Past NEET Entrance Papers

Hint

(b) The magnetic potential energy \(U\) of a magnetic dipole moment \(\vec{m}\) in a magnetic field \(\vec{B}\) is given by the formula:
\(
U=-\vec{m} \cdot \vec{B}
\)
Here, the dot product \(\vec{m} \cdot \vec{B}\) represents the scalar product of the vectors. When the magnetic moment \(\vec{m}\) is placed perpendicular to the magnetic field \(\vec{B}\), the angle \(\theta\) between these two vectors is 90 degrees. The dot product in this case can be written as:
\(
\vec{m} \cdot \vec{B}=m B \cos \theta
\)
Since \(\theta=90^{\circ}\), we have \(\cos 90^{\circ}=0\). Therefore, the magnetic potential energy \(U\) becomes:
\(
\begin{aligned}
& U=-m B \cos 90^{\circ} \\
& U=-m B \cdot 0 \\
& U=0
\end{aligned}
\)
So, the magnetic potential energy when a magnetic bar of magnetic moment \(\vec{m}\) is placed perpendicular to the magnetic field \(\vec{B}\) is zero.

Hint

(d)

\(
\begin{aligned}
&\text { For diamagnetic material, }\\
&\begin{aligned}
& 0 \leq \mu_r<1 \\
& \chi=\mu_r-1 \\
& \Rightarrow-1 \leq \chi<0 \\
& \mu_r=\frac{\mu}{\mu_0} \\
& \Rightarrow \mu<\mu_0 \quad\left(\because \mu_r<1\right)
\end{aligned}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\text { Time period of oscillation of magnet inside the magnetic field } T=2 \pi \sqrt{\frac{I}{M B}}\\
&\begin{aligned}
& T=\frac{t}{n}=\frac{10}{10}=1 \mathrm{~s} \\
& 1=2 \pi \sqrt{\frac{10^{-6}}{\pi^2 \times 1.0 \times 10^{-2} \times B}} \\
& \frac{1}{4}=\frac{10^{-4}}{B} \Rightarrow B=0.4 \mathrm{mT}
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { We have } 60^{\circ}=\frac{\pi}{3} \text { radian }\\
&\Rightarrow L=\frac{\pi}{3} R \Rightarrow R=\frac{3 L}{\pi}
\end{aligned}
\)
\(
M^{\prime}=m(A B)=m \times R=\frac{m 3 L}{\pi}=\frac{3}{\pi} m L=\frac{3}{\pi} M
\)

Hint

(c)

\(
\begin{aligned}
&\text { Time period of Oscillation, } T=2 \pi \sqrt{\frac{I}{M B}}\\
&\begin{aligned}
& \Rightarrow \frac{1}{4}=2 \pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}} \\
& \Rightarrow \quad \frac{1}{16}=4 \pi^2 \times \frac{9.8 \times 10^{-6}}{M \times 49 \times 10^{-3}} \\
& \Rightarrow \quad M=\frac{4 \pi^2 \times 9.8 \times 10^{-6}}{49 \times 10^{-3}} \times 16 \\
& =\frac{4 \pi^2 \times 9.8 \times 16 \times 10^{-3}}{49} \\
& =12.8 \pi^2 \times 10^{-3} \times 10^{-2} \times 10^2 \\
& =1280 \pi^2 \times 10^{-5} \mathrm{Am}^2
\end{aligned}
\end{aligned}
\)

Hint

(d)

North of magnet is moving away from solenoid 1 so end \(B\) of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end \(C\) of solenoid 2 is South.

Hint

(a) To match the magnetic materials with their respective magnetic susceptibilities, we must understand the characteristics of each type of material regarding its magnetic behavior. Here are the implications of each type:

Diamagnetic Materials: These materials are repelled by magnetic fields. The susceptibility \((\chi)\) of diamagnetic materials is negative but very small, near zero. Therefore, \(\chi>-1\) and closer to zero, reflects this property.

Ferromagnetic Materials: They have a very high, positive susceptibility because these materials can be permanently magnetized. Their susceptibility \((\chi)\) is much greater than one ( \(\chi \gg 1\) ).

Paramagnetic Materials: These materials are weakly attracted by a magnetic field, and their susceptibility is positive. However, the value of \(\chi\) for paramagnetic materials is small but greater than zero \((0<\chi<\varepsilon)\), where \(\varepsilon\) represents a small positive value.

Non-magnetic Materials: For materials that are considered non-magnetic, the susceptibility \((\chi)\) is zero \((\chi=0)\), as they are neither attracted nor repelled by magnetic fields.

Let’s match these descriptions with the given List II in your question:
Diamagnetic (A) matches with \(0>\chi \geq-1\) (II).
Ferromagnetic (B) corresponds to \(\chi \gg 1\) (III).
Paramagnetic (C) should be aligned with \(0<\chi<\varepsilon\) (IV).
Non-magnetic (D) clearly fits \(\chi=0\) (I).
Therefore, the correct answer matches:
A-II, B-III, C-IV, D-I.

Hint

(b) When the bar is bent at an angle \(\theta=60^{\circ}\), the effective magnetic moment is:
\(
M_{n e w}=M \cos \frac{\theta}{2}
\)
Substitute \(\theta=60^{\circ}\) :
\(
M_{\text {new }}=M \cos 30^{\circ}=M \cdot \frac{\sqrt{3}}{2} .
\)
The effective length is halved, so:
\(
M_{\text {final }}=\frac{M}{2}
\)

Hint

(c) 1. Understand the Initial Magnetic Moment:
The magnetic moment \(M\) of the straight rod can be expressed as:
\(
M=m \cdot L
\)
where \(m\) is the pole strength and \(L\) is the length of the rod.
2. Determine the New Shape:
When the rod is bent into a semicircle, its length remains the same, but its shape changes. The length of the rod \(L\) will now correspond to the arc length of the semicircle.
3. Relate Length to Radius:
The length of the semicircle can be expressed in terms of its radius \(R\) :
\(
L=\pi R
\)
This is because the circumference of a full circle is \(2 \pi R\), and for a semicircle, it is half of that.
4. Solve for Radius:
From the equation \(L=\pi R\), we can solve for \(R\) :
\(
R=\frac{L}{\pi}
\)
5. Calculate the New Magnetic Moment:
The magnetic moment of the semicircular rod can be expressed as:
\(
M^{\prime}=m \cdot(2 R)
\)
Here, \(2 R\) is the diameter of the semicircle.
6. Substitute for Radius:
Substitute \(R\) into the equation for \(M^{\prime}\) :
\(
M^{\prime}=m \cdot\left(2 \cdot \frac{L}{\pi}\right)=\frac{2 m L}{\pi}
\)
7. Relate New Magnetic Moment to Original:
Since \(m L=M\), we can substitute this into our equation for \(M^{\prime}\) :
\(
M^{\prime}=\frac{2 M}{\pi}
\)

Hint

(d) A-II, B-III, C-IV, D-I

Hint

(b) Step 1: Understand the relationship between susceptibility and temperature
Paramagnetic susceptibility ( \(\chi\) ) is known to be inversely proportional to the absolute temperature (T). This means that as the temperature increases, the susceptibility decreases.
Step 2: Write the mathematical expression
From the relationship, we can express this as:
\(
\chi \propto \frac{1}{T}
\)
This can also be written as:
\(
\chi=\frac{k}{T}
\)
where \(k\) is a constant.
Step 3: Analyze the graph of the relationship
The equation \(\chi=\frac{k}{T}\) indicates that the graph of \(\chi\) versus \(T\) will be a hyperbola. Specifically, it will be a rectangular hyperbola that approaches the axes but never touches them.
Step 4: Identify the characteristics of the graph
In a graph of \(\chi\) versus \(T\) :
As T increases, \(\chi\) decreases.
The graph will not intersect the axes, indicating that \(\chi\) approaches zero as T approaches infinity.
Step 5: Select the correct graph
Given the options available, we need to identify which graph represents a rectangular hyperbola. The correct graph will show a curve that decreases as Tincreases, demonstrating the inverse relationship.

Hint

(b) To understand which statement(s) is/are correct, we must explore the interactions between the magnetic field and different types of materials (magnetic, nonmagnetic, conducting, non-conducting).

Statement A: If a sheet is magnetic and placed near a strong magnetic pole, it will experience a magnetic force due to the magnetic field. This force could either attract or repel the sheet depending on the polarity of the magnetic pole and the induced or inherent poles in the sheet. A force is needed to hold the sheet stationary against this magnetic force. Hence, statement A is true.

Statement B: If the sheet is non-magnetic, it will not experience any magnetic force because non-magnetic materials do not respond to magnetic fields in a manner where a force would be exerted on them. Thus, no external force is needed to hold a non-magnetic sheet in place near a magnetic pole. Statement B is false.

Statement C: If the sheet is conducting and you move it away from the magnetic pole, it will experience a change in the magnetic flux through it. According to Faraday’s Law of Electromagnetic Induction, a change in magnetic flux induces an electromotive force (EMF) and consequently, currents known as eddy currents in the conductor. These induced currents produce their own magnetic fields, which interact with the original magnetic field, leading to a force (Lenz’s Law). To move the sheet away with uniform velocity, a force must counteract this magnetic interaction. Therefore, statement C is true.

Statement D: If the sheet is non-conducting and non-polar, it will neither induce currents (since it’s non-conducting) nor experience magnetic forces (since it’s nonmagnetic). Thus, moving such a sheet away from a magnetic pole with uniform velocity does not require overcoming any electromagnetic forces. Statement D is false.

Given these explanations, the correct statements are A and C only.

Hint

(b) 1. Understanding Magnetic Flux:
Magnetic flux (Ф) through a surface is defined as the integral of the magnetic field ( \(B\) ) over that surface. Mathematically, it is expressed as:
\(
\Phi=\int \vec{B} \cdot d \vec{S}
\)
where \(d \vec{S}\) is a differential area vector on the surface.
2. Applying Gauss’s Law for Magnetism:
According to Gauss’s Law for magnetism, the net magnetic flux through any closed surface is always zero. This is expressed as:
\(
\oint \vec{B} \cdot d \vec{S}=0
\)
This law states that there are no magnetic monopoles; hence, magnetic field lines that enter a closed surface must also exit it.
3. Conclusion:
Since the net magnetic flux through any closed surface is zero, we conclude that:
Net magnetic flux \(=0\)

Hint

(a) By magnetic property
\(
\chi \propto \frac{1}{T}
\)
\(\chi\) vs \(\frac{1}{T}\) graph will be straight line.

Hint

(c) Gauss’s law for magnetism, also known as Gauss’s law for magnetic fields, states that the net magnetic flux through any closed surface is always zero.
Mathematically, Gauss’s law for magnetism is given by: \(\oint \vec{B} \cdot \vec{dA}=0\) where,
” \(\oint\) ” represents the closed surface integral,
\(B\) is the magnetic field vector,
\(\vec{dA}\) is the infinitesimal area vector.
In simple terms, Gauss’s law for magnetism tells us that the total number of magnetic field lines entering a closed surface is equal to the total number of magnetic field lines exiting the surface. This is because magnetic field lines always form closed loops and do not have a starting or ending point is a crucial aspect of Gauss’s law for magnetism. Unlike electric charges, which can exist as isolated positive or negative charges (monopoles), magnetic charges (magnetic monopoles) have never been observed in nature. If magnetic monopoles were to exist, Gauss’s law for magnetism would take a different form. However, so far, all magnetic field configurations have been found to satisfy Gauss’s law for magnetism with a net magnetic flux of zero over any closed surface.

Hint

(a) Statement I is correct.
Magnetic field on the axis at a distance \(x\) from the centre of a circular current-carrying loop is given by,
\(
\mathrm{B}=\frac{\mu_0 \mathrm{IR}^2}{2 \mathrm{R}^2+\mathrm{x}^{2^{\frac{3}{2}}}}
\)
At a very far away point on the axial line means \(x \gg R\).
\(
\text { Then, } B=\frac{\mu_0 {R}^2}{2 R^2+x^2 2^{\frac{3}{2}}} \simeq \frac{\mu_0 {R}^2}{2 x^{2 \frac{3}{2}}}=\frac{2 \mu_0 I \pi R^2}{4 \pi x^3}=\frac{2 \mu_0 M}{4 \pi x^3}
\)
Where, magnetic Moment is, \(M=I\left(\pi R^2\right)\).
Statement II is incorrect.
The magnetic field due to magnetic dipole varies inversely with the cube of the distance from the centre on the axial line.
\(
\begin{aligned}
& B=\frac{\mu_0 M}{2 \pi x^3} \\
& B \propto \frac{1}{x^3}
\end{aligned}
\)

Hint

(c) 

\(
\begin{aligned}
& \phi_B=N B A \cos \omega t \\
& \varepsilon=\frac{-d \phi_B}{d t}=-N B A \omega(-\sin \omega t) \\
& \varepsilon=N B A \omega \sin \omega t \\
& i_{\max }=\frac{\varepsilon_{\max }}{R}-\frac{N B A \omega}{R} \\
& =\frac{1000 \times 2 \times 10^{-5} \times \pi(10)^2 \times 2}{12.56} \\
& =1 \mathrm{~A}
\end{aligned}
\)

Hint

(c) Given \({q}=1\) and \(\vec{v}=2 \hat{i}+4 \hat{j}+6 \widehat{k}\) and \(\vec{F}=4 \hat{i}-20 \hat{j}+12 \widehat{k}\)
Also given, \(\vec{F}=q(\vec{v} \times \vec{B})\)
\(
\begin{aligned}
& =\vec{q} \times\left(B \hat{i}+B \hat{j}+B_0 \widehat{k}\right) \\
& \Rightarrow(4 \hat{i}-20 \hat{j}+12 \widehat{k})=-1 \times\left[(2 \hat{i}+4 \hat{j}+6 \widehat{k}) \times\left(B \hat{i}+B \hat{j}+B_0 \widehat{k}\right)\right]
\end{aligned}
\)
Thus, calculating values of RHS,
\(
\vec{v} \times \vec{B}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 4 & 6 \\
B & B & B_0
\end{array}\right|
\)
\(
=\hat{i}\left(4 \mathrm{~B}_0-6 \mathrm{~B}\right)-\hat{j}\left(2 \mathrm{~B}_0-6 \mathrm{~B}\right)+\widehat{k}(2 \mathrm{~B}-4 \mathrm{~B})
\)
\(
\begin{aligned}
&\text { Comparing L.H.S and R.H.S, }\\
&\begin{aligned}
& 4 B_0-6 B=4 \ldots . .(1) \\
& -\left(2 B_0-6 B\right)=-20 \ldots . .(2)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&2 B-4 B=12 \Rightarrow B=-6 \dots(3)\\
&\text { From (2) and (3) } B=-6 \text { and } B_0=-8\\
&\therefore \vec{B}=-6 \hat{i}-6 \hat{j}-8 \widehat{k}
\end{aligned}
\)

Hint

(b) Magnetic dipole moment is associated with magnetic dipole, current-carrying conductors, etc. It is vector quantity and is equal to amount of electric current in the wire multiplied by the area of the loop.
The strength of the magnet or the magnetic dipole moment is represented by the equation:
\(
M=N I A
\)
Where M represents magnetic dipole moment.
Calculation:
Current in the loop will be \(\frac{V}{R}=I\) which is same for both loops.
Now magnetic moment of Triangle loop \(=NIA\)
\(
M_1=\left(\frac{12 a}{3 a}\right) \cdot I \cdot \frac{\sqrt{3}}{4} a^2=\sqrt{3} I a^2
\)
and magnetic moment of square loop \(=NIA \)
\(
\begin{aligned}
& =\left(\frac{12 a}{4 a}\right) \cdot I \cdot a^2 \\
& \mathrm{M}_2=3I \mathrm{a}^2
\end{aligned}
\)

Hint

(d) Given, \(\chi_m=599\)
Also, \(\mu_r=1+\chi_m=600\)
We know, \(\mu=\mu_r \mu_o\)
\(
\begin{aligned}
& \mu=600 \times 4 \pi \times 10^{-7} \\
& \mu=2400 \pi \times 10^{-7} \\
& \mu=2.4 \pi \times 10^{-4} T m A^{-1}
\end{aligned}
\)

Hint

(a) As the sourceof current is switched on, a magnetic field sets up in between the poles of the electromagnet. As we know that a diamagnetic substances when placed in a magnetic field acquires a feeble magnetism opposite to the direction of the magnetic field.
Also, in the presence of the field (non-uniform), these substances are attracted towards the weaker field, i.e. they move from stronger to weaker magnetic field. Due to these reasons, the rod is repelled by the field produced to the current source. Hence, it is pushed up, out of horizontal field and gains gravitational potential energy.

Hint

(d) Work done in a coil
\(
W=m B\left(\cos \theta_1-\cos \theta_2\right)
\)
When it is rotated by angle \(180^{\circ}\) then
\(
W=2 m B=2(N I A) B \dots(i)
\)
Given: \(N=250, I=85 \mu \mathrm{~A}=85 \times 10^{-6} \mathrm{~A}\)
\(
A=1.25 \times 2.1 \times 10^{-4} \mathrm{~m}^2 \approx 2.6 \times 10^{-4} \mathrm{~m}^2
\)
\(
\begin{aligned}
&B=0.85 \mathrm{~T}\\
&\text { Putting these values in eqn. (i), we get }\\
&\begin{aligned}
W & =2 \times 250 \times 85 \times 10^{-6} \times 2.6 \times 10^{-4} \times 0.85 \\
& \approx 9.1 \times 10^{-6} \mathrm{~J}=9.1 \mu \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

(d) Magnetic susceptibility \(\chi\) for dia-magnetic materials only is negative and low \(|\chi|=-1\); for paramagnetic substances low but positive \(|\chi|=1\) and for ferromagnetic substances positive and high \(|\chi|=10^2\).

Hint

(b) At equilibrium, potential energy of dipole
\(
U_i=-M B_H
\)
Final potential energy of dipole,
\(
\begin{aligned}
& U_f=-M B_H \cos 60^{\circ}=-\frac{M B_H}{2} \\
& \begin{aligned}
W= & U_f-U_i=-\frac{M B_H}{2}-\left(-M B_H\right)=\frac{M B_H}{2} \dots(i) \\
& \tau=M B_H \sin 60^{\circ}=2 W \times \frac{\sqrt{3}}{2} \text { [Using eqn. (i)] } \\
& =\sqrt{3} W
\end{aligned}
\end{aligned}
\)

Hint

(c)

The direction of magnetic dipole moment is from south to north pole of the magnet.In configuration (1),
\(
\begin{aligned}
m_{\text {net }} & =\sqrt{m^2+m^2+2 m m \cos 90^{\circ}} \\
& =\sqrt{m^2+m^2}=m \sqrt{2}
\end{aligned}
\)

In configuration (2),

\(
m_{\mathrm{net}}=m-m=0
\)

In configuration (3),

\(
\begin{aligned}
& m_{\mathrm{net}}=\sqrt{m^2+m^2+2 m m \cos 30^{\circ}} \\
& =\sqrt{2 m^2+2 m^2\left(\frac{\sqrt{3}}{2}\right)}=m \sqrt{2+\sqrt{3}}
\end{aligned}
\)

In configuration (4),

\(
\begin{aligned}
& \quad m_{\text {net }}=\sqrt{m^2+m^2+2 m m \cos 60^{\circ}} \\
& =\sqrt{2 m^2+2 m^2\left(\frac{1}{2}\right)}=m \sqrt{3}
\end{aligned}
\)

Hint

(d)

Let \(m\) be strength of each pole of bar magnet of length \(l\). Then
\(
M=m \times l \dots(i)
\)
When the bar magnet is bent in the form of an arc as shown in the figure. Then
\(
l=\frac{\pi}{3} \times r=\frac{\pi r}{3} \text { or } r=\frac{3 l}{\pi}
\)
New magnetic dipole moment
\(
M^{\prime}=m \times 2 r \sin 30^{\circ}
\)
\(
=m \frac{3 l}{\pi}=\frac{3 M}{\pi}
\)

Hint

\(
\text { (b) Torque } F L=\mathrm{MB} \Rightarrow L=\frac{M B}{F}
\)

Hint

(b) Work done in changing the orientation of a magnetic needle of magnetic moment \(M\) in a magnetic field \(B\) from position \(\theta_1\) to \(\theta_2\) is given by (According to work energy theorem)
\(
W=U_{\text {final }}-U_{\text {initial }}=M B\left(\cos \theta_1-\cos \theta_2\right)
\)
\(
\begin{aligned}
&\text { Here, } \theta_1=0^{\circ}, \theta_2=60^{\circ}\\
&=M B\left(1-\frac{1}{2}\right)=\frac{M B}{2} \dots(i)
\end{aligned}
\)
The torque on the needle is \(\vec{\tau}=\vec{M} \times \vec{B}\) In magnitude,
\(
\tau=M B \sin \theta=M B \sin 60^{\circ}=\frac{\sqrt{3}}{2} M B \dots(ii)
\)
Dividing (ii) by (i), we get
\(
\frac{\tau}{W}=\sqrt{3} \text { or } \tau=\sqrt{3} W=\sqrt{3} \times \sqrt{3} \mathrm{~J}=3 \mathrm{~J}
\)

Hint

(b) Here, Magnetic moment, \(M=0.4 \mathrm{~J} \mathrm{~T}^{-1}\)
Magnetic field, \(B=0.16 \mathrm{~T}\)
When a bar magnet of magnetic moment is placed in a uniform magnetic field, its potential energy is
\(
U=-\vec{M} \cdot \vec{B}=-M B \cos \theta
\)
For stable equilibrium, \(\theta=0^{\circ}\)
\(
\therefore \quad U=-M B=-\left(0.4 \mathrm{~J} \mathrm{~T}^{-1}\right)(0.16 \mathrm{~T})=-0.064 \mathrm{~J}
\)

Hint

(b) A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable.
When a current loop is placed in a magnetic field it experiences a torque. It is given by
\(
\vec{\tau}=\vec{M} \times \vec{B}=\mathrm{MB} \sin \theta
\)
where, \(\vec{M}\) is the magnetic moment of the loop and \(\vec{B}\) is the magnetic field.
If \(\theta=0^{\circ} \Rightarrow \tau=0\) (stable equilibrium)
If \(\theta=180^{\circ} \Rightarrow \tau=0\) (unstable equilibrium)

Hint

(d) Concept:
Magnetic materials are broadly classified into
1. Paramagnetic
2. Diamagnetic
3. Ferromagnetic based on the intensity of magnetization
The given table below shows the behavior of each material in the external magnetic field.

Hint

(d) The time period \(T\) of oscillation of a magnet is given by
\(
T=2 \pi \sqrt{\frac{I}{M B}}
\)
where,
\(I=\) Moment of inertia of the magnet about the axis of rotation
\(M=\) Magnetic moment of the magnet
\(B=\) Uniform magnetic field
As \(I\) and \(M\) remain the same
\(
\therefore \quad T \propto \frac{1}{\sqrt{B}} \text { or } \frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}
\)
According to given problem,
\(
\begin{aligned}
& B_1=24 \mu \mathrm{~T}, B_2=24 \mu \mathrm{~T}-18 \mu \mathrm{~T}=6 \mu \mathrm{~T}, T_1=2 \mathrm{~s} \\
\therefore \quad & T_2=(2 \mathrm{~s}) \sqrt{\frac{(24 \mu \mathrm{~T})}{(6 \mu \mathrm{~T})}}=4 \mathrm{~s}
\end{aligned}
\)

Hint

(c) The magnetic moment of a diamagnetic atom is equal to zero.

Hint

(a) Magnetic field due to bar magnets exerts force on moving charges only. Since the charge is at rest, no force acts on it.

Hint

\(
\begin{aligned}
&\text { (c)  Magnetic moment of the loop }\\
&\begin{aligned}
M= & \text { NIA }=2000 \times 2 \times 1.5 \times 10^{-4}=0.6 \mathrm{~J} / \mathrm{T} \\
\text { Torque } \tau & =M B \sin 30^{\circ} \\
& =0.6 \times 5 \times 10^{-2} \times \frac{1}{2}=1.5 \times 10^{-2} \mathrm{Nm}
\end{aligned}
\end{aligned}
\)

Hint

(d) Diamagnetic substances do not have any unpaired electron. And they are magnetised in direction opposite to that of magnetic field. Hence, when they are brought to north or south pole of a bar magnet, they are repelled by poles.

Hint

\(
\begin{aligned}
& \text { (b) Here, } M=2 \times 10^4 \mathrm{~J} \mathrm{~T}^{-1} \\
& B=6 \times 10^{-4} \mathrm{~T}, \theta_1=0^{\circ}, \theta_2=60^{\circ} \\
& W=M B\left(\cos \theta_1-\cos \theta_2\right)=M B\left(1-\cos 60^{\circ}\right) \\
& W=2 \times 10^4 \times 6 \times 10^{-4}\left(1-\frac{1}{2}\right)=6 \mathrm{~J}
\end{aligned}
\)

Hint

(a) Curie temperature is the temperature above which ferromagnetic material becomes paramagnetic material.

Hint

(d) Beyond Curie temperature, ferromagnetic substances behaves like a paramagnetic substance.

Hint

(d)

\(
\begin{aligned}
&\text { Magnetic moment } \mu=\mathrm{IA}\\
&\begin{aligned}
& \mathrm{T}=\frac{2 \pi R}{v} \text { and } \mathrm{I}=\frac{q}{T}=\frac{q v}{2 \pi R} \\
& \therefore \mu=\left(\frac{q v}{2 \pi R}\right)\left(\pi R^2\right)=\frac{q v R}{2}
\end{aligned}
\end{aligned}
\)

Hint

(d) The magnetic dipole moment of diamagnetic material is zero as each of its pair of electrons have opposite spins, i.e., \(\mu_d=0\).
Paramagnetic substances have dipole moment \(>0\), i.e. \(\mu_{\mathrm{p}} \neq 0\), because of excess of electrons in its molecules spinning in the same direction.
Ferro-magnetic substances are very strong magnets and they also have permanent magnetic moment, i.e. \(\mu_{\mathrm{f}} \neq 0\).

Hint

(b)

The current flowing clockwise in the equilateral triangle has a magnetic field in the direction \(\widehat{k}\).
\(
\tau=\mathrm{BiNA} \sin \theta=\mathrm{BiA} \sin 90^{\circ}
\)
\(
\therefore \mathrm{A}=\frac{\tau}{i B}
\)
Area of equilateral triangle
\(
\begin{aligned}
& \mathrm{A}=\frac{1}{2}(l) \sqrt{l^2-\left(\frac{l}{2}\right)^2}=\frac{\sqrt{3}}{4} l^2 \\
& \Rightarrow \frac{\sqrt{3}}{4} l^2=\frac{\tau}{i B} \\
& \Rightarrow l=2\left(\frac{\tau}{\sqrt{3} B i}\right)^{\frac{1}{2}}
\end{aligned}
\)

Hint

(b) A diamagnetic material in a magnetic field moves from stronger to the weaker parts of the field.

Hint

(b) Initial mass of the magnet \(\left(m_1\right)=m\) and final mass of the magnet \(\left(m_2\right)=4 m\).
The time period, \(T=2 \pi \sqrt{\frac{I}{M B}}=2 \pi \sqrt{\frac{m k^2}{M B}} \propto \sqrt{m}\)
Therefore \(\frac{T_1}{T_2}=\frac{\sqrt{m_1}}{\sqrt{m_2}}=\frac{\sqrt{m}}{\sqrt{4 m}}=\frac{1}{2}\)
or \(T_2=2 T_1=2 T\)

Hint

\(
\text { (b) According to Curie’s law } \chi \propto \frac{1}{T}
\)

Hint

(a)

(i)
\(
\begin{aligned}
& M=M_1+M_2 \\
& I=I_1+I_2
\end{aligned}
\)
(ii)
\(
\begin{aligned}
& M=M_1-M_2 \\
& I=I_1+I_2
\end{aligned}
\)
(i) Similar poles are placed at the same side (sum position)
(ii) Opposite poles are placed at the same side (difference position)
\(I_1\) and \(I_2\) are the moments of inertia of the magnets and \(M_1\) and \(M_2\) are the moments of the magnets.
Here \(M_1=M\) and \(M_2=2 M, I_1=I_2=I\) (say), for same geometry.
For sum position
\(
T_1=2 \pi \sqrt{\frac{I_1+I_2}{\left(M_1+M_2\right) H}}=2 \pi \sqrt{\frac{2 I}{(M+2 M) H}}
\)
For difference position.
\(
\begin{aligned}
& T_2=2 \pi \sqrt{\frac{I_1+I_2}{\left(M_2-M_1\right) H}}=2 \pi \sqrt{\frac{2 I}{(2 M-M) H}} \\
& \therefore \quad \frac{T_1}{T_2}=\sqrt{\frac{M}{3 M}}=\frac{1}{\sqrt{3}}<1 \quad \text { or } \quad T_1<T_2
\end{aligned}
\)

Hint

(a) Diamagnetism does not depened on the temperature

Hint

(b) When a current is passed through the galvanomejer coil, then a magnetic fieid B is produced at right angles to the plane of the coil, ie, at right angles to the horizontal component of earth’s magnetic field \(H\). Under the influence of two crossed magnetic fields \(B\) and \(H\), the magnetic needle of galvanometer undergoes a deflection \(\theta\) which is given by the tangent law. Using tangent law, we can find a relation
\(
I \propto \tan \theta
\)
Which clearly indicates that tangent. galvanometer is an instrument used for detection of electric current in a circuit.

Hint

(c) Step-1: Understanding the Vibration Magnetometer: A vibration magnetometer is used to measure the horizontal component of the Earth’s magnetic field. It consists of a bar magnet suspended freely so that it can oscillate.
Step-2: Time Period Formula: The time period \(T\) of the oscillation of the bar magnet is given by the formula:
\(
T=2 \pi \sqrt{\frac{I}{m H}}
\)
where: \(I=\) moment of inertia of the magnet, \(-m=\) mass of the magnet,
\(H\) = horizontal component of the Earth’s magnetic field.
Step-3: Horizontal Component of Magnetic Field: The horizontal component \(H\) of the Earth’s magnetic field can be expressed as:
\(
H=B \cos \theta
\)
where \(B\) is the total magnetic field and \(\theta\) is the dip angle.
Step-4: Analyzing the Relationship: From the formula for the time period, we can see that \(T\) is inversely proportional to \(H\) :
\(
T \propto \frac{1}{H}
\)
This means that as \(H\) increases, \(T\) decreases.
Step-5: Effect of Dip Angle: Since \(H=B \cos \theta\), if \(\theta\) decreases (moving towards the equator where \(\theta=0\) ), \(\cos \theta\) increases.
At the equator, \(\cos (0)=1\), which maximizes \(H\) and minimizes \(T\).
Step-6: Conclusion: Therefore, the time period \(T\) of the suspended bar magnet can be reduced by moving towards the equator, where the dip angle \(\theta\) is zero.
Final Answer: The time period of a suspended bar magnet can be reduced by moving towards the equator

Hint

(c) We know that when a bar magnet is placed in the magnetic field at an angle \(\theta\), then torque acting on the bar magnet ( \(\tau\) )
\(
=M B \sin \theta=\vec{M} \times \vec{B}
\)

Note: The direction of torque \((\tau)\) is perpendicular to the plane containing M and B and is given by right hand screw rule.

Hint

(b) The iron can produces a magnetic screening for the equipment as lines of magnetic force can not enter iron enclosure.

Note:When two magnets are placed in a vibration magnetometer such that identical poles in same direction then net magnetic moment \(\mathrm{M}_{\mathrm{s}}=\mathrm{M}_1+\mathrm{M}_2\) And when two magnets placed such that north facing south of other and vice versa then net magnetic moment \(\mathrm{M}_{\mathrm{d}}=\mathrm{M}_1-\mathrm{M}_2\)

Hint

\(
\begin{aligned}
&\text { (d)  Magnetic moment }=\text { pole strength } \times \text { length }\\
&M^{\prime}=M / 2=0.5 M
\end{aligned}
\)

Hint

(a) Angle of magnet \((\theta)=90^{\circ}\) and \(60^{\circ}\).
Work done in turning the magnet through \(90^{\circ}\)
\(
W_1=M B\left(\cos 0^{\circ}-\cos 90^{\circ}\right)=M B(1-0)=M B .
\)
Similarly
\(
W_2=M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right)=M B\left(1-\frac{1}{2}\right)=\frac{M B}{2}
\)
Therefore, \(W_1=2 W_2\) or \(n=2\)

Hint

(b) Magnetic moment of current carrying circular loop \(=I A\)
\(
\begin{aligned}
& M=I A \\
& M \propto A \quad[I-\text { Same }] \\
& \frac{M_1}{M_2}=\frac{A_1}{A_2}=\frac{\pi r_1^2}{\pi r_2^2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}
\end{aligned}
\)