Magnetic Properties of Matter

Ferromagnetic materials (like iron, nickel, cobalt) are characterized by the fact that their individual atoms possess a permanent magnetic dipole moment. Due to strong quantum mechanical interactions between neighbouring atoms, these dipoles spontaneously align parallel to each other within small macroscopic regions.

These aligned regions are indeed called magnetic domains. Within a domain, the magnetization is saturated, even in the absence of an external field.

Hence, the Assertion (A) is True.

Temperature represents thermal agitation. As the temperature of a ferromagnetic material increases, the thermal energy tends to randomize the orientation of the magnetic dipoles.

At a specific critical temperature called the Curie Temperature ( $\mathrm{T}_{\mathrm{c}}$ ), the thermal agitation becomes strong enough to overcome the exchange coupling that holds the domains together. The domain structure breaks down (disintegrates), and the material transitions from ferromagnetic to paramagnetic, losing its spontaneous magnetization.

So, the Reason (R) is also True.

The formation of domains (A) is caused by the minimization of exchange energy and magnetostatic energy. The destruction of domains at high temperature ( R ) is a property of the material's phase transition.

Knowing that domains disappear at high heat does not explain the fundamental physical mechanism (exchange interaction) that causes them to align spontaneously at lower temperatures.

Both statements are correct facts, but R is not the functional explanation for A , i.e., both $(\mathrm{A})$ and $(\mathrm{R})$ are true but $(\mathrm{R})$ is not the correct explanation of (A).

Hence, the correct option is (D).

For a magnetic material,

$ B=\mu H $

where

$ \mu=\mu_r \mu_0 $

Given:

$ B=1.0\ \text{T}, \quad \mu_r=400, \quad \mu_0=4\pi \times 10^{-7} $

So,

$ H=\frac{B}{\mu_r \mu_0} $

Substituting the values,

$ H=\frac{1}{400 \times 4\pi \times 10^{-7}} $

$ H=\frac{1}{1600\pi \times 10^{-7}} $

$ H=\frac{10^7}{1600\pi} $

$ H=\frac{10^4}{1.6\pi} $

$ H=\frac{6250}{\pi} $

Now write it in the form $\alpha \times 10^5$:

$ H=\frac{6250}{\pi}=\frac{1}{16\pi}\times 10^5 $

Therefore,

$ \alpha=\frac{1}{16\pi} $

So, the correct option is:

$ \boxed{\frac{1}{16\pi}} $

Option B

The assertion states that magnetic monopoles do not exist. This is consistent with current experimental findings and the standard formulation of Maxwell's equations.

The reason provided is that magnetic field lines are continuous and form closed loops. This is mathematically expressed by the Maxwell equation:

$\nabla \cdot B = 0$

This equation tells us that there are no "sources" or "sinks" of magnetic fields (unlike electric fields, which can originate or terminate at charges). In other words, magnetic field lines neither begin nor end; they loop continuously.

Since the fact that the magnetic field lines form closed loops is a direct consequence of the absence of isolated magnetic charges (monopoles), the reason correctly explains the assertion.

Thus, both (A) and (R) are correct, and (R) is the correct explanation of (A).

The most appropriate answer is Option A.

The percentage change in the magnetic field (B) when the space within a current-carrying solenoid is filled with magnesium is calculated as follows:

We start by using the formula for percentage change in the magnetic field:

$ \% \text{ change in } B = \frac{B_{\text{new}} - B_{\text{old}}}{B_{\text{old}}} \times 100\% $

Substituting the expressions for the magnetic field, we have:

$ = \frac{\mu n i - \mu_0 n i}{\mu_0 n i} \times 100\% = \frac{\mu - \mu_0}{\mu_0} \times 100\% $

Where:

$\mu$ is the permeability of the material inserted (here, magnesium).

$\mu_0$ is the permeability of free space.

$n$ is the number of turns per unit length of the solenoid.

$i$ is the current through the solenoid.

Given that the relationship between the permeability of the medium ($\mu$) and relative permeability ($\mu_r$) is:

$ \mu = \mu_0 \mu_r $

We can simplify this to:

$ = \frac{\mu_0 \mu_r - \mu_0}{\mu_0} \times 100\% $

Which reduces to:

$ = (\mu_r - 1) \times 100\% $

The relative permeability $\mu_r$ is equal to $1 + \chi$, where $\chi$ is the magnetic susceptibility. Thus, we have:

$ = \chi \times 100\% $

When magnesium is used, with a magnetic susceptibility $\chi_{\text{Mg}} = 1.2 \times 10^{-5}$, the calculation becomes:

$ = 1.2 \times 10^{-5} \times 100\% = 1.2 \times 10^{-3}\% $

Therefore, the percentage increase in the magnetic field when the solenoid is filled with magnesium is $1.2 \times 10^{-3}\%$.

To determine the potential energy of a magnetic dipole in a uniform magnetic field, we start with the torque equation:

$ \tau = M \times B = MB \sin 60^\circ $

Given that $\tau = 80 \sqrt{3} \, \text{N m}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we can equate and solve for $MB$:

$ MB \frac{\sqrt{3}}{2} = 80 \sqrt{3} $

Solving for $MB$, we find:

$ MB = 160 $

The potential energy $U$ of the dipole in the magnetic field is given by:

$ U = -M \cdot B = -MB \cos 60^\circ $

Substituting in $\cos 60^\circ = \frac{1}{2}$, we calculate:

$ U = -160 \times \frac{1}{2} = -80 \, \text{J} $

Thus, the potential energy of the dipole is $-80 \, \text{J}$.

The relationship between magnetic susceptibility $(\chi)$ and magnetic permeability $(\mu)$ is expressed as follows:

Relative Permeability ($\mu_r$): This is defined as the ratio of the permeability of a material ($\mu$) to the permeability of free space ($\mu_0$).

$ \mu = \mu_0 \mu_r \Rightarrow \mu_r = \frac{\mu}{\mu_0} $

Magnetic Susceptibility ($\chi$): This relates to relative permeability as follows:

$ \mu_r = 1 + \chi \Rightarrow \chi = \mu_r - 1 $

Substituting the expression for $\mu_r$ from above:

$ \chi = \left(\frac{\mu}{\mu_0} - 1\right) $

Thus, the equation shows how magnetic susceptibility is derived from the magnetic permeability relative to free space.

(A) The unit "gauss" is used to measure magnetic induction, also known as magnetic flux density, within the CGS system of units.

$B = {\phi \over A} = {{\max well} \over {c{m^2}}} = gauss\,(CGS)$

(B) Magnetic intensity (H) is defined as the product of the number of turns per unit length in a coil (n) and the current that it carries (I).

i.e. $H = nI = {N \over L}I$

Unit : A/m or Am$^{-1}$

(C) Magnetic flux, $\phi = BA$

unit : weber

(D) Magnetic moment (M) : magnetic moment of a current loop, $M = NIA$

where, N = number of turns

I = current

A = Area

So, unit = Am$^2$

Hence, option A is correct.

Paramagnetic substances exhibit specific characteristics in the presence of an external magnetic field. Understanding these characteristics will help us choose the most appropriate answer. Let's discuss each statement individually:

A. align themselves along the directions of external magnetic field.
This is true. Paramagnetic substances have unpaired electrons, and under the influence of an external magnetic field, the atomic dipoles (due to those unpaired electrons) tend to align themselves in the direction of the magnetic field.

B. attract strongly towards external magnetic field.
This statement is the cause of some confusion. Paramagnetic substances are attracted towards an external magnetic field, but the keyword here is strongly. Compared to ferromagnetic substances, the attraction is relatively weak. Given the context of the options provided, saying they are "strongly" attracted could be misleading, but it is undeniable they are attracted nevertheless.

C. has susceptibility little more than zero.
This is correct. The magnetic susceptibility of paramagnetic materials is positive, meaning they are attracted to magnetic fields, but it is small in magnitude, typically greater than zero but far less than the susceptibility of ferromagnetic materials.

D. move from a region of strong magnetic field to weak magnetic field.
This statement is incorrect for paramagnetic substances. They are attracted to magnetic fields, implying they move from a region of weaker magnetic field to a stronger magnetic field, not the other way around.

Taking these points into account, the most accurate answer would be:

Options B: A, C Only
A) They indeed align along the direction of an external magnetic field, and C) Their susceptibility is indeed a little more than zero, indicating a weak attraction to magnetic fields. Statement B can be considered incorrect not in its entirety that they are attracted, but in the use of the term "strongly," and D is certainly incorrect regarding their movement in magnetic fields.

In para and dia materials, susceptibility is independent of magnetic field.

Inside a current carrying wire, $B \propto X$

Outside a current carrying wire, $B \propto \frac{1}{x}$

Inside a solenoid, B is constant upto ends.

$\begin{aligned} W & =\int_0^{180} d \tau \cdot d \theta \\ & =m B(\cos 0-\cos 180) \\ & =0.5 \times 8 \times 10^{-2}(2) \\ & =8 \times 10^{-2} \mathrm{~J} \end{aligned} $

When a bar magnet is released from rest along the axis of a very long vertical copper tube, it will move down with an almost constant speed after some time.

As the magnet falls, it moves through the copper tube, inducing eddy currents in the tube. These eddy currents, in turn, create an opposing magnetic field that opposes the motion of the magnet. The opposing force generated by the eddy currents acts as a damping force, which slows down the magnet's acceleration. Eventually, the magnet reaches a terminal velocity, at which point the gravitational force pulling the magnet downward is balanced by the opposing force from the eddy currents.

At this terminal velocity, the magnet moves down with an almost constant speed.

Both Statement I and Statement II are true.

Statement I: For diamagnetic substances, the magnetic susceptibility (χ) lies between -1 and 0. This is because diamagnetic substances have a negative magnetic susceptibility, which means they have a tendency to oppose the applied magnetic field.

Statement II: Diamagnetic substances, when placed in an external magnetic field, experience a force that tends to move them from stronger to weaker parts of the field. This is due to the fact that diamagnetic substances oppose the applied magnetic field and try to minimize their exposure to it.

Both Assertion A and Reason R are correct, and R is indeed the correct explanation of A.

Electromagnets are commonly made of soft iron because of its high permeability, which allows it to easily magnetize in response to an external magnetic field. Its low retentivity is also desirable because it allows the magnetization to be easily reversed or removed once the external field is removed. This combination of properties makes soft iron ideal for electromagnets, which require rapid and efficient changes in magnetization.

$\therefore$ $\sin 37^\circ = {{{B_v}} \over {{B_{net}}}}$

$ \Rightarrow {B_v} = {B_{net}}\sin 37^\circ $

$ \Rightarrow {B_{net}} = {{{B_v}} \over {\sin 37^\circ }}$

$ = {{6 \times {{10}^{ - 5}}} \over {{3 \over 5}}}$

$ = 10 \times {10^{ - 5}}\,T$

$ = 1 \times {10^{ - 4}}\,T$

$T \propto {1 \over {\sqrt {B\cos \delta } }}$

$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{B_2}\cos {\delta _2}} \over {{B_1}\cos {\delta _1}}}} $

$ \Rightarrow {{3\,s} \over {6\,s}} = \sqrt {{{{B_1}} \over {{B_2}}} \times {{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}} $

$ \Rightarrow {{{B_2}} \over {{B_1}}} = {\left( {{1 \over 2}} \right)^2} \times \sqrt 3 = {{\sqrt 3 } \over 4}$

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

$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}} \sqrt {{{{M_2}} \over {{M_1}}}} $

$ \Rightarrow {3 \over 4} = \sqrt {{3 \over 2}} \sqrt {{{{M_2}} \over {{M_1}}}} $

$ \Rightarrow {{{M_1}} \over {{M_2}}} = {3 \over 2} \times {{16} \over 9} = {8 \over 3}$

$\tan 60^\circ = {{{B_0}\sin \delta } \over {{B_0}\cos \delta \cos 45^\circ }}$

$ \Rightarrow \tan \delta = \sqrt {{3 \over 2}} $

$ \Rightarrow \delta = {\tan ^{ - 1}}\left( {\sqrt {{3 \over 2}} } \right)$

$T = {{2\pi m} \over {Bq}}$

$\Rightarrow$ Frequency $f = {{Bq} \over {2\pi m}}$

$ = {{{{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}}} \over {2\pi \times 9 \times {{10}^{ - 31}}}}$

$ \simeq 2.8 \times {10^6}$ Hz

${B_H} = B\cos 30^\circ $

$ \Rightarrow B = {1 \over {\sqrt 3 }}G$

$\overrightarrow {B'} = {\mu _0}(1 + X)ni$ (in the material)

$\overrightarrow {B'} = {\mu _0}ni$ (without material)

So fractional increase is

${{B' - B} \over B} = X = 1.2 \times {10^{ - 5}}$

$\mu$_r = x + 1

= 99 + 1 = 100

$\Rightarrow$ $\mu$ = $\mu$_r$\mu$₀ = 100 $\times$ 4$\mu$ $\times$ 10^$-$7 Wb/Am

= 4$\mu$ $\times$ 10^$-$5 Wb/Am

$U = - \overrightarrow M \,.\,\overrightarrow B $

So ${U_f} - {U_i} = - MB(1 - \cos \theta )$

$ = - 14J$

So $W = - \Delta U = 14J$

Statement I is true as susceptibility of ferromagnetic and paramagnetic materials is inversely related to temperature.

Statement II is true as because of orbital motion of electrons the diamagnetic material is able to oppose external magnetic field.

Electromagnet requires high permeability and low retentivity.

For diamagnetic material

$\chi $ is independent of temperature and negative.

Magnetisation (M) is directly proportional to H (The relation between magnetisation and magnetising field, M = –mH)

Option (a)

To determine the work done in reversing the direction of a magnetic moment in a time-varying magnetic field, we'll follow these steps:

Given:

Magnetic moment: $\mathbf{m} = 10^{-2} \hat{i}$ A·m²

Magnetic field: $\mathbf{B}(t) = B \cos(\omega t) \hat{i}$, where $B = 1$ T and $\omega = 0.125$ rad/s

Time at which reversal occurs: $t = 1$ s

Step 1: Calculate the Magnetic Field at $t = 1$ s

$ B(t) = B \cos(\omega t) = 1 \times \cos(0.125 \times 1) = \cos(0.125 \text{ rad}) $

Compute $\cos(0.125 \text{ rad})$:

$ \cos(0.125 \text{ rad}) \approx 0.9922 $

So,

$ B(t) \approx 0.9922 \text{ T} $

Step 2: Calculate the Work Done

The potential energy $U$ of a magnetic dipole in a magnetic field is:

$ U = -\mathbf{m} \cdot \mathbf{B} = -mB \cos\theta $

The work done $W$ in reversing the magnetic moment from $\theta = 0^\circ$ to $\theta = 180^\circ$ is:

$ W = U_{\text{final}} - U_{\text{initial}} = [-mB \cos(180^\circ)] - [-mB \cos(0^\circ)] = mB (\cos 0^\circ - \cos 180^\circ) $

Simplify:

$ W = mB (1 - (-1)) = 2mB $

Substitute the values:

$ W = 2 \times (10^{-2} \text{ A·m}^2) \times (0.9922 \text{ T}) \approx 0.01984 \text{ J} $

Step 3: Approximate Using RMS Value

Given that the magnetic field is time-varying, we can consider the root mean square (RMS) value of $\cos(\omega t)$ over a complete cycle:

$ \text{RMS of } \cos(\omega t) = \frac{1}{\sqrt{2}} $

Thus, the RMS value of the magnetic field is:

$ B_{\text{RMS}} = B \times \frac{1}{\sqrt{2}} = 1 \times \frac{1}{\sqrt{2}} \approx 0.7071 \text{ T} $

Now, calculate the work done using $B_{\text{RMS}}$:

$ W_{\text{RMS}} = 2mB_{\text{RMS}} = 2 \times (10^{-2}) \times 0.7071 \approx 0.01414 \text{ J} $