Magnetic Properties of Matter

When a magnetic dipole like a bar magnet is placed in a magnetic field (B) at an angle ( $\theta$ ), it experiences a torque.

$ \tau=\mathrm{M} \mathrm{~B} \sin \theta $

Where, $\mathrm{M}=$ magnetic moment of the magnet, $\mathrm{B}=$ magnetic field strength, $\theta=$ Angle between the magnet's axis and the field.

The magnetic field is given in Gauss. We must convert it to the standard SI unit, Tesla (T).

1 Tesla $=10000$ Gauss $\left(10^4 \mathrm{G}\right)$

So, $\mathrm{B}=800$ Gauss $=800 \times 10^{-4} \mathrm{~T}=0.08 \mathrm{~T}$

Using $\tau=\mathrm{MB} \sin \theta$ :

$ 0.016=M \times(0.08) \times \sin \left(30^{\circ}\right) $

$\Rightarrow $ $0.016=\mathrm{M} \times 0.08 \times 0.5$

$\Rightarrow $ $\mathrm{M}=\frac{0.016}{0.04}$

$\Rightarrow $ $ \mathrm{M}=0.4 \mathrm{~A} \cdot \mathrm{~m}^2 $

The potential energy of the magnetic dipole is given as $\mathrm{U}=-\mathrm{MB} \cos \theta$

The magnet is aligned with the magnetic field, i.e., $\theta_1=0^{\circ}$. The potential energy is minimum.

The magnet is aligned against (opposite to) the magnetic field, i.e., $\theta_2=180^{\circ}$. The potential energy is maximum.

The work done to rotate a magnet from an initial angle $\theta_1$ to a final angle $\theta_2$ is equal to the change in its potential energy.

$ \mathrm{W}=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) $

• 1. $\theta_1=0^{\circ}$

• 2. $\theta_2=180^{\circ}$

So, the work done is,

$ \mathrm{W}=(0.4) \times(0.08) \times\left[\cos \left(0^{\circ}\right)-\cos \left(180^{\circ}\right)\right] $

$\Rightarrow $ $\mathrm{W}=0.032 \times[1-(-1)]=0.032 \times 2$

$\Rightarrow $ $\mathrm{W}=0.064 \mathrm{~J}=64 \times 10^{-3} \mathrm{~J}=\alpha \times 10^{-3} \mathrm{~J}$

$\Rightarrow $ $\alpha=64$

Coercivity is a measure of the resistance of a ferromagnetic material to becoming demagnetized. It is defined as the intensity of the applied magnetic field required to reduce the magnetization of a material to zero after the magnetization of the sample has been driven to saturation. In this case, coercivity $H_c$ is given to be $5 \times 10^3 \mathrm{~A/m}$.

The relationship between the magnetic field $H$ inside a solenoid and the current $I$ passed through it is given by the formula:

$H = \frac{N \cdot I}{L}$

where:

• $H$ is the magnetic field strength inside the solenoid in amperes per meter (A/m),
• $N$ is the total number of turns of wire,
• $I$ is the current in amperes (A), and
• $L$ is the length of the solenoid in meters (m).

Given that the number of turns of the solenoid $N = 150$ and the length of the solenoid $L = 30 \, \text{cm} = 0.3 \, \text{m}$, we can rearrange the formula to solve for $I$:

$I = \frac{H \cdot L}{N}$

Substituting the given values:

$I = \frac{(5 \times 10^3) \times 0.3}{150}$

Simplifying, we get:

$I = \frac{1500}{150}$

$I = 10 \, \text{A}$

Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is 10 A.

$\begin{aligned} & B_H=3.5 \times 10^{-5} T \\ & F=i \ell B \sin \theta, \quad \mathrm{i}=\sqrt{2} \mathrm{~A} \\ & \frac{F}{\ell}=i B \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \\ & =35 \times 10^{-6} \mathrm{~N} / \mathrm{m} \end{aligned}$

$\begin{aligned} & \mathrm{V}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^2} \\ & \Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \frac{\mathrm{M}}{\left(20 \times 10^{-2}\right)^2} \\ & \Rightarrow \mathrm{M}=\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}} \\ & \mathrm{M}=1.5 \times 4=6 \end{aligned}$

$ \begin{aligned} & \text {Period of oscillation } \alpha \frac{1}{\sqrt{B_{\mathrm{H}}}} \\\\ & \mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{B} \cos \theta}} \Rightarrow \frac{T_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}} \\\\ & \Rightarrow \frac{60 / 20}{60 / 30}=\sqrt{\frac{\mathrm{B}_2}{\mathrm{~B}_1} \frac{\cos 60^{\circ}}{\cos 30^{\circ}}} \Rightarrow \frac{3}{2}=\sqrt{\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1}} \\\\ & \Rightarrow \frac{9}{4}=\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1} \Rightarrow \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{4}{9 \sqrt{3}}=\frac{4}{\sqrt{243}} \end{aligned} $

We know that the magnetizing field (H) inside a solenoid is given by the formula :

$ H = \frac{N \cdot I}{L} $

where (N) is the number of turns, (I) is the current in Amperes, and (L) is the length of the solenoid in meters.

To demagnetize a bar magnet that has a magnetic intensity (H) of ($2.4 \times 10^3 \, \text{Am}^{-1}$), you need to create a magnetizing field in the solenoid that is equal in magnitude but opposite in direction.

Given:

• ($H = 2.4 \times 10^3 \, \text{Am}^{-1}$)
• (N = 60) turns
• ($L = 15 \, \text{cm} = 0.15 \, \text{m}$) (since $(1\, \text{m} = 100 \, \text{cm})$)

You can rearrange the formula for (H) to solve for (I) :

$ I = \frac{H \cdot L}{N} $

Substituting the given values, you get :

$ I = \frac{(2.4 \times 10^3 \, \text{Am}^{-1}) \cdot 0.15 \, \text{m}}{60} $

$ I = 6 \, \text{Amperes} $

So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $\, \text{A}$).

In a solenoid, the magnetic field intensity ($H$) is given by the product of the number of turns per unit length ($n$) and the current ($I$) flowing through the solenoid. This can be represented mathematically as:

$H = nI$

This is actually derived from Ampere's law applied to the special case of a solenoid, where the magnetic field is uniform and directed along the axis of the solenoid.

If we rearrange this equation to solve for the current ($I$), we get:

$I = \frac{H}{n}$

In this problem, we're given that the magnetic field intensity ($H$) at the center of the solenoid is $1.6 \times 10^{3} \, \text{Am}^{-1}$ and the number of turns per unit length ($n$) is 8 per cm (which is equal to $8 \times 10^{-2}$ per meter, as there are 100 cm in a meter).

Substituting these values into the equation gives:

$I = \frac{1.6 \times 10^{3} \, \text{Am}^{-1}}{8 \times 10^{-2} \, \text{turns/m}} = 2 \, \text{A}$

So, the current flowing through the solenoid is $2 \, \text{A}$.

$T = 2\pi \sqrt {{I \over {MB}}} $

B = 80 $\times$ 10^$-$4 = 8 mT

$B = \mu .(H + I)$

$B = \mu .H\left( {1 + {1 \over H}} \right)$

$B = {B_0}(1 + x)$

$B - {B_0} = {B_0}x$

${{B - {B_0}} \over {{B_0}}} = x$

${{B - {B_0}} \over {{B_0}}} \times 100 = 100x$

$ = 2.2 \times {10^{ - 3}} = {{22} \over {{{10}^4}}}$