Magnetism and Matter

Coercivity is equal to magnetising field intensity required to reduce the magnetic field of a ferromagnetic substance to zero after it has been magnetised to saturation. Thus, point R represents coercivity of the material.

Total number of atoms in sample $N=$ Number of atom per cubic metre × Volume of the sample

$ \begin{aligned} & =8.7 \times 10^{28} \times\left(1 \times 10^{-6}\right)^3 \\ & =8.7 \times 10^{10} \text { atoms } \end{aligned} $

Maximum possible dipole moment

$ \begin{aligned} M_{\max }= & N \times \text { magnetic dipole moment of } \\ & \text { each atom } \\ = & 8.7 \times 10^{10} \times 9.3 \times 10^{-24} \\ = & 8.09 \times 10^{-13}=8.1 \times 10^{-13} \\ & \approx 81 \times 10^{-14} \mathrm{Am}^2 \end{aligned} $

We use Curie's law to understand how the magnetism of a paramagnetic material changes with temperature and magnetic field.

According to Curie's law, the magnetisation ($M$) is directly proportional to the magnetic field ($B_0$) and inversely proportional to the temperature ($T$):

$M \propto \frac{B_0}{T}$ or, $M=C \cdot \frac{B_0}{T}$ where $C$ (Curie's constant) stays the same for the same substance.

Curie's constant can be written as: $C = \frac{MT}{B_0}$ where:

• $T$ = temperature in Kelvin
• $B_0$ = external magnetic field
• $M$ = magnetisation (total dipole moment)

Since $C$ (Curie's constant) is the same for all measurements of the same sample, we can relate two cases as:

$ \frac{M_1 T_1}{B_{0_1}} = \frac{M_2 T_2}{B_{0_2}} $ So, $ M_2 = M_1 \cdot \frac{T_1}{T_2} \cdot \frac{B_{0_2}}{B_{0_1}} $

Now, let's fill in the given values:

• Number of dipoles $= 2 \times 10^{24}$
• Dipole moment per dipole $= 15 \times 10^{-23}~\mathrm{JT}^{-1}$

Step 1: Find Initial Total Dipole Moment

Total dipole moment possible (if 100% aligned): $2 \times 10^{24} \times 15 \times 10^{-23} = 3 \times 10^{2}~\mathrm{JT}^{-1}$ But, at first, only 20% saturation is achieved: $M_1 = 0.2 \times 3 \times 10^{2} = 60~\mathrm{JT}^{-1}$

Step 2: Use Curie’s Law for New Conditions

Magnetic field changes from $B_{0_1} = 0.6~T$ to $B_{0_2} = 0.9~T$ and temperature from $T_1 = 4.2~K$ to $T_2 = 2.8~K$.

Now, use the formula: $M_2 = M_1 \times \frac{T_1}{T_2} \times \frac{B_{0_2}}{B_{0_1}}$

Substitute the numbers in: $M_2 = 60 \times \frac{4.2}{2.8} \times \frac{0.9}{0.6}$

Calculate step by step:

• $\frac{4.2}{2.8} = 1.5$
• $\frac{0.9}{0.6} = 1.5$

$W=M B\left(\cos \theta_1-\cos \theta_2\right)$

Initially, $\theta_1=0^{\circ}, \theta_2=45^{\circ}$

$ \begin{aligned} \therefore W_1 & =M B\left(\cos 0-\cos 45^{\circ}\right) \\ & =M B\left(1-\frac{1}{\sqrt{2}}\right) \\ & =\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) M B \end{aligned} $

Again, $\theta_2=45^{\circ}+15^{\circ}=60^{\circ}$

$ \begin{aligned} \therefore W_2 & =M B\left(\cos 45^{\circ}-\cos 60^{\circ}\right) \\ & =M B\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right) \\ & =M B\left(\frac{\sqrt{2}-1}{2}\right) \end{aligned} $

$ \begin{aligned} & \therefore \frac{W_2}{W_1}=\frac{M B\left(\frac{\sqrt{2}-1}{2}\right)}{M B\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}=\frac{1}{\sqrt{2}} \\ & \Rightarrow W_2=\frac{W_1}{\sqrt{2}}=\frac{W}{\sqrt{2}} \end{aligned} $

Materials suitable for permanent magnets should have high retentivity and high coercivity.

Given data

• Magnetic moment of the bar magnet:

  $ M = 10^4 \ \text{J T}^{-1} $

• Magnetic field:

  $ B = 4 \times 10^{-5} \ \text{T} $

• Initial orientation: $ \theta_1 = 0^\circ $ (parallel to $ B $)

• Final orientation: $ \theta_2 = 60^\circ $

Relevant concept

The potential energy of a magnetic dipole in a magnetic field is

$ U = - M B \cos\theta $

When the magnet is rotated slowly (quasi-statically), the work done by an external agent equals the change in potential energy of the system:

$ W = U_2 - U_1 = [-M B \cos\theta_2] - [-M B \cos\theta_1] $

$ \Rightarrow W = M B (\cos\theta_1 - \cos\theta_2) $

Substitute the values

$ W = (10^4)(4 \times 10^{-5})(\cos 0^\circ - \cos 60^\circ) $

Compute step by step:

$ \cos 0^\circ = 1, \quad \cos 60^\circ = \tfrac{1}{2} $

So,

$ W = (10^4)(4 \times 10^{-5})(1 - 0.5) $

$ W = (10^4)(4 \times 10^{-5})(0.5) $

$ W = (10^4 \times 4 \times 10^{-5} \times 0.5) $

$ W = (4 \times 10^{-1} \times 0.5) = 0.2\ \text{J} $

✅ Final Answer:

$ \boxed{W = 0.2\ \text{J}} $

Correct option: (A) 0.2 J

$ \text { } \begin{aligned} B & =\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r^3}=10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} \\ & =0.96 \times 10^{-4} \mathrm{~T}=0.96 \text { Gauss } \end{aligned} $

$ \begin{aligned} &\text { Torque on bar magnet, }\\ &\begin{aligned} \tau & =M B \sin \theta \\ \Rightarrow \quad M & =\frac{\tau}{B \sin \theta}=\frac{0.36 \sqrt{2}}{2 \sin 45^{\circ}} \\ & =\frac{0.36 \sqrt{2}}{2 \times \frac{1}{\sqrt{2}}}=\frac{0.36 \sqrt{2}}{\sqrt{2}}=0.36 \mathrm{~J} / \mathrm{T} \end{aligned} \end{aligned} $

To find the value of $ H $ that produces a flux density of $ 1 \, \text{T} $, we are given:

$ \mu = \left[\frac{0.4}{H} + 12 \times 10^{-4}\right] \, \text{H/m} $

We also know the relationship between $ B $ and $ \mu H $:

$ B = \mu H $

Given $ B = 1 \, \text{T} $, substitute into the equation:

$ \mu = \frac{0.4}{H} + 12 \times 10^{-4} $

Multiply both sides by $ H $:

$ \mu H = 0.4 + 12 \times 10^{-4} H $

Since $ B = \mu H = 1 \, \text{T} $, substitute the expression for $ \mu H $:

$ 1 = 0.4 + 12 \times 10^{-4} H $

Rearrange to solve for $ H $:

$ 1 - 0.4 = 12 \times 10^{-4} H $

$ 0.6 = 12 \times 10^{-4} H $

$ H = \frac{0.6}{12 \times 10^{-4}} $

$ H = \frac{6000}{12} $

$ H = 500 \, \text{A/m} $

Thus, the value of $ H $ is $ 500 \, \text{Am}^{-1} $.

To calculate the torque experienced by a compass needle due to the Earth's magnetic field, we use the following information:

The angle of declination is $ \alpha = 30^{\circ} $.

The magnetic moment of the compass is $ M = 18 \, \mathrm{Am}^2 $.

The magnetic field is $ B = 3 \times 10^{-5} \, \mathrm{T} $.

The formula to calculate torque ($\tau$) is:

$ \tau = M B \sin \theta $

where $\theta$ is the angle between the magnetic moment and the magnetic field, which in this case is the angle of declination, $30^{\circ}$. Substituting the given values into the formula:

$ \tau = 18 \times 3 \times 10^{-5} \times \sin 30^{\circ} $

Since $\sin 30^{\circ} = \frac{1}{2}$, the calculation becomes:

$ \tau = 18 \times 3 \times 10^{-5} \times \frac{1}{2} $

$ \tau = 27 \times 10^{-5} \, \mathrm{Nm} $

Therefore, the torque experienced by the compass needle is $ 27 \times 10^{-5} \, \mathrm{Nm} $.

Given the vertical component of the Earth's magnetic field as $ B_V = 0.45 \, \mathrm{G} $ and the angle of dip $\delta = 60^\circ$, we can determine the Earth's magnetic field at this location.

Calculate the Horizontal Component ($ B_H $):

The formula for the angle of dip is:

$ \tan \delta = \frac{B_V}{B_H} $

Given $\delta = 60^\circ$, we find that:

$ B_H = \frac{B_V}{\tan 60^\circ} = \frac{0.45}{\sqrt{3}} \approx 0.26 \, \mathrm{G} $

Calculate the Earth's Magnetic Field ($ B_e $):

The Earth's total magnetic field can be found using the horizontal component and the angle of dip:

$ \cos \delta = \frac{B_H}{B_e} $

With $\cos 60^\circ = 0.5$, we have:

$ \begin{aligned} B_e &= \frac{B_H}{\cos 60^\circ} \\ &= \frac{0.26}{0.5} = 0.52 \, \mathrm{G} \end{aligned} $

Thus, the magnetic field of Earth at this location is $0.52 \, \mathrm{G}$.

Given:

Earth's magnetic field, $ B_e = 4 \times 10^{-5} \, \text{T} $

Distance from the magnet, $ r = 0.4 \, \text{m} $

The magnetic field due to a short bar magnet at a point on its normal bisector is given by:

$ B_m = \frac{\mu_0}{4\pi} \frac{M}{r^3} $

For a $ 45^\circ $ angle of inclination of the resultant magnetic field with Earth's field, we have:

$ B_m = B_e $

Substituting the values, we get:

$ 4 \times 10^{-5} = \frac{\mu_0}{4\pi} \frac{M}{(0.4)^3} $

This simplifies to:

$ 4 \times 10^{-5} = \frac{10^{-7} \times M}{(0.4)^3} $

Solving for $ M $, the magnetic moment:

$ M = 2.56 \times 10^1 $

Therefore, the magnetic moment of the magnet is:

$ M = 25.6 \, \text{Am}^2 $

The correct answer is Option D: heated to high temperature.

Here's why:

Magnetic materials (especially ferromagnetic ones like iron) have a special temperature called the Curie temperature.

When heated above this Curie temperature, the thermal agitation disrupts the alignment of the magnetic domains.

As a result, the material loses its permanent magnetic properties.

Thus, heating the material to a high temperature causes it to lose its magnetism.

To calculate the magnetization of a ferromagnetic material domain, follow these steps:

Determine the Volume of the Domain:

The domain is a cube with each side measuring $2 \, \mu\mathrm{m}$. First, convert this measurement to meters:

$ 2 \, \mu\mathrm{m} = 2 \times 10^{-6} \, \mathrm{m} $

Therefore, the volume $V$ of the cube is:

$ V = (2 \times 10^{-6})^3 = 8 \times 10^{-18} \, \mathrm{m}^3 $

Use Given Values:

Number of atoms in the domain, $ N = 9 \times 10^{10} $.

Dipole moment of each atom, $ M = 9 \times 10^{-24} \, \mathrm{Am}^2 $.

Calculate the Total Dipole Moment:

The total dipole moment $ M' $ for all the atoms is given by:

$ M' = N \times M = 9 \times 10^{10} \times 9 \times 10^{-24} = 81 \times 10^{-14} \, \mathrm{Am}^2 $

Find the Magnetization:

Magnetization $ \text{Magnetization} $ is the total dipole moment $ M' $ divided by the volume $ V $:

$ \text{Magnetization} = \frac{M'}{V} = \frac{81 \times 10^{-14}}{8 \times 10^{-18}} = 9 \times 10^4 \, \mathrm{Am}^{-1} $

Thus, the magnetization of the domain is approximately $ 9 \times 10^4 \, \mathrm{Am}^{-1} $.

We know that the time period for an oscillating magnet is given by:

$ T = 2 \pi \sqrt{\frac{I}{M B_H}} $

where:

$ M $ is the magnetic moment,

$ I $ is the moment of inertia, and

$ B_H $ is the horizontal component of the Earth's magnetic field.

Given that the magnetic moment decreases by 19%, we can express the initial magnetic moment as $ M_1 = m $ and the reduced magnetic moment as:

$ M_2 = m - \frac{19 \times m}{100} = 0.81m $

From the equation for the time period, we derive:

$ \frac{T_1}{T_2} = \sqrt{\frac{M_2}{M_1}} $

Substituting the values we have:

$ \frac{T_1}{T_2} = \sqrt{\frac{0.81m}{m}} = \sqrt{0.81} = \frac{9}{10} $

Rearranging gives:

$ T_2 = \frac{10}{9} T_1 $

This implies:

$ T_2 = 1.11 T_1 $

So, the new time period $ T_2 $ is $ T_1 + 0.11 T_1 $, which indicates that:

The time period increases by 11%.

When two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres. Then their resultant dipole moment is given as

$\begin{aligned} M^{\prime} & =\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos \theta} \\ & =\sqrt{M^2+M^2+2 M M \cos 90^{\circ}} \\ \Rightarrow M^{\prime} & =M \sqrt{2} \quad \text{.... (i)} \end{aligned}$

The magnitude of the magnetic field at a distance $d$ from the centre on the bisector i.e at a distance $d$ on the axial position of dipole,

$\begin{aligned} B & =\frac{\mu_0}{4 \pi} \cdot \frac{2 M^{\prime}}{d^3} \\ & =\frac{\mu_0}{4 \pi} \cdot \frac{2 \sqrt{2} M}{d^3} \quad \text{[from Eq. (i)]} \end{aligned}$

Length of steel wire $=l$

Magnetic moment, $M=m l\quad \text{.... (i)}$

Where, $m$ is pole strength when steel wire is bent into semicircular wire, then

$\begin{aligned} & l=\pi R \\ & \Rightarrow \quad R=\frac{l}{\pi} \\ \end{aligned}$

$\therefore$ length of wire $=$ diameter

$\begin{array}{ll} \Rightarrow & l^{\prime}=2 R=\frac{2 l}{\pi} \\ \Rightarrow & l^{\prime}=\frac{2 l}{\pi} \end{array}$

New magnetic dipole moment,

$\begin{aligned} M^{\prime} & =m l^{\prime} \\ & =m \frac{2 l}{\pi}=\frac{2}{\pi}(m l)=\frac{2 M}{\pi} \end{aligned}$

Given:

$B_H=0.3 \mathrm{G}$

Angle of dip:

$\delta=30^{\circ}$

We know the relationship between the horizontal component and the total magnetic field:

$B_H = B \cos \delta$

Solving for $B$:

$ \begin{aligned} \Rightarrow B & = \frac{B_H}{\cos \delta} \\ & = \frac{0.3}{\cos 30^{\circ}} \\ & = \frac{0.3}{\sqrt{3} / 2} \\ & = 2 \times \frac{0.3}{\sqrt{3}} \\ & = \frac{0.6}{\sqrt{3}} \\ & = \frac{2}{10} \times \frac{3}{\sqrt{3}} \\ & = \frac{1}{5} \sqrt{3} \\ & = \frac{\sqrt{3}}{5} \mathrm{G} \end{aligned} $

When compass needle oscillates 20 times per minute, then its time period,

$T_1=\frac{1}{f}=\frac{1}{\left(\frac{20}{60}\right)}=3 \mathrm{~s}$

Angle of dip, $\delta_1=45^{\circ}$

When the same oscillates 30 times per min then, its time period.

$T_2=\frac{1}{f_2}=\frac{1}{30 / 60}=2 \mathrm{~s}$

and angle of dip, $\delta_2=30^{\circ}$

we know that, time period of compass needle

$T=2 \pi \sqrt{\frac{I}{M B_H}}$

where, $I=$ moment of inertia $B_H=$ horizontal component of Earth's magnetic field.

$\begin{aligned} & \Rightarrow \quad T \propto \frac{1}{\sqrt{B_H}} \\ & \Rightarrow \quad T \propto \frac{1}{\sqrt{B \cos \delta}} \Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{B_1 \cos \delta_1}{B_2 \cos \delta_2}} \end{aligned}$

$=\sqrt{\frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}}=\sqrt{\frac{B_1}{B_2}}=\sqrt{\frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2}{\sqrt{6}}} \cdot \sqrt{\frac{B_1}{B_2}}$

$\begin{aligned} & \Rightarrow \quad \frac{T_2^2}{T_1^2}=\frac{2}{\sqrt{6}} \cdot \frac{B_1}{B_2} \Rightarrow \frac{2^2}{3^2}=\frac{2}{\sqrt{6}} \frac{B_1}{B_2} \\ & \Rightarrow \quad \frac{4}{9}=\frac{2}{\sqrt{6}} \frac{B_1}{B_2} \Rightarrow \frac{B_1}{B_2}=\frac{2 \sqrt{6}}{9} \\ & \Rightarrow \quad \frac{B_1}{B_2}=\frac{2 \sqrt{2}}{3 \sqrt{3}} \Rightarrow B_1: B_2=2 \sqrt{2}: 3 \sqrt{3} \end{aligned}$

Given,

Initial magnetisation, $m_1=0.8 \mathrm{~Am}^{-1}$

Magnetic field strength, $B=0.8 \mathrm{~T}$

Initial temperature, $T_1=5 \mathrm{~K}$

Final temperature, $T_2=20 \mathrm{~K}$

Let, final magnetisation be $m_2$.

Since, $m=C B / T$

where, $C$ is Curie's temperature.

$\begin{aligned} & \Rightarrow \quad m \propto \frac{1}{T} \\ & \therefore \quad \frac{m_1}{m_2}=\frac{T_2}{T_1} \\ & \Rightarrow \quad m_2=\frac{m_1 T_1}{T_2}=\frac{0.8 \times 5}{20}=0.2 \mathrm{~Am}^{-1} \\ \end{aligned}$

Given, first apparent $\operatorname{dip}=\delta_1$

Second apparent dip $=\delta_2$

Declination $=\theta$

$\begin{aligned} & \text { As we know that, } \tan \theta=\frac{\tan _1}{\tan _2} \\ & \therefore \quad \theta=\tan ^{-1}\left(\frac{\tan \delta_1}{\tan \delta_2}\right) \end{aligned}$