Magnetism and Matter

$ \text { } \begin{aligned} B & =2 B v \\ B & =2 B \sin \theta \\ \Rightarrow \sin \theta & =\frac{1}{2} \Rightarrow \theta=30^{\circ} \end{aligned} $

$ \begin{aligned} & \therefore \quad \frac{B_H}{B}=\frac{B \cos \theta}{B}=\cos \theta=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \\ & \therefore B_H: B=\sqrt{3}: 2 \end{aligned} $

Angle for

Stable position $=0^{\circ}$

Unstable position $=180^{\circ}$

$ \begin{aligned} W & =m B\left[\cos \theta_1-\cos \theta_2\right] \\ & =2.5 \times 4 \times 10^{-5}\left[\cos 0^{\circ}-\cos 180^{\circ}\right] \\ & =10^{-4}[1-(-1)]=2 \times 10^{-4} \mathrm{~J} \\ & =20 \times 10^{-5} \mathrm{~J} \end{aligned} $

On the axial position of bar magnet

$ B_1=\frac{\mu_0}{4 \pi} \times \frac{2 M}{r^3} $

Similarly,

$ B_2=\frac{\mu_0}{4 \pi} \times \frac{M^{\prime}}{r^3} $

here $m^{\prime}=2 m$

$ \begin{aligned} & \therefore \quad B_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 m}{r^3} \\ & \text { Since, } B_1=B_2=B \\ & \begin{aligned} \therefore B_{\text {resultant }} & =\sqrt{B_1^2+B_2^2} \\ & =\sqrt{B^2+B^2}=\sqrt{2} B \end{aligned} \end{aligned} $

$ \begin{aligned} & B_H=B \cos 60^{\circ} \\ & B_H=\frac{B}{2} \\ & B_V=B \sin 60^{\circ} \\ & B_V=\frac{\sqrt{3}}{2} B \Rightarrow B=\frac{2}{\sqrt{3}} B_V \end{aligned} $

Given:

Magnetic susceptibility $ \chi = 0.6 $

We need the ratio $ \dfrac{\mu}{\mu_0} $, where

$\mu$ = permeability of the substance

$\mu_0$ = permeability of free space

Relation between $\mu$ and $\chi$:

$ \mu = \mu_0 (1 + \chi) $

Substitute the value:

$ \frac{\mu}{\mu_0} = 1 + \chi = 1 + 0.6 = 1.6 $

$ \frac{\mu}{\mu_0} = \frac{1.6}{1} = \frac{8}{5} $

✅ Final Answer:

$ \boxed{\frac{\mu}{\mu_0} = \frac{8}{5}} $

Option (C) → $8:5$

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} $

Given,

Magnetic flux, $ \phi=22 \times 10^{-6} \mathrm{~Wb} $

Cross-sectional area, $ A=2 \times 10^{-5} \mathrm{~m}^{2} $

Magnetic susceptibility of the metal, $ \chi=699 $

We now that,

$ \begin{array}{l} \phi=B \cdot A \\\\ B=\frac{\phi}{A}=\frac{22 \times 10^{-6}}{2 \times 10^{-5}} \\\\ B=1.1 \mathrm{~T} \end{array} $

Relation between magnetic field and magnetizing field is given by,

$ B=\mu_{0}(H+M) $

where, $ H= $ magnetising field, $ \mu_{0}= $ permeability of free space and $ M= $ magnetisation of the material but, $ M=\chi H $

Putting value in Eq. (i), we get

$ \begin{aligned} B & =\mu_{0}(H+\chi H)=\mu_{0} H(1+\chi) \\\\ 1.1 & =4 \pi \times 10^{-7} \times H(1+699) \\\\ H & =\frac{1.1 \times 10^{7}}{4 \pi \times 700} \\\\ H & =0.01250 \times 10^{5} \\\\ H & =1250 \mathrm{~A} / \mathrm{m} \end{aligned} $

When a short bar magnet is placed in a uniform magnetic field at an angle $ \theta $, it experiences a torque. The formula for the torque $ \tau $ is given by:

$ \tau = M B \sin \theta $

where $ M $ is the dipole moment and $ B $ is the magnetic field strength.

Case 1: When the angle $ \theta = 30^\circ $,

$ \tau_1 = M B \sin 30^\circ = \frac{M B}{2} $

Case 2: When the angle $ \theta = 45^\circ $,

$ \tau_2 = M B \sin 45^\circ = \frac{M B}{\sqrt{2}} $

The change in torque is given by:

$ \tau_2 - \tau_1 = \frac{M B}{\sqrt{2}} - \frac{M B}{2} $

Since $ \tau_2 - \tau_1 $ is positive, the torque increases. The percentage increase can be calculated as follows:

$ \frac{\tau_2 - \tau_1}{\tau_1} \times 100 $

$ = \frac{M B(2 - \sqrt{2})}{2\sqrt{2} \times M B} \times 100 $

$ = 41.4\% $

Thus, the torque on the magnet increases by 41.4%.

Torque $=M B \sin \theta$

$ \begin{aligned} & =10^{-4} \times 12 \times 10^{-3} \times \sin 30^{\circ} \\\\ & =10^{-4} \times 12 \times 10^{-3} \times \frac{1}{2} \\\\ & =6 \times 10^{-7} \mathrm{~N}-\mathrm{m} \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%.

Given,

Magnetic moment, $M=2 \mathrm{Am}^2$

Magnetic field, $B=0.3 \mathrm{~T}$

Work done by external torque to rotate the bar magnet by $90^{\circ}$

$ \begin{aligned} & W=-M B\left(\cos \theta_2-\cos \theta_1\right) \\ & W=-M B\left[\cos 90^{\circ}-\cos 0^{\circ}\right] \\ & W=-2 \times 0.3 \times-1 \\ & W=0.6 \mathrm{~J} \end{aligned} $

Hence work done is 0.6 J .

Ferromagnetic substance are those substance, when placed in magnetic field it can get highly magnetised.

So, susceptibility of ferromagnetic substance is much greater than 1 .

$ \chi \gg 1 $

We know that, period of oscillation of bar magnet

$ \begin{aligned} & T=2 \pi \sqrt{\frac{I}{M B_H}} \\ \Rightarrow & T \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{M_1}{M_2}} \end{aligned} $

Here, $T_1=2 \mathrm{~s}, M_2=4 M_1$

$ \begin{aligned} \frac{T_2}{2} & =\sqrt{\frac{M_1}{4 M_1}} \\ \Rightarrow \quad \frac{T_2}{2} & =\frac{1}{2} \Rightarrow T_2=1 \mathrm{~s} \end{aligned} $

Material suitable for making permanent magnets must have the following properties.

(i) High retentivity so that it produces a strong magnetic field.

(ii) High coercivity so that its magnetisation is not destroyed by stray magnetic fields, temperature variations or minor mechanical damage.

(iii) High permebility. Inspite of its slightly smaller retentivity than soft iron, steel is favoured for making permanent magnets. Steel has much higher coercivity than soft iron. The magnetisation of steel is not easily destroyed by stray fields. Once magnetised under a strong field, it retains magnetisation for a long duration.

We know that, magnetic energy stored in a solenoid per unit volume

$ \begin{aligned} \frac{U}{V} & =\frac{1}{2} \frac{B^2}{\mu_0} \\ U & =\frac{1}{2} \cdot \frac{B^2 V}{\mu_0} \\ & =\frac{B^2 A L}{2 \mu_0} \quad(\because V=A L) \end{aligned} $

Magnetic susceptibility is a measure of how much a material becomes magnetized in the presence of an external magnetic field.

The formula for magnetic susceptibility is given by:

$ \chi = \frac{M}{H} = \frac{\text{Magnetization (M)}}{\text{Magnetic field intensity (H)}} $

Ferromagnetic materials are substances that can be highly magnetized when exposed to a magnetic field. As a result, the magnetic susceptibility of ferromagnetic materials is significantly greater than 1.

To solve this, we first recall that the magnetic field intensity inside a solenoid is given by:

$H = \frac{NI}{l}$

where:

• $ N $ is the number of turns (60 turns),

• $ I $ is the current in amperes, and

• $ l $ is the length of the solenoid in meters (12 cm = 0.12 m).

The bar magnet has a coercivity of $4 \times 10^3 \, \mathrm{A/m}$, which means to demagnetize the magnet, the solenoid needs to produce an opposing field of this magnitude.

Now, set the solenoid’s field equal to the coercivity:

$4 \times 10^3 = \frac{60 \times I}{0.12}$

Now, solve for $I$:

Multiply both sides by 0.12:

$4 \times 10^3 \times 0.12 = 60I$

Compute the left side:

$480 = 60I$

Divide both sides by 60:

$I = \frac{480}{60} = 8 \, \text{A}$

Thus, the current that should be passed through the solenoid to demagnetize the bar magnet is 8 A.

The correct answer is:

Option D: 8 A.

Given, horizontal component of earth's magnetic field, $B_H=86.6 \mathrm{G}$

Magnetic field of earth,

$ B=100 \mathrm{G} $

We know that, $B_H=B \cos \delta$ where $\delta$ is the angle of dip.

$ \begin{array}{rlrl} & \therefore \cos \delta =\frac{B_H}{B}=\frac{86.6}{100}=0.866 \\ & \therefore \delta=\cos ^{-1}(0.866)=30^{\circ} \end{array} $

Given, cross-sectional area of iron bar, $A=2 \times 10^{-5} \mathrm{~m}^2$

Magnetising field, $H=2400 \mathrm{~A} / \mathrm{m}$

Magnetic flux, $\phi=2.4 \pi \times 10^{-5} \mathrm{~Wb}$

We know that,

$ \begin{array}{rlrl} \phi & =B A \\ \Rightarrow \quad B & =\frac{\phi}{A}=\frac{2.4 \pi \times 10^{-5}}{2 \times 10^{-5}} \\ & =1.2 \pi \mathrm{~T} \\ \therefore & \text { Again, } B =\mu H \\ \Rightarrow \quad \mu & =\frac{B}{H}=\frac{1.2 \pi}{2400} \\ \Rightarrow \quad \mu & =5 \pi \times 10^{-4} \end{array} $

Again, we know that, magnetic susceptibility,

$ \begin{aligned} \chi & =\mu_r-1=\frac{\mu}{\mu_0}-1 \\ & =\frac{5 \pi \times 10^{-4}}{4 \pi \times 10^{-7}}-1=1.25 \times 10^3-1 \\ & =1250-1=1249 \end{aligned} $

Given, $r=1 \mathrm{~m}, B_{\text {axial }}=5 \times 10^{-8} \mathrm{~T}$

We know that, magnitude of axial field due to bar magnet.

$ \begin{array}{rlrl} & B_{\text {axial }} =\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r^3} \\ & \Rightarrow 5 \times 10^{-8} =\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 M}{(1)^3} \\ & \Rightarrow 5 \times 10^{-8} =2 \times 10^{-7} M \\ & \Rightarrow M =\frac{5}{2} \times 10^{-1}=0.25 \mathrm{~A} \cdot \mathrm{~m}^2 \end{array} $

Given, torque, $\tau=0.012 \mathrm{~N}-\mathrm{m}$

$ =1.2 \times 10^{-2} \mathrm{~m} $

Magnetic field, $B=200 \mathrm{G}=2 \times 10^{-2} \mathrm{~T}$

$ \left[\because 1 \mathrm{G}=10^{-4} \mathrm{~T}\right] $

$ \begin{aligned} &\theta=30^{\circ}\\ &\text { We know that, }\\ &\begin{aligned} \tau & =M B \sin \theta \\ \Rightarrow \quad M & =\frac{\tau}{B \sin \theta}=\frac{1.2 \times 10^{-2}}{2 \times 10^{-2} \sin 30^{\circ}} \\ & =\frac{1.2 \times 10^{-2}}{2 \times 10^{-2} \times(1 / 2)} \\ & =1.2 \mathrm{~A}-\mathrm{m}^2 \end{aligned} \end{aligned} $

Magnetic lines form continuous closed loops because magnetic monopoles do not exist in nature. We always find magnetic North and South pole, which are coupled together in such a way that the field lines originate from one pole and end at another pole.

Magnetic dipole moment of planet,

$ M=27 \times 10^{22} \mathrm{~A}-\mathrm{m}^2 $

Radius of planet,

$ \begin{aligned} & R=300 \mathrm{~km}=3 \times 10^5 \mathrm{~m} \\ & \frac{\mu_0}{4 \pi}=10^{-7} \end{aligned} $

Magnetic field at the equator,

$ \begin{aligned} B & =\frac{\mu_0}{4 \pi} \times \frac{M}{R^3}=10^{-7} \times \frac{27 \times 10^{22}}{\left(3 \times 10^5\right)^3} \\ & =10^{-7} \times \frac{27 \times 10^{22}}{27 \times 10^{15}}=1 \mathrm{~T} \end{aligned} $

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 $=\theta$

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

Given, relative permeability,

$ \mu_r=\frac{800}{\pi} $

Current, $I=2 \mathrm{~A} \Rightarrow n=1000 \mathrm{~m}^{-1}$

As we know that,

$ \begin{aligned} H & =n I=1000 \times 2=2 \times 10^3 \mathrm{Am}^{-1} \\ \text { Since, } B & =\mu H=\mu_0 \mu_r H \quad\left(\because \mu=\mu_0 \mu_r\right) \\ & =4 \pi \times 10^{-7} \times \frac{800}{\pi} \times 2 \times 10^3 \\ & =0.64 \mathrm{~T}=640 \mathrm{mT} \end{aligned} $

The material used for making permanent magnet must have high retentivity, so that it produces a strong magnetic field and must have high coercivity, so that its magnetisation is not destroyed by stray magnetic fields, temperature variations or minor mechanical damage.

Two magnet are joined as shown in figure. The magnetic field at point $P$ due to magnet $M_1$ is on axial line, while that due to $M_2$ is on equatorial line.

So, net magnetic field at a point $P$ at a distance $R$ on $Y$-axis,

$ \begin{aligned} B_{\text {net }} & =\sqrt{B_1^2+B_2^2} \\ & =\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 M}{R^3}\right)^2+\left(\frac{\mu_0}{4 \pi} \frac{M}{R^3}\right)^2} \\ B_{\text {net }} & =\frac{\mu_0}{4 \pi R^3} \cdot \sqrt{5} M \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { Given, } B_{\text {net }} & =\frac{\mu_0}{4 \pi R^3} \cdot M_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

By comparision Eq. (i) and Eq. (ii), we get

$ M=M_0 / \sqrt{5} $

We have following relation between relative permeability and susceptibility,

$ \begin{aligned} \mu_r & =1+\chi_B \Rightarrow \chi_B=\mu_r-1=5500-1 \\ & =5499 \end{aligned} $