 = 10^{-6} \, m^3$.
Given magnetic field strength $H = 0.5 \times 10^4 \, Am^{-1}$ and magnetic moment $M = 5 \, Am^2$.
Intensity of magnetization $I = \frac{M}{V} = \frac{5}{10^{-6}} = 5 \times 10^6 \, Am^{-1}$.
Magnetic induction $B$ is given by $B = \mu_0(I + H)$.
Since $I \gg H$,we have $B \approx \mu_0 I$.
$B = 4\pi \times 10^{-7} \times 5 \times 10^6 = 20\pi \times 10^{-1} = 2\pi \approx 6.28 \, T$.

(A) The magnetic field intensity $H$ and magnetization $M$ are related to the magnetic flux density $B$ by the equation: $B = \mu_0(H + M)$.
Dividing by $H$, we get $B/H = \mu_0(1 + M/H)$.
Since $B/H = \mu$ (absolute permeability) and $M/H = \chi$ (magnetic susceptibility), we have $\mu = \mu_0(1 + \chi)$.
Dividing by $\mu_0$, we get $\mu/\mu_0 = 1 + \chi$.
Since the relative permeability $\mu_r = \mu/\mu_0$, we obtain the relation $\mu_r = 1 + \chi$, which can be rearranged as $\chi = \mu_r - 1$.

(B) According to Curie's law for ferromagnetic materials above the Curie temperature,the magnetic susceptibility $\chi_m$ is inversely proportional to the absolute temperature $T$,given by $\chi_m = \frac{C}{T - T_c}$. However,for general problems involving temperature dependence of susceptibility where the Curie temperature is not specified,we use the relation $\chi_m \propto \frac{1}{T}$.
Given:
$T_1 = 30\,^oC = 30 + 273 = 303\,K$
$T_2 = 333\,^oC = 333 + 273 = 606\,K$
$\chi_1 = \chi$
Using the relation $\frac{\chi_1}{\chi_2} = \frac{T_2}{T_1}$:
$\frac{\chi}{\chi_2} = \frac{606}{303} = 2$
Therefore,$\chi_2 = \frac{\chi}{2} = 0.5\chi$.

(D) The magnetic field intensity $H$ is independent of the material of the core and is given by:
$H = nI = 1000 \times 2.0 = 2 \times 10^3 \; A/m$.
$(b)$ The magnetic field $B$ is given by $B = \mu_r \mu_0 H$:
$B = 400 \times (4\pi \times 10^{-7} \; T \cdot m/A) \times 2 \times 10^3 \; A/m \approx 1.0 \; T$.
$(c)$ Magnetisation $M$ is given by $M = \chi_m H = (\mu_r - 1)H$:
$M = (400 - 1) \times 2 \times 10^3 = 399 \times 2000 = 7.98 \times 10^5 \; A/m \approx 8 \times 10^5 \; A/m$.
$(d)$ The magnetising current $I_m$ is the additional current required to produce the same magnetic field $B$ in the absence of the core. The relation is $B = \mu_0 n (I + I_m)$.
Since $B = \mu_r \mu_0 n I$,we have $\mu_r \mu_0 n I = \mu_0 n (I + I_m)$,
$\mu_r I = I + I_m \implies I_m = I(\mu_r - 1)$,
$I_m = 2 \times (400 - 1) = 2 \times 399 = 798 \; A$.

(N/A) The volume of the cubic domain is $V = (10^{-6}\; m)^3 = 10^{-18}\; m^3 = 10^{-12}\; cm^3$.
The mass of the domain is $\text{volume} \times \text{density} = 7.9\; g/cm^3 \times 10^{-12}\; cm^3 = 7.9 \times 10^{-12}\; g$.
Using Avogadro's number $(N_A = 6.023 \times 10^{23}\; mol^{-1})$,the number of iron atoms $N$ is given by $N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{7.9 \times 10^{-12}\; g}{55\; g/mol} \times 6.023 \times 10^{23}\; mol^{-1} \approx 8.65 \times 10^{10}$ atoms.
The maximum possible dipole moment $m_{\max}$ is achieved when all atomic moments are perfectly aligned: $m_{\max} = N \times (9.27 \times 10^{-24}\; A m^2) = (8.65 \times 10^{10}) \times (9.27 \times 10^{-24}) \approx 8.0 \times 10^{-13}\; A m^2$.
The magnetisation $M_{\max}$ is $m_{\max} / V = (8.0 \times 10^{-13}\; A m^2) / (10^{-18}\; m^3) = 8.0 \times 10^5\; A/m$.

(C) The mean radius of the Rowland ring is $r = 15\; cm = 0.15\; m$.
The number of turns on the ferromagnetic core is $N = 3500$.
The relative permeability of the core material is $\mu_r = 800$.
The magnetizing current is $I = 1.2\; A$.
The magnetic field $B$ inside a toroid (Rowland ring) is given by the formula:
$B = \frac{\mu_r \mu_0 N I}{2 \pi r}$
Where $\mu_0$ is the permeability of free space,$\mu_0 = 4 \pi \times 10^{-7}\; T\cdot m/A$.
Substituting the values:
$B = \frac{800 \times (4 \pi \times 10^{-7}) \times 3500 \times 1.2}{2 \pi \times 0.15}$
$B = \frac{800 \times 2 \times 10^{-7} \times 3500 \times 1.2}{0.15}$
$B = \frac{800 \times 2 \times 10^{-7} \times 4200}{0.15} = \frac{0.672}{0.15} = 4.48\; T$.
Thus,the magnetic field in the core is $4.48\; T$.

(N/A) Magnetisation $M$ of a sample is defined as its net magnetic moment per unit volume.
$\vec{M} = \frac{\vec{m}_{\text{net}}}{V}$ $(1)$
$M$ is a vector quantity.
The dimension of $M$ is $[L^{-1} A]$.
The $SI$ unit of magnetisation is $A/m$.

(N/A) Consider a long solenoid with $n$ turns per unit length carrying a current $I$.
The magnetic field in the interior of the solenoid due to the current is:
$\vec{B}_{0} = \mu_{0} n I \quad \dots (1)$
If the interior of the solenoid is filled with a magnetic material,the total magnetic field $\vec{B}$ inside the solenoid is the sum of the field due to the current $(\vec{B}_{0})$ and the field due to the magnetization of the material $(\vec{B}_{m})$:
$\vec{B} = \vec{B}_{0} + \vec{B}_{m} \quad \dots (2)$
The additional field $\vec{B}_{m}$ is proportional to the magnetization $\vec{M}$ of the material:
$\vec{B}_{m} = \mu_{0} \vec{M} \quad \dots (3)$
We define the magnetic intensity $\vec{H}$ as:
$\vec{H} = \frac{\vec{B}}{\mu_{0}} - \vec{M} \quad \dots (4)$
Rearranging this equation,we get the relation for the total magnetic field $\vec{B}$:
$\vec{B} = \mu_{0}(\vec{H} + \vec{M}) \quad \dots (5)$
Here,$\vec{H}$ represents the contribution from external currents,and $\vec{M}$ represents the contribution from the magnetic material.

(N/A) The magnetization $M$ of a material is proportional to the magnetic intensity $(H)$.
$\overrightarrow{M} = \chi \overrightarrow{H}$....$(1)$
Here,$\chi$ is a dimensionless quantity called magnetic susceptibility. It measures how a magnetic material responds to an external field.
For paramagnetic materials,$\chi$ is small and positive.
For diamagnetic materials,$\chi$ is small and negative because $\overrightarrow{M}$ and $\overrightarrow{H}$ are in opposite directions.
Consider a magnetic material placed inside a solenoid. The total magnetic field $B$ when a current $I$ flows through the solenoid is:
$\overrightarrow{B} = \mu_{0}(\overrightarrow{H} + \overrightarrow{M})$....$(2)$
Substituting equation $(1)$ into $(2)$:
$\overrightarrow{B} = \mu_{0}(\overrightarrow{H} + \chi \overrightarrow{H}) = \mu_{0}(1 + \chi) \overrightarrow{H}$....$(3)$
Here,$\mu = \mu_{0}(1 + \chi)$ is defined as the magnetic permeability of the material.
Thus,$\mu = \mu_{0}(1 + \chi)$....$(4)$
Dividing by $\mu_{0}$,we get the relative magnetic permeability $\mu_{r}$:
$\mu_{r} = \frac{\mu}{\mu_{0}} = 1 + \chi$....$(5)$
Therefore,the relation is $\mu = \mu_{0} \mu_{r}$....$(6)$
Substituting this into equation $(3)$,we get:
$\overrightarrow{B} = \mu \overrightarrow{H}$....$(7)$
$\mu_{r}$ is a dimensionless quantity,analogous to the dielectric constant in electrostatics.

(N/A) Magnetisation $(M)$ of a substance is defined as the net magnetic dipole moment per unit volume of the material.
Mathematically,it is expressed as:
$M = \frac{m_{net}}{V}$
where $m_{net}$ is the net magnetic dipole moment and $V$ is the volume of the substance.
The $SI$ unit of magnetisation is $A/m$ (ampere per meter).

(N/A) Magnetisation $(M)$ is defined as the magnetic moment per unit volume of a material.
Mathematically,$M = \frac{m_{net}}{V}$,where $m_{net}$ is the net magnetic moment and $V$ is the volume.
The $SI$ unit of magnetic moment $(m_{net})$ is $A \cdot m^2$ and the unit of volume $(V)$ is $m^3$.
Therefore,the $SI$ unit of magnetisation is $\frac{A \cdot m^2}{m^3} = A \cdot m^{-1}$ (Ampere per meter).
The dimensional formula of magnetic moment is $[M^0 L^2 T^0 I^1]$ and the dimensional formula of volume is $[M^0 L^3 T^0 I^0]$.
Thus,the dimensional formula of magnetisation is $\frac{[I^1 L^2]}{[L^3]} = [M^0 L^{-1} T^0 I^1]$.

(N/A) The total magnetic field $\vec{B}$ inside the core of a solenoid is the sum of the magnetic field due to the current in the solenoid windings $(\vec{B}_0)$ and the magnetic field due to the magnetization of the material $(\vec{B}_m)$.
Mathematically,this is expressed as:
$\vec{B} = \vec{B}_0 + \vec{B}_m$
Since $\vec{B}_0 = \mu_0 \vec{H}$ and $\vec{B}_m = \mu_0 \vec{M}$,where $\vec{H}$ is the magnetic intensity and $\vec{M}$ is the magnetization of the material,we can write:
$\vec{B} = \mu_0 \vec{H} + \mu_0 \vec{M}$
$\vec{B} = \mu_0 (\vec{H} + \vec{M})$

(A) Magnetic susceptibility,denoted by the symbol $\chi_m$,is a dimensionless proportionality constant that indicates the degree of magnetization of a material in response to an applied magnetic field.
Mathematically,it is defined as the ratio of the intensity of magnetization $(M)$ to the magnetic intensity $(H)$: $\chi_m = \frac{M}{H}$.
Therefore,it shows how easily a material can be magnetized when placed in an external magnetic field.

(B) Given: Intensity of magnetization $M = 10^6$ $A/m$. Length of the cylinder $l = 10$ $cm = 0.1$ $m = 10^{-1}$ $m$.
Magnetization current $I_M$ is related to magnetization $M$ by the formula:
$M = \frac{I_M}{l}$
Therefore,the magnetization current is:
$I_M = M \times l$
Substituting the given values:
$I_M = (10^6$ $A/m) \times (10^{-1}$ $m)$
$I_M = 10^5$ $A$.

(A) Magnetic susceptibility $\chi$ is defined as the ratio of magnetization $M$ to magnetic field intensity $H$. Since both $M$ and $H$ have the same units $(A/m)$,$\chi$ is a dimensionless quantity,i.e.,$[M^0 L^0 T^0]$.
To construct an expression for $\chi$ using $e, m, v, R, \mu_0$,we note that $\mu_0$ has dimensions $[M L T^{-2} Q^{-2}]$ and $e^2$ has dimensions $[Q^2]$. Thus,$\mu_0 e^2$ has dimensions $[M L T^{-2}]$.
Let $\chi = \mu_0 e^2 m^a v^b R^c$.
Substituting dimensions: $[M^0 L^0 T^0] = [M L T^{-2}] [M]^a [L T^{-1}]^b [L]^c$.
Equating powers of $M, L, T$:
For $M$: $1 + a = 0 \implies a = -1$.
For $T$: $-2 - b = 0 \implies b = -2$.
For $L$: $1 + b + c = 0 \implies 1 - 2 + c = 0 \implies c = 1$.
Thus,$\chi \propto \frac{\mu_0 e^2 R}{m v^2}$.
Using Bohr's model,$v^2 = \frac{e^2}{4 \pi \epsilon_0 m R}$,so $\frac{e^2}{m R} \approx v^2$. Substituting this,$\chi \approx \mu_0 \epsilon_0 v^2 \approx \frac{v^2}{c^2}$.
Given $v \approx \alpha c$ (where $\alpha \approx 1/137$),$\chi \approx \alpha^2 \approx (1/137)^2 \approx 5 \times 10^{-5}$,which is consistent with the order of $10^{-5}$ for diamagnetic materials.

(D) The magnetic field inside the solenoid is $B_0 = \mu_0 n i$.
When an iron rod is placed inside,the magnetic field becomes $B = \mu_r B_0 = \mu_r \mu_0 n i$.
The magnetization $M_z$ is given by $M_z = \frac{B - \mu_0 H}{\mu_0} = (\mu_r - 1) H$.
Since $\mu_r = 1000 \gg 1$,we can approximate $M_z \approx \mu_r H = \mu_r n i$.
The magnetic moment $m$ is given by $m = M_z \times V$,where $V$ is the volume of the rod.
Given: $V = 10^{-3} \ m^3$,$\mu_r = 1000$,$n = 10 \ turns/cm = 1000 \ turns/m$,$i = 0.5 \ A$.
$m = (\mu_r n i) V = 1000 \times 1000 \times 0.5 \times 10^{-3}$.
$m = 10^6 \times 0.5 \times 10^{-3} = 0.5 \times 10^3 = 500 \ Am^2$.
$m = 5 \times 10^2 \ Am^2$.

(B) The relative permeability $\mu_{r}$ is related to magnetic susceptibility $\chi_{m}$ by the formula: $\mu_{r} = 1 + \chi_{m}$.
Given $\chi_{m} = 599$,we have $\mu_{r} = 1 + 599 = 600$.
The absolute permeability $\mu$ of the material is given by $\mu = \mu_{0} \mu_{r}$.
Substituting the values,$\mu = (4 \pi \times 10^{-7} \, T m A^{-1}) \times 600$.
$\mu = 2400 \pi \times 10^{-7} \, T m A^{-1} = 2.4 \pi \times 10^{-4} \, T m A^{-1}$.

(B) The total magnetic field $B$ inside a toroid is given by $B = \mu_{0}(H + I)$,where $H$ is the magnetic field intensity and $I$ is the intensity of magnetization.
The magnetic field intensity $H$ for a toroid is given by $H = \frac{N i}{2 \pi r}$,where $N = 400$,$i = 2\,A$,and the radius $r = \frac{2.5}{2} = 1.25\,m$.
$H = \frac{400 \times 2}{2 \pi \times 1.25} = \frac{800}{2.5 \pi} = \frac{320}{\pi} \approx 101.86\,A/m$.
Given $B = 10\,T$ and $\mu_{0} = 4 \pi \times 10^{-7}\,T\cdot m/A$,we have:
$10 = 4 \pi \times 10^{-7} (H + I)$
$H + I = \frac{10}{4 \pi \times 10^{-7}} = \frac{10^{8}}{4 \pi}$
Since $H$ is very small compared to $I$ (as $I \approx \frac{10^{8}}{4 \pi} \approx 7.96 \times 10^{6}$),the intensity of magnetization $I \approx \frac{10^{8}}{4 \pi}\,A/m$.

(C) The absolute permeability $\mu$ is given by the formula: $\mu = \mu_0(1 + \chi_m)$,where $\mu_0$ is the permeability of free space and $\chi_m$ is the magnetic susceptibility.
Given: $\mu_0 = 4\pi \times 10^{-7} \text{ H/m}$ and $\chi_m = 499$.
Substituting the values:
$\mu = 4\pi \times 10^{-7} \times (1 + 499)$
$\mu = 4\pi \times 10^{-7} \times 500$
$\mu = 4\pi \times 10^{-7} \times 5 \times 10^2$
$\mu = 20\pi \times 10^{-5} \text{ H/m}$
$\mu = 2\pi \times 10^{-4} \text{ H/m}$.
Comparing this with the given format $....\pi \times 10^{-4} \text{ H/m}$,the missing value is $2$.

(A) The magnetic field $B$ inside a material is given by $B = \mu_0(H + M) = \mu_0 H(1 + \chi)$,where $\chi$ is the magnetic susceptibility.
The magnetic field in vacuum is $B_0 = \mu_0 H$.
Thus,$B = B_0(1 + \chi)$.
The increase in the magnetic field is $\Delta B = B - B_0 = B_0 \chi$.
The percentage increase is given by $\frac{\Delta B}{B_0} \times 100 = \chi \times 100$.
Given $\chi = 2.2 \times 10^{-5}$,the percentage increase is $(2.2 \times 10^{-5}) \times 100 = 2.2 \times 10^{-3} = \frac{2.2}{10^3} = \frac{22}{10^4}$.
Comparing this with $\frac{x}{10^4}$,we get $x = 22$.

(B) The magnetic force on a current-carrying conductor is given by $F = BIl \sin \theta$. Thus,the magnetic field $B = F / (Il)$.
The dimensions of force $F$ are $[MLT^{-2}]$.
The dimensions of current $I$ are $[A]$.
The dimensions of length $l$ are $[L]$.
Therefore,the dimensions of $B$ are $[MLT^{-2}] / ([A][L]) = [MT^{-2}A^{-1}]$.
The magnetic permeability $\mu$ is related to $B$ and $H$ by $B = \mu H$,where $H$ has dimensions $[IL^{-1}] = [AL^{-1}]$.
Thus,$\mu = B / H = [MT^{-2}A^{-1}] / [AL^{-1}] = [MLT^{-2}A^{-2}]$.
Hence,the dimensions $[MLT^{-2}A^{-2}]$ correspond to magnetic permeability.

(C) Given,magnetic susceptibility $\chi = 99$.
The relative permeability $\mu_{r}$ is related to susceptibility by the formula $\mu_{r} = 1 + \chi$.
Substituting the value of $\chi$,we get $\mu_{r} = 1 + 99 = 100$.
The absolute permeability $\mu$ is given by $\mu = \mu_{0} \mu_{r}$.
Given $\mu_{0} = 4 \pi \times 10^{-7} \ Wb / A-m$.
Therefore,$\mu = (4 \pi \times 10^{-7}) \times 100$.
$\mu = 4 \pi \times 10^{-5} \ Wb / A-m$.

(A) The magnetic field inside an air-cored solenoid is given by $B_0 = \mu_0 n I$.
When the solenoid is filled with a magnetic material of relative permeability $\mu_r$,the magnetic field becomes $B = \mu_0 \mu_r n I$.
The magnetic susceptibility $\chi$ is related to relative permeability by the formula $\mu_r = 1 + \chi$.
Given $\chi = 1.2 \times 10^{-5}$,we have $\mu_r = 1 + 1.2 \times 10^{-5}$.
The fractional increase in the magnetic field is given by $\frac{\Delta B}{B_0} = \frac{B - B_0}{B_0} = \frac{\mu_0 \mu_r n I - \mu_0 n I}{\mu_0 n I} = \mu_r - 1$.
Substituting $\mu_r = 1 + \chi$,we get $\frac{\Delta B}{B_0} = (1 + \chi) - 1 = \chi$.
Therefore,the fractional increase is $1.2 \times 10^{-5}$.

(B) The magnetic induction $B$ in a material is given by the relation:
$B = \mu_0(M + H)$,where $M$ is the intensity of magnetization and $H$ is the magnetising field intensity.
We know that the magnetic susceptibility $\chi_m$ is defined as the ratio of magnetization to the magnetising field intensity:
$M = \chi_m H$.
Substituting this into the first equation:
$B = \mu_0(\chi_m H + H) = \mu_0 H(\chi_m + 1)$.
Rearranging the terms to solve for $\chi_m$:
$\frac{B}{\mu_0 H} = \chi_m + 1$.
Therefore,$\chi_m = \frac{B}{\mu_0 H} - 1$.

(A) The magnetic permeability $\mu$ of a substance is defined as the ratio of the magnetic flux density $B$ to the magnetizing field $H$,given by $\mu = \frac{B}{H}$.
For a ferromagnetic material,as the magnetizing field $H$ increases,the magnetic flux density $B$ increases initially but eventually reaches a state of magnetic saturation.
Since $B$ approaches a constant saturation value while $H$ continues to increase,the ratio $\mu = \frac{B}{H}$ decreases as $H$ increases. Therefore,the permeability of a ferromagnetic material decreases with an increase in the magnetizing field.