Magnetic Materials (Diamagnetic, Paramagnetic and Ferromagnetic) Questions in English

(A) Electromagnets require materials that can be easily magnetized and demagnetized. Soft iron has high permeability and low retentivity,which makes it easy to magnetize. Additionally,soft iron has low coercivity,meaning it can be demagnetized easily by a small reverse magnetic field. Therefore,both the assertion and the reason are true,and the reason correctly explains why soft iron is preferred for electromagnets.

(C) Diamagnetic substances are weakly repelled by magnetic fields.
When placed in a non-uniform magnetic field,they experience a force that pushes them from the stronger region of the field toward the weaker region.
This behavior is a characteristic property of diamagnetic materials.

(A) Ferromagnetic materials are classified based on their retentivity and coercivity.
Soft ferromagnetic materials have low retentivity and low coercivity,meaning they are easily magnetized and demagnetized.
Therefore,when the external magnetic field is removed,their magnetization disappears.
In contrast,hard ferromagnetic materials retain their magnetization even after the external field is removed.

(C) The Curie temperature $T_{C}$ is the critical temperature above which a ferromagnetic material loses its spontaneous magnetization and behaves as a paramagnetic material.
When a ferromagnetic substance is heated above its Curie temperature,the thermal agitation of the atoms overcomes the alignment of magnetic dipoles,resulting in a transition to a paramagnetic state.
Therefore,the Curie temperature represents the transition from ferromagnetic to paramagnetic.

(B) According to Curie's law, the magnetization $M$ of a paramagnetic material is given by $M = C \frac{B}{T}$, where $B$ is the external magnetic field, $T$ is the absolute temperature, and $C$ is Curie's constant.
Given $M_1 = 0.8 \text{ A m}^{-1}$, $B_1 = 0.8 \text{ T}$, and $T_1 = 5 \text{ K}$.
Substituting these values into the formula: $0.8 = C \frac{0.8}{5} \Rightarrow C = 5 \text{ K}$.
Now, for the new temperature $T_2 = 20 \text{ K}$ with the same magnetic field $B_2 = 0.8 \text{ T}$, the new magnetization $M_2$ is:
$M_2 = C \frac{B_2}{T_2} = 5 \times \frac{0.8}{20} = \frac{4}{20} = 0.2 \text{ A m}^{-1}$.

(D) For a material to be suitable for a permanent magnet,it must retain its magnetism even after the external magnetic field is removed.
$1$. High Retentivity: This ensures that the material remains strongly magnetized after the removal of the magnetizing field.
$2$. High Coercivity: This ensures that the magnetization is not easily destroyed by stray magnetic fields,temperature fluctuations,or minor mechanical impacts.
Therefore,materials for permanent magnets must have both high retentivity and high coercivity.

(B) The total dipole moment of the sample at saturation is $M_{\text{sat}} = N \mu = (2 \times 10^{24}) \times (1.5 \times 10^{-23} \text{ J T}^{-1}) = 30 \text{ J T}^{-1}$.
According to Curie's law for paramagnetism, the magnetization $M$ is proportional to $B/T$, i.e., $M \propto B/T$.
In the first case, the degree of saturation is $20 \%$, so the magnetic moment is $M_1 = 0.20 \times M_{\text{sat}} = 0.20 \times 30 = 6 \text{ J T}^{-1}$.
The ratio of magnetic moments is given by $\frac{M_2}{M_1} = \frac{B_2 / T_2}{B_1 / T_1}$.
Substituting the values: $\frac{M_2}{6} = \frac{0.9 / 2.8}{0.6 / 4.2} = \frac{0.9}{2.8} \times \frac{4.2}{0.6} = \frac{0.9}{0.6} \times \frac{4.2}{2.8} = 1.5 \times 1.5 = 2.25$.
Therefore, $M_2 = 6 \times 2.25 = 13.5 \text{ J T}^{-1}$.

(D) When any magnetic material is heated to a high temperature,it loses its magnetic property. This temperature is known as the Curie temperature. Above this temperature,the thermal agitation of the atoms overcomes the magnetic alignment,resulting in the loss of magnetism.

(C) According to Curie's Law,the magnetic susceptibility $\chi_m$ of a paramagnetic substance is inversely proportional to its absolute temperature $T$,i.e.,$\chi_m \propto \frac{1}{T}$.
Given:
Initial temperature $T_1 = -173^{\circ} C = (-173 + 273) K = 100 K$.
Initial susceptibility $\chi_{m1} = 1.5 \times 10^{-2}$.
Final susceptibility $\chi_{m2} = 0.5 \times 10^{-2}$.
Using the relation $\chi_{m1} T_1 = \chi_{m2} T_2$:
$(1.5 \times 10^{-2}) \times 100 = (0.5 \times 10^{-2}) \times T_2$.
$T_2 = \frac{1.5 \times 100}{0.5} = 300 K$.
Converting $T_2$ to Celsius: $T_2 = 300 K - 273 = 27^{\circ} C$.
The change in temperature $\Delta T = T_2 - T_1 = 27^{\circ} C - (-173^{\circ} C) = 200^{\circ} C$.

(A) Superconductors are materials that exhibit perfect diamagnetism below a critical temperature,known as the Meissner effect. In this state,the magnetic field inside the material is zero,making them the most exotic diamagnetic materials.

(B) The degree to which a material can be magnetised by applying an external magnetic field is called magnetic susceptibility,denoted by $\chi_m$.
For ferromagnetic materials,the magnetic susceptibility is very large and positive,which means $\chi_m > 1$.
This high value indicates that these materials are strongly attracted by an external magnetic field.
Therefore,the correct option is $B$.

(D) Materials suitable for permanent magnets must possess high retentivity so that they can retain the magnetic field strongly.
They must have high coercivity so that the magnetization is not easily destroyed by external magnetic fields,temperature fluctuations,or minor mechanical damage.
They should also have high permeability to allow for easy magnetization.

(D) The relationship between relative permeability $\mu_r$ and magnetic susceptibility $\chi$ is given by the formula: $\mu_r = 1 + \chi$.
For ferromagnetic substances,the magnetic susceptibility $\chi$ is a large positive value,typically $\chi >> 1$.
Consequently,the relative permeability $\mu_r$ is also much greater than $1$ (i.e.,$\mu_r >> 1$).
Therefore,the correct condition for a ferromagnetic substance is $\chi >> 1$.

(C) The material used for making permanent magnets must have high retentivity so that it produces a strong magnetic field.
Additionally,it must have high coercivity so that its magnetization is not easily destroyed by stray magnetic fields,temperature variations,or minor mechanical damage.
Therefore,a material with both high retentivity and high coercivity is desirable.

(A) The magnetic susceptibility $(\chi)$ of a diamagnetic material is a negative quantity,$(\chi < 0)$,whereas paramagnetic and ferromagnetic materials have positive susceptibility $(\chi > 0)$. Therefore,statement $(A)$ is incorrect.
Paramagnetic materials obey Curie's law,which states that susceptibility is inversely proportional to absolute temperature.
Ferromagnetic materials consist of small regions called magnetic domains,each containing approximately $10^{17}$ atoms,which act as permanent magnetic dipoles.
Soft ferromagnetic materials have low retentivity,meaning their magnetisation disappears easily upon the removal of the external magnetic field.
Thus,statement $(A)$ is the incorrect statement.

(D) Statement $(A)$ is wrong because domain theory is used to explain ferromagnetism,not paramagnetism.
Statement $(B)$ is correct because the magnetic susceptibility of diamagnetic materials is independent of temperature,as it arises from the orbital motion of electrons which is not significantly affected by thermal agitation.

(B) The relationship between magnetic susceptibility $\chi$ and relative permeability $\mu_r$ is given by $\chi = \mu_r - 1$.
Also,the absolute permeability $\mu$ is related to relative permeability by $\mu = \mu_r \mu_0$.
For a paramagnetic substance,$\chi$ is small and positive $(\chi > 0)$,which implies $\mu_r > 1$,and therefore $\mu > \mu_0$.
For a diamagnetic substance,$\chi$ is small and negative $(\chi < 0)$,which implies $\mu_r < 1$,and therefore $\mu < \mu_0$.
For a ferromagnetic substance,$\chi$ is very large and positive $(\chi \gg 1)$,which implies $\mu_r \gg 1$,and therefore $\mu \gg \mu_0$.
Comparing these with the given options,both $(b)$ and $(d)$ are scientifically correct statements.

(A) $1$. For paramagnetic materials,according to Curie's Law,$\chi = \frac{C}{T}$,which implies $\frac{1}{\chi} = \frac{T}{C}$. This represents a straight line passing through the origin with a positive slope.
$2$. For ferromagnetic materials above the Curie temperature $(T_c)$,the susceptibility follows the Curie-Weiss Law: $\chi = \frac{C}{T - T_c}$,which implies $\frac{1}{\chi} = \frac{T - T_c}{C}$. This represents a straight line with a positive slope that does not pass through the origin.
$3$. For diamagnetic materials,the susceptibility $\chi$ is small,negative,and independent of temperature. Thus,$\frac{1}{\chi}$ is a constant negative value,resulting in a horizontal line (zero slope) in the $\frac{1}{\chi}$ vs $T$ graph.
$4$. In the given figure,line $A$ passes through the origin with a positive slope,indicating it is paramagnetic. Line $B$ is a horizontal line (zero slope),indicating it is diamagnetic.
$5$. Therefore,$A$ is paramagnetic and $B$ is diamagnetic.

(C) Assertion $(A)$ is true because ferromagnetic materials consist of small regions called domains where atomic magnetic dipoles are aligned due to strong exchange interactions.
Reason $(R)$ is also true because thermal agitation at high temperatures disrupts the alignment of these magnetic dipoles,causing the domain structure to break down.
When the temperature exceeds the Curie temperature $(T_c)$,the material transitions from ferromagnetic to paramagnetic behavior,meaning the spontaneous magnetization disappears.
Since the disintegration of the domain structure at high temperatures is the fundamental reason why the ferromagnetic property (and thus spontaneous magnetization) is lost,Reason $(R)$ correctly explains Assertion $(A)$.

(B) Paramagnetic substances possess a small positive magnetic susceptibility. When placed in a non-uniform magnetic field,they experience a net force directed towards the region of higher magnetic field intensity. Therefore,a paramagnetic substance tends to move from a region of weak magnetic field to a region of strong magnetic field.

(D) Superconductors exhibit the Meissner effect,which means they are perfect diamagnetic materials.
For a perfect diamagnetic material,the internal magnetic field $B$ is equal to $0$.
Since $B = \mu_0(H + M) = 0$,we have $M = -H$.
The magnetic susceptibility $\chi$ is defined as the ratio of magnetization $M$ to the magnetic field intensity $H$,i.e.,$\chi = M/H$.
Substituting $M = -H$,we get $\chi = -H/H = -1$.