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) → $8:5$

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} $

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} $

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} $