exemplar-phy-12-magnetism-and-matter

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. A toroid of $n$ turns, mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $x y$-plane. Its magnetic moment $\text{m}$
<div class="fragment">
(a) is non-zero and points in the $z$-direction by symmetry
(b) points along the axis of the toroid $(m=m \hat\phi)$
(c) is zero, otherwise there would be a field falling as $\frac{1}{r^{3}}$ at large distances outside the toroid
(d) is pointing radially outwards
</div>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. Thus, in such a case magnetic field is only confined inside the body of toroid.
Ans. (c)
In case of toroid, the magnetic field is only confined inside the body of toroid in the form of concentric magnetic lines of force and there is no magnetic field outside the body of toroid. This is because the loop encloses no current. Thus, the magnetic moment of toroid is zero.
In general, if we take $r$ as a large distance outside the toroid, then $m \propto \frac{1}{r^{3}}$. But this case is not possible here.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. The magnetic field of the earth can be modelled by that of a point dipole placed at the centre of the earth. The dipole axis makes an angle of $11.3^{\circ}$ with the axis of the earth. At Mumbai, declination is nearly zero. Then,
<div class="fragment">
(a) the declination varies between $11.3^{\circ} W$ to $11.3^{\circ} E$
(b) the least declination is $0^{\circ}$
(c) the plane defined by dipole axis and the earth axis passes through Greenwich
(d) declination averaged over the earth must be always negative
</div>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: center;font-size:20px'>
Ans. (a)
For the earth’s magnetism, the magnetic field lines of the earth resemble that of a hypothetical magnetic dipole located at the centre of the earth.
The axis of the dipole does not coincide with the axis of rotation of the earth but is presently tilted by approxmately $11.3^{\circ}$ with respect to the later. This results into two situations as given in the figure ahead.
<img src="https://cdn.mathpix.com/snip/images/cRp7-zAFWFoqyGuQuuYhK-nzTvzMqjI5Cb-blvyOqQQ.original.fullsize.png" height="100px">
Hence, the declination varies between $11.3^{\circ} W$ to $11.3^{\circ} E$.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
Permanent magnet at room temperature behave as a ferromagnetic substance for a long period of time.
Ans. (d)
As we know a permanent magnet is a substance which at room temperature retain ferromagnetic property for a long period of time.
The individual atoms in a ferromagnetic material possess a dipole moment as in a paramagnetic material.
However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. Thus, we can say that in a permanent magnet at room temperature, domains are all perfectly aligned.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. Consider the two idealised systems (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length L » R, radius of cross-section. In (i) $\text{E}$ is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below
<div class="fragment">
(a) case (i) contradicts Gauss’ law for electrostatic fields
(b) case (ii) contradicts Gauss’ law for magnetic fields
(c) case (i) agrees with $\oint E.dl = 0$
(d) case (ii) contradicts $\oint H.dl =$ $I_{en}$
</div>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
The electric field lines, do not form a continuous closed path while the magnetic field lines form the closed paths.
Ans. (b)
As Gauss’ law states, $\int_s E.d s=\frac{q}{\varepsilon_0}$ for electrostatic field. It does not contradict for electrostatic fields as the electric field lines do not form continuous closed path.
According to Gauss’ law in magnetic field,
$\int_s E . ds=0$
It contradicts for magnetic field, because there is a magnetic field inside the solenoid and no field outside the solenoid carrying current but the magnetic field lines form the closed path.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. paramagnetic sample shows a net magnetisation of 8 Am$^{-1}$ when placed in an external magnetic field of 0.6 T at a temperature of 4K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetisation will be
<div class="fragment">
(a) $\frac{32}{3} Am^{-1}$
(b) $\frac{2}{3} Am^{-1}$
(c) $6 Am^{-1}$
(d) $2.4 Am^{-1}$
</div>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (Single correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
From Curie law, we know that magnetisation is directly proportional to the magnetic field induction and inversely proportional to the temperature in kelvin.
Ans. (b)
As Curie law explains, we can deduce a formula for the relation between magnetic field induction, temperature and magnetisation.
i.e., $\quad I$ (magnetisation) $\propto \frac{B \text{ (magnetic field induction })}{t \text{ (temperature in kelvin })}$
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
According to the properties of magnetic field lines (B), for any magnet, it forms continuous closed loops. This is unlike the electric dipole where these field lines begin from a positive charge and end on the negative charge or escape to infinity.
Also, magnetic intensity $(H)$ outside any magnet is $H=B / \mu_0$ and for inside the magnet $H=B / \mu_0 \mu_r$, where $\mu_r$ is the relative permeability of material (magnetic).
Ans. (a, d)
Magnetic field lines for magnetic induction (B) form continuous lines. So, lines of B are necessarily continuous across $S$.
Also, magnetic intensity $\text{(H)}$ varies for inside and outside the lump. So, lines of $\text{H}$ cannot all be continuous across $S$.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. A long solenoid has 1000 turns per metre and carries a current of 1 A. It has a soft iron core of $\mu_r=1000$. The core is heated beyond the Curie temperature, $T_c$.
<div class="fragment">
(a) The $\text{H}$ field in the solenoid is (nearly) unchanged but the $\text{B}$ field decreases drastically
(b) The $\text{H}$ and $\text{B}$ fields in the solenoid are nearly unchanged
(c) The magnetisation in the core reverses direction
(d) The magnetisation in the core diminishes by a factor of about $10^{8}$
</div>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
The magnetic intensity $\text{H}$ field $=n \text{I}$, where $n=$ number of turns per metre of a solenoid
and $I=$ current and $B=\mu_0 \mu_r n I$.
Also, at normal temperature, a solenoid behave as a ferromagnetic substand and at the temperature beyond the Curie temperature, it behaves as a paramagnetic substance.
Ans. (a, d)
Here, for solenoid $\text{H}=n I$.
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thus, from above equation, we find that $\text{B}$ varies with the variation in $\mu_r$.
Now, for magnetisation in the core, when temperature of the iron core of solenoid is raised beyond Curie temperature, then it behave as paramagnetic material, where
$ and (\chi_m) _{\text{Fero }} \approx 10^3 $
and $(\chi_m) _{\text{Para }} \approx 10^{-5} $
$\Rightarrow \frac{(\chi_mt) _{\text{Fero }}}{(\chi_m)} =\frac{10^3}{10^{-5}}=10^8 $
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to
<div class="fragment">
(a) electrostatic field lines can end on charges and conductors have free charges
(b) lines of $\text{B}$ can also end but conductors cannot end them
(c) lines of $\text{B}$ cannot end on any material and perfect shielding is not possible
(d) shells of high permeability materials can be used to divert lines of $\text{B}$ from the interior region
</div>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans. (a, c, d)
Electrostatic shielding is the phenomenon to block the effects of an electric field. The conducting shell can block the effects of an external field on its internal content or the effect of an internal field on the outside environment.
Magnetostatic shielding is done by using an enclosure made of a high permeability magnetic material to prevent a static magnetic field outside the enclosure from reaching objects inside it or to confine a magnetic field within the enclosure.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">MCQ (More than one correct option) </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
Angle of inclination or dip is the angle that the total magnetic field of the earth makes with the surface of the earth.
Ans. (b, c, d)
If the total magnetic field of the earth is modelled by a point magnetic dipole at the centre, then it is in the same plane of geographical equator, thus the angle of dip at a point on the geographical equator is bounded in a range from positive to negative value.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Very short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
Mass of a proton is very larger than the mass of an electron, so its spinning is negligible as compared to that of electron spin.
Ans.
The comparison between the spinning of a proton and an electron can be done by comparing their magnetic dipole moment which can be given by
$ M =\frac{e h}{4 \pi m} \text{ or } M \propto \frac{1}{m} $
$\therefore \frac{M_p}{M_e} =\frac{m_e}{m_p} $
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Very short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
$\text{ density of copper } \rho _{Cu}=\frac{8 ~g}{22.4 L}=\frac{8 ~g}{22400 cc} $
Now, comparing both densities
Also given
$\frac{\rho _{N_2}}{\rho _{{Cu}}} =\frac{28}{22400} \times \frac{1}{8}=1.6 \times 10^{-4} $
$\frac{\chi _{N_2}}{\chi _{{Cu}}} =\frac{5 \times 10^{-9}}{10^{-5}}=5 \times 10^{-4} $
$\chi =\frac{\text{ Magnetisation }(M)}{\text{ Magnetic intensity }(H)} $
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Very short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
$ =\frac{\text{ Magnetic moment }(M) / \text{ Volume }(V)}{H} $
$ =\frac{M}{H V}=\frac{M}{H \text{ (mass / density) }}=\frac{M \rho}{H m} $
We know that, $\quad \chi=\frac{\text{ Magnetisation }(M)}{\text{ Magnetic intensity }(H)}$
$\therefore \quad \chi \propto \rho \quad(\because \frac{M}{H m}=$ constant $)$
Hence,
$\frac{{N_2}}{{{Cu}}}=\frac{\rho _{N_2}}{{Cu}}=1.6 \times 10^{-4}$
Thus, we can say that magnitude difference or major difference between the diamagnetic susceptibility of $N_2$ and ${Cu}$.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Very short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans.
Susceptibility of magnetic material $\chi=\frac{I}{H}$, where $I$ is the intensity of magnetisation induced in the material and $H$ is the magnetising force.
Diamagnetism is due to orbital motion of electrons in an atom developing magnetic moments opposite to applied field. Thus, the resultant magnetic moment of the diamagnetic material is zero and hence, the susceptibility $\chi$ of diamagnetic material is not much affected by temperature.
Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature is raised, the alignment is disturbed, resulting decrease in susceptibility of both with increase in temperature.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Very short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
A superconducting material and nitrogen are diamagnetic in nature.
Ans.
When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.
(i) Thus, it will be repelled.
(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans.
<div style='float: left; width: 60%'>
Let us draw the figure for given situation,
We have to prove that $g \text{B}.d \text{S}=0$. This is called Gauss’s law in magnetisation.
</div>
<div style='float: right; width: 38%'>
<img src="https://cdn.mathpix.com/snip/images/zx8UC7ZvN_vAcJipUeJrz5P0LELrWeg6P7SOlcHpU2M.original.fullsize.png" height="180px">
</div>
 
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
According to question,
Magnetic moment of dipole at origin $O$ is
$\text{M}=M \hat{k}$
Let $P$ be a point at distance $r$ from $O$ and $O P$ makes an angle $\theta$ with $z$-axis. Component of $\text{M}$ along $O P=M \cos \theta$.
Now, the magnetic field induction at $P$ due to dipole of moment $\text{M} \cos \theta$ is
$\text{B}=\frac{\mu_0}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{r}$
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
From the diagram, $r$ is the radius of sphere with centre at $O$ lying in yz-plane.
Take an elementary area $\text{S}$ of the surface at $P$. Then,
$d \text{S} =r(r \sin \theta d \theta) \hat{\text{r}}=r^{2} \sin \theta d \theta \hat{\text{r}} $
$ \text{B} . d \text{S} =\oint \frac{\mu_0}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{\text{r}}(r^{2} \sin \theta d \theta \hat{\text{r}}) $
$ =\frac{\mu_0}{4 \pi} \frac{M}{r} \int_0^{2 \pi} 2 \sin \theta . \cos \theta d \theta $
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: center;font-size:22px'>
Ques. Three identical bar magnets are rivetted together at centre in the same plane as shown in figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the figure. Determine the poles of the remaining two.
<img src="https://cdn.mathpix.com/snip/images/qAZB6IZw2vTQqI_MvUoycZPWepCwL-GsAe2jMP2NzGg.original.fullsize.png" height="190px">
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole ${p}$ in an electrostatic field ${E}$ and (ii) magnetic dipole ${M}$ in a magnetic field ${B}$. Write down a set of conditions on ${E}, {B}, {p}, {M}$ so that the two motions are verified to be identical. (Assume identical initial conditions).
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
$E(r)=c B(r)$, suppose the angle between ${p}$ and ${E}$ is ${\theta}$. Torque on electric dipole of moment
${p}$ in electric field ${E}, \tau=p E \sin \theta$.
Ans.
Now, suppose that the angle between ${M}$ and $B$ is $\theta$.
Torque on magnetic dipole moment ${M}$ in magnetic field ${B}$,
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans.
Given, $I=$ moment of inertia of the bar magnet
$m=$ mass of bar magnet
$l=$ length of magnet about an any passing through its centre and perpendicular to its length
$M=$ magnetic moment of the magnet
$B=$ uniform magnetic field in which magnet is oscillating, we get time period of oscillation is,
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Let $C$ be the amperian loop. Then,
$\int_Q^{P} H . d l=\int_Q^{P} \frac{B}{m_0} . d l$
We know that the angle between $\mathbf{B}$ and $\mathbf{d I}$ is less than $90^{\circ}$ inside the bar magnet. So, it is positive.
i.e.,
$\int_Q^{P} H . dl=\int_Q^{P} \frac{B}{\mu_0} . dl>0$
Hence, the lines of $B$ must run from south pole $(S)$ to north pole $(N)$ inside the bar magnet.
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Short answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
According to Ampere’s law,
$\therefore \oint _{P Q P} H . dl=0 $
$\therefore \oint _{P Q P} H . dl=\int_P^{Q} H . dl+\int_Q^{P} H . dl=0 $
$\text{ As } \int_Q^{P} H . dl>0, so, \int_P^{Q} H . dl<0 $
It will be so if angle between ${H}$ and ${~ d l}$ is more than $90^{\circ}$, so that $\cos \theta$ is negative. It means the line of $H$ must run from $N$-pole to $S$-pole inside the bar magnet.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. Verify the Ampere’s law for magnetic field of a point dipole of dipole moment $\text{m}=m \hat{\text{k}}$. Take $C$ as the closed curve running clockwise along
(i) the $z$-axis from $z=a>0$ to $z=R$,
(ii) along the quarter circle of radius $R$ and centre at the origin in the first quadrant of x-z plane,
(iii) along the $x$-axis from $x=R$ to $x=a$, and
(iv) along the quarter circle of radius $a$ and centre at the origin in the first quadrant of x-z plane
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans.
From $P$ to $Q$, every point on the $z$-axis lies at the axial line of magnetic dipole of moment ${M}$. Magnetic field induction at a point distance $z$ from the magnetic dipole of moment is
$|{B}|=\frac{\mu_0}{4 \pi} \frac{2|{M}|}{z^{3}}=\frac{\mu_0 M}{2 \pi z^{3}}$
(i) Along z-axis from $P$ to $Q$.
$\int _{\rho}^{Q} B \cdot d l =\int _{\rho}^{Q} B \cdot d l \cos 0^{\circ}=\int_a^{B} B d z $
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
<div style='float: left; width: 60%'>
$ =\int_a^{R} \frac{\mu_0}{2 \pi} \frac{M}{z^{3}} d z=\frac{\mu_0 M}{2 \pi} \frac{-1}{2} \frac{1}{R^{2}}-\frac{1}{a^{2}} $
$ =\frac{\mu_0 M}{4 \pi} \frac{1}{a^{2}}-\frac{1}{R^{2}} $
(ii) Along the quarter circle $Q S$ of radius $R$ as given in the figure below
</div>
<div style='float: right; width: 37%'>
<img src="https://cdn.mathpix.com/snip/images/rKHyr1Cj3OH6UG7unmGJ3Df-HrujZFXKP0d8SmZvyXM.original.fullsize.png" height="200px">
</div>
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
The point $A$ lies on the equatorial line of the magnetic dipole of moment $M \sin \theta$. Magnetic field at point $A$ on the circular arc is
$B =\frac{\mu_0}{4 \pi} \frac{M \sin \theta}{R^{3}} ; dl=R d \theta $
$\therefore \int B . dl=\int B \cdot dl \cos \theta=\int_0^{\frac{\pi}{2}} \frac{\mu_0}{4 \pi} \frac{M \sin \theta}{R^{3}} R d \theta $
Circular arc $=\frac{\mu_0}{4 \pi} \frac{M}{R}(-\cos \theta)_0^{\pi / 2}=\frac{\mu_0}{4 \pi} \frac{M}{R^{2}}$
(iii) Along $x$-axis over the path $S T$, consider the figure given ahead
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
$ \therefore \int _{\beta}^{T} . {d l}=\int_R^{a}-\frac{\mu_0 {M}}{4 \pi x^{3}} . d {d l}=0[\because \text{ angle between }(-{M}) \text{ and } {d l} \text{ is } 90^{\circ}]$
(iv) Along the quarter circle TP of radius a. Consider the figure given below
<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/6705ad3b-566f-414e-ffbf-09d9e352d600/public" height="170px">
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
From case (ii), we get line integral of ${B}$ along the quarter circle $T P$ of radius $a$ is circular arc TP
$ \int B. dl \int _{\pi / 2}^0 \frac{\mu_0}{4 \pi} \frac{M \sin \theta}{a^{3}} ad \theta $
$ =\frac{\mu_0}{4 \pi} \frac{M}{a^{2}} \int _{\pi / 2}^0 \sin \theta d \theta=\frac{\mu_0}{4 \pi} \frac{M}{a^{2}}[-\cos \theta] _{\pi / 2}^0 $
$ =\frac{-\mu_0}{4 \pi} \frac{M}{a^{2}} $
$ =\frac{\mu_0 M}{4} \frac{1}{a^{2}}-\frac{1}{R^{2}}+\frac{\mu_0}{4 \pi} \frac{M}{R^{2}}+0+-\frac{\mu_0}{4 \pi} \frac{M}{a^{2}}=0 $
$\therefore \quad \oint _{P Q S T} B. d l=\int_P^{Q} B. d l+\int_Q^{S} B. d l+\int_S^{T} B. d l+\int_T^{P} B. d l$
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. What are the dimensions of $\chi$, the magnetic susceptibility? Consider an $H$-atom. Gives an expression for $\chi$, upto a constant by constructing a quantity of dimensions of $\chi$, out of parameters of the atom $e, m, v, R$ and $\mu_0$. Here, $m$ is the electronic mass, $v$ is electronic velocity, $R$ is Bohr radius. Estimate the number so obtained and compare with the value of $|\chi| \sim 10^{-5}$ for many solid materials.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Thinking process :
Magnetic susceptibility is a measure of how a magnetic material responds to an external field. i.e., magnetic susceptibility
$\chi_m=\frac{I}{H}=\frac{(\text{ Intensity of magnetisation) }}{\text{ (Magnetising force) }}$
Ans.
As $I$ and $H$ both have same units and dimensions, hence, $\chi$ has no dimensions. Here, in this question, $\chi$ is to be related with $e, m, v, R$ and $\mu_0$. We know that dimensions of $\mu_0=[ML \theta^{-2}]$
From Biot-Savart’s law,
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:20px'>
$d B =\frac{\mu_0}{4 \pi} \frac{I d l \sin \theta}{r^{2}} $
$\Rightarrow \quad \mu_0 =\frac{4 \pi r^{2} d B}{I d l \sin \theta}=\frac{4 \pi r^{2}}{I d l \sin \theta} \times \frac{f}{q v \sin \theta} \quad \because d B=\frac{F}{q v \sin \theta} $
$\therefore \quad \text{ Dimensions of } \mu_0 =\frac{L^{2} \times(MLT^{-2})}{(QT^{-1})(L) \times 1 \times(Q)(LT^{-1}) \times(1)} \newline=[MLQ^{-2}] $
where $Q$ is the dimension of charge.
As $\chi$ is dimensionless, it should have no involvement of charge $Q$ in its dimensional formula. It will be so if $\mu_0$ and $e$ together should have the value $\mu_0 e^{2}$, as $e$ has the dimensions of charge.
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:20px'>
Let
$\chi=\mu_0 e^{2} m^{a} v^{b} R^{c}$
where $a, b, c$ are the power of $m, v$ and $R$ respectively, such that relation (i) is satisfied.
Dimensional equation of (i) is
${[M^{0} L^{0} T^{0} Q^{0}] } =[ML Q^{-2}] \times[Q^{2}][M^{a}] \times(LT^{-1})^{b} \times[L]^{c}$
$ =[M^{1+a}+L^{1+b+c} T^{-b} Q^{0}] $
Equating the powers of $M, L$ and $T$, we get
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ques. Assume the dipole model for the earth’s magnetic field $B$ which is given by $B_V=$ vertical component of magnetic field $=\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^{3}}$
$B_H=$ horizontal component of magnetic field $=\frac{\mu_0}{4 \pi} \frac{\sin \theta m}{r^{3}}$
$\theta=90^{\circ}$-lattitude as measured from magnetic equator.
Find loci of points for which (a) $|B|$ is minimum (b) dip angle is zero and (c) dip angle is $45^{\circ}$.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans.
(a) $ B_V =\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^{3}} $
$ B_H =\frac{\mu_0}{4 \pi} \frac{\sin \theta m}{r^{3}} $
Squaring both the equations and adding, we get
$ B_V^{2}+B_H^{2} =\frac{\mu_0}{4 \pi} \frac{m^{2}}{r^{6}}[4 \cos ^{2} \theta+\sin ^{2} \theta] $
$B =\sqrt{B_V^{2}+B_H^{2}}=\frac{\mu_0}{4 \pi} \frac{m}{r^{3}}[3 \cos ^{2} \theta+1]^{1 / 2} $
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
From Eq. (iii), the value of $B$ is minimum, if $\cos \theta=\frac{\pi}{2}$
$\theta=\frac{\pi}{2}$. Thus, the magnetic equator is the locus.
(b) Angle of dip,
$ \tan \delta=\frac{B_V}{B_H}=\frac{\frac{\mu_0}{4 \pi} \cdot \frac{2 m \cos \theta}{r^{3}}}{\frac{\mu_0}{4 \pi} \cdot \frac{\sin \theta \cdot m}{r^{3}}}=2 . \theta $
$ \tan \delta=2 . \theta $
For dip angle is zero i.e., $\delta=0$
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: center;font-size:20px'>
Ans.
$P$ is in the plane $S$, needle is in north, so the declination is zero.
<img src="https://cdn.mathpix.com/snip/images/kORYQZRbpgLEwg2RGBsWfr-qbZgopnhYp2fUsd20N84.original.fullsize.png" height="170px">
$P$ is also on the magnetic equator, so the angle of dip $=0$, because the value of angle of dip at equator is zero. $Q$ is also on the magnetic equator, thus the angle of dip is zero. As earth tilted on its axis by $11.3^{\circ}$, thus the declination at $Q$ is $11.3^{\circ}$.
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>

<Success style="float:center; text-align:center;font-size:28px">MAGNETISM AND MATTER </Success>
<Primary style="float:center; text-align:center;font-size:24px">Long answer type questions </Primary>
<hr>
<main>
<div style='float: width:100%; text-align: left;font-size:22px'>
Ans.
 $C_1=$ circular coil of radius $R$, length $L$, number of turns per unit length 
$n_1=\frac{L}{2 \pi R}$ 
$C_2=$ square of side $a$ and perimeter $L$, number of turns per unit length $n_2=\frac{L}{4 a}$
<img src="https://cdn.mathpix.com/snip/images/SVVI2tN3l_2aC1d0hHzkxHu3NxhNhUXTOeOWdBG3ksU.original.fullsize.png" height="120px">
<Success>Sol. to be continued …</Success>
</main>
<hr>
<footer> MCQ (Single correct option) $\rarr$ MCQ (More than one correct option) $\rarr$ Very short answer type questions $\rarr$ Short answer type questions $\rarr$ Long answer type questions</footer>