NEET Practice Q & A

Hint

(d) Magnetic pole strength is stronger at end part of magnet, so maximum iron powder is collected at the end point of magnet.

Hint

(b) A permanent magnet has large retentivity and coercivity, so it attracts only magnetic substances.

Hint

(d) fluxmeter.
Explanation:
Fluxmeter: A fluxmeter is an instrument used to measure magnetic flux, which is related to the strength of a magnetic field. It works by measuring the voltage induced in a coil when it is moved through a magnetic field.
Pyrometer: A pyrometer is used to measure high temperatures, such as those found in furnaces.
Hydrometer: A hydrometer is used to measure the specific gravity of liquids.
Thermometer: A thermometer is used to measure temperature.

Hint

 (c) isoclinic lines. Isoclinic lines represent places of constant angle of dip on the Earth’s magnetic field.

Explanation:
Isoclinic lines: These are lines on a map that connect points where the angle of dip (the angle between the Earth’s magnetic field and the horizontal) is the same.
Isobaric lines: These are lines connecting points of equal atmospheric pressure.
Isogonic lines: These are lines connecting points of equal magnetic declination (the angle between true north and magnetic north).
Isodynamic lines: These are lines connecting points of equal magnetic intensity.

Hint

(d) A line passing through places having zero value of magnetic dip is called aclinic line. At all places upon this line, a freely suspended magnet will remain horizontal.

Hint

(a) zero dip. Aclinic lines are the lines on a map that join places where the magnetic dip is zero.

Hint

(b) north-south.
Explanation:
In a deflection magnetometer, the “tan B” position refers to the alignment of the magnetometer’s arms along the direction of the Earth’s horizontal magnetic field. Since the Earth’s magnetic field primarily points north-south, placing the arms in the tan B position means aligning them with the northsouth axis.

Hint

The current and deflection dependence of a moving coil galvanometer is given by \(i=\frac{k}{n A B} \theta \Rightarrow i \propto \theta\)
Therefore, if we double the current, the deflection also gets doubled.
However, in a tangent galvanometer, i \(\propto \tan \theta\); that is, there is no direct relation between \(\theta\) and current.

Hint

(d) Bismuth. 
Explanation: Diamagnetic materials are those that are weakly repelled by a magnetic field because all of their electrons are paired, meaning they have no unpaired electrons to contribute to a magnetic field. Bismuth is a known diamagnetic element.

Why other options are incorrect:
(a) Aluminium: Aluminium is paramagnetic, meaning it has a weak attraction to a magnetic field.
(b) Quartz: Quartz is also diamagnetic.
(c) Nickel: Nickel is ferromagnetic, meaning it is strongly attracted to a magnetic field and can retain its magnetization even after the external magnetic field is removed.

Hint

(a) slightly more than vacuum.
Explanation:
Paramagnetic materials: These are substances that are weakly attracted to a magnetic field. When placed in a magnetic field, they develop a slight magnetization in the same direction as the applied field.
Permeability: This is a measure of how easily a magnetic field can penetrate a material.
Relative permeability: This is the ratio of the permeability of a material to the permeability of vacuum. For a paramagnetic material, the relative permeability is slightly greater than 1, meaning the material allows magnetic fields to pass through it more easily than a vacuum.

Why other options are incorrect:
(b) slightly less than vacuum:

Diamagnetic materials have a relative permeability slightly less than vacuum, meaning they are weakly repelled by magnetic fields.
(c) much more than vacuum:

This is characteristic of ferromagnetic materials, which have a significantly higher permeability than vacuum due to their strong magnetic response.
(d) None of the above:

Since paramagnetic materials have a permeability slightly more than vacuum, option (a) is correct, making this option incorrect.

Hint

(d) SI unit of magnetic induction is
\(
B=\frac{\text { Unit of } \phi}{\text { Unit of } \Delta A}=\frac{\mathrm{Wb}}{\mathrm{~m}^2}=\text { tesla }
\)
Also, SI unit of \(B\) is weber \(/\) metre \(^2\) and newton/ampere-metre.

Hint

(c) Magnetic moment per unit volume is intensity of magnetisation, and number of lines of force crossing per unit area is intensity of magnetic field. 

Hint

(b) Permeability is defined as the ratio of magnetic induction to the magnetizing field(\(\mu=B / H\)).

Hint

(c) greater than.
Explanation: Steel generally exhibits a larger hysteresis loop area compared to iron. This larger area signifies a greater amount of energy lost during each magnetization and demagnetization cycle, thus resulting in higher hysteresis loss.

Hint

(b) ferromagnetic.
Explanation
Hysteresis is a property observed in certain materials where their state depends on their history, meaning they don’t immediately return to their original state after a force or field is removed.
Magnetic hysteresis refers to the lagging of magnetic flux density (B) behind the magnetic field strength \((\mathrm{H})\) in magnetic materials when an external magnetic field is applied and then removed or reversed.
Ferromagnetic materials exhibit hysteresis, meaning they can retain magnetization even after the external magnetic field is removed. This is due to the presence of magnetic domains that align with the field and can stay aligned even after the field is removed.
Paramagnetic materials are weakly attracted to magnetic fields but lose their magnetism once the external field is removed.
Diamagnetic materials are weakly repelled by magnetic fields and do not retain any magnetization.

Hint

(d) Hard iron.
Explanation:
Retentivity refers to a material’s ability to retain magnetization even after the external magnetic field is removed. Hard iron is known for its high retentivity, meaning it can hold onto its magnetism for a long time.

Why other options are incorrect:
(a) Copper: Copper is not magnetic. It has low retentivity because it does not retain any magnetization after the external field is removed.
(b) Zinc: Zinc is also non-magnetic and therefore has low retentivity.
(c) Soft iron: While soft iron has high permeability (meaning it easily becomes magnetized), it has low retentivity. This means it readily loses its magnetism once the external field is removed.

Hint

(d) energy loss per cycle.
Explanation: A hysteresis loop shows the relationship between the magnetic field strength \((\mathrm{H})\) and the magnetic flux density \((\mathrm{B})\) in a material as the magnetic field is cycled. The area enclosed by this loop represents the energy dissipated as heat during each cycle of magnetization and demagnetization.

Why other options are incorrect:
(a) retentivity:

Retentivity is the ability of a material to retain a magnetic field even after the external magnetizing field is removed. It’s measured by the residual magnetization the material retains when the magnetizing field is reduced to zero. The area under the hysteresis loop is related to retentivity but doesn’t directly measure it.
(b) susceptibility:

Susceptibility is a measure of how easily a material is magnetized, or the ratio of the magnetization to the applied magnetic field. It’s a property of the material itself, independent of the energy loss during magnetization.
(c) permeability:

Permeability is the measure of a material’s ability to allow magnetic flux to pass through it. It’s related to susceptibility but doesn’t directly represent the energy loss per cycle.

Hint

(a) Steel has more retentivity and coercivity, so it is used for making permanent magnet.

Hint

(b) The materials suitable for making electromagnets should have low retentivity and low coercivity.

Explanation:
Retentivity:
Refers to a material’s ability to retain magnetism after the magnetizing force is removed. For electromagnets, you want a material that loses its magnetism quickly when the current is turned off, so low retentivity is desired.

Coercivity:
Refers to the magnetic field strength needed to demagnetize a material. For electromagnets, you want a material that can be easily demagnetized, so low coercivity is desirable.

Soft iron:
A common material used for electromagnets, possessing low retentivity and coercivity.

Why other options are incorrect:
(a) high retentivity and high coercivity: This describes the properties of permanent magnets, not electromagnets.
(c) high retentivity and low coercivity: While this might be a specific case for certain applications, it’s not the general characteristic of electromagnets.
(d) low retentivity and high coercivity: This is also not ideal for electromagnets as it would be difficult to turn the magnet off.

Hint

(c) The most suitable material for the core of an electromagnet is Soft iron.

Explanation: Soft iron is preferred because it has low retentivity and coercivity, meaning it easily gains magnetization when a current is applied and loses it quickly when the current is removed, which is crucial for the efficient functioning of an electromagnet.

Why other options are incorrect:
(a) Iron:

While iron is generally ferromagnetic (meaning it can be magnetized), “soft iron” specifically refers to a type of iron with low hysteresis, making it more suitable for electromagnet cores.
(b) Steel:

Steel often contains alloying elements that can increase its hardness and resistance to deformation, but also raise its retentivity and coercivity, making it less ideal for electromagnet cores.
(d) Cu-Ni alloy:

Copper-nickel alloys are not ferromagnetic and therefore not suitable for electromagnet cores. They are primarily used for their electrical conductivity and resistance to corrosion.

Hint

(a) A magnetic needle placed in a non-uniform magnetic field experiences both a force and a torque. This is because the magnetic field strength and direction vary across the needle, causing unequal forces on its poles. These unequal forces result in a net force and a torque that tends to rotate the needle.

Hint

(b) For a diamagnetic substance, \(\chi\) is small, negative and independent of temperature.

Hint

(b) The angle between the Earth’s magnetic axis and the Earth’s geographic axis is approximately \(11.5^{\circ}\).

Hint

(a) A freely hanged magnet stays with its magnetic axis parallel to magnetic meridian.

Explanation: When a freely suspended magnet is allowed to rotate, its magnetic axis aligns itself with the Earth’s magnetic field, which defines the magnetic meridian.

Hint

(a) A dip needle in a plane perpendicular to magnetic meridian will always remain vertical.

Explanation: The magnetic meridian is the plane where the magnetic field lines of the Earth are vertical. A dip needle, when in this plane, aligns with the vertical component of the Earth’s magnetic field, and thus points downwards. If the needle is placed in a plane perpendicular to the magnetic meridian, the horizontal component of the Earth’s magnetic field is ineffective. This is because the axis of rotation of the needle lies in the direction of the horizontal component. Therefore, the needle remains vertical.

Hint

(d) The dip circle is oriented perpendicular to the magnetic meridian.
This means it is aligned with the vertical component of the Earth’s magnetic field.
When aligned with the vertical component, the dip needle will point straight down. This corresponds to an angle of \(90^{\circ}\) with the horizontal.

Hint

 (d) stay in any position.
Explanation: At the geomagnetic pole, the Earth’s magnetic field is vertical, meaning there is no horizontal component to the field. Since a compass needle aligns itself with the horizontal component of the magnetic field, at the geomagnetic pole, it can freely rotate in any direction within the horizontal plane. –

Hint

(a) If the \(N\)-pole of a bar magnet points south, the fields of the magnet and the earth will point in opposite directions along the axis of the magnet. So, two neutral points are obtained which are equidistant from the magnet on its axis.

Hint

(d) Due to magnetic force the charged particles get deviated for positively, charged particles, due to east-west asymmetry of particles, more particles get deviated towards east, then towards west, but since they are deviated, hardly any charged particles will reach equator.

Hint

(c) the horizontal component of the Earth’s magnetic field.

Explanation:
When a magnetic needle is suspended horizontally, it aligns itself with the Earth’s magnetic field. If the needle is displaced from this equilibrium position, the horizontal component of the Earth’s magnetic field exerts a torque on the needle, forcing it back to its original position. This torque acts as the restoring force, causing the needle to oscillate.

Why other options are incorrect:
(a) the torsion of the silk fibre:

While the silk fiber provides the means to suspend the needle, its torsion (twisting) contributes very little to the restoring force. The torque due to the Earth’s magnetic field is significantly larger.
(b) the force of gravity:

Gravity acts vertically downward on the needle. This force does not affect the needle’s horizontal oscillations, as it is perpendicular to the plane of movement.
(d) All the above factors:

Although the silk fiber’s torsion and gravity might play minor roles, the primary restoring force is undeniably the horizontal component of the Earth’s magnetic field. Therefore, option (d) is incorrect.

Hint

(b) An electron moving around the nucleus has a magnetic moment given by \(\mu=\frac{e}{2 m} L\)
where, \(L\) is the magnitude of the angular momentum of the circulating electron around the nucleus. The smallest value of \(\mu\) is called the Bohr magneton \(\mu_B\) and its value is
\(
\mu_B=9.27 \times 10^{-24} \mathrm{JT}^{-1} .
\)

Hint

(b) The time period of a vibration magnetometer placed at the south pole (or any magnetic pole) is infinity. This is because the horizontal component of the Earth’s magnetic field is zero at the poles, and the time period of oscillation in a vibration magnetometer is inversely proportional to the horizontal component of the magnetic field, according to a physics education website.

Hint

The true statement about the magnetic susceptibility \(\left(\chi_m\right)\) of a paramagnetic substance is:
(a) Value of \(\chi_m\) is inversely proportional to the absolute temperature of the sample.

This relationship is described by Curie’s Law, which states that \(\boldsymbol{\chi}_{\boldsymbol{m}}=\frac{\boldsymbol{C}}{\boldsymbol{T}}\), where C is the Curie constant and T is the absolute temperature.

Explanation:
Paramagnetic substances are weakly attracted by an external magnetic field. They have unpaired electrons, which align their spins with the applied field.
However, thermal vibrations tend to disrupt this alignment.
As the temperature increases, thermal agitation becomes stronger, making it harder for the magnetic moments to align with the applied field.
This results in a decrease in the magnetic susceptibility.

Why other options are incorrect:
(b) \(\chi_m\) is zero at all temperature: This is incorrect. Paramagnetic materials have a positive magnetic susceptibility, although it’s small.
(c) \(\chi_m\) is negative at all temperature: This describes diamagnetic materials, not paramagnetic ones.
(d) \(\chi_m\) does not depend on the temperature of the sample: This is incorrect. Paramagnetic susceptibility is highly dependent on temperature, following Curie’s Law.

Hint

(a) From the stronger to the weaker part of the magnetic field.

Explanation: A diamagnetic material is repelled by a magnetic field, meaning it will move away from the stronger part of the field and towards the weaker part.

Why the other options are incorrect:
(b) from weaker to the stronger part of the magnetic field:

This is the opposite of the actual behavior of a diamagnetic material. Diamagnetic materials are repelled by magnetic fields, so they move away from the stronger field and towards the weaker field.
(c) perpendicular to the magnetic field:

The force on a diamagnetic material is parallel to the magnetic field gradient, not perpendicular to it. The force is directed from the stronger to the weaker region of the field, which is along the gradient.
(d) in the direction making \(60^{\circ}\) to the magnetic field:

This is not the correct direction. The force is aligned with the magnetic field gradient, not at a \(60^{\circ}\) angle.

Hint

(a) From Gauss’s law \(\oint \mathbf{B} \cdot d \mathbf{S}=0\) or divergence of \(B=\nabla \cdot \mathbf{B}=0\)

Hint

(d) do not intersect at all.
Explanation: Magnetic lines of force due to a bar magnet never intersect because if they did, it would mean that at that point, the magnetic field has two directions at once, which is not possible.

Hint

(b) Let pole strength be \(m_1\) and \(m_2\) and distance between them is \(r\), then force between magnetic poles, \(F \propto \frac{m_1 m_2}{r^2}\)
In second case, \(F^{\prime} \propto \frac{\left(2 m_1\right)\left(2 m_2\right)}{(2 r)^2}, \propto \frac{m_1 m_2}{r^2}\)
\(
\therefore \quad F^{\prime}=F
\)

Hint

(b)

The situation is summarised in figure, we see that magnetic pole strength of each part becomes half, i.e. \(m / 2\).

Note: The bar magnet is cut into four parts as shown in the figure. The pole strength is affected when it is cut parallel to its axis. It is not affected when it is cut parallel to its equator.
\(\therefore\) Pole strength of each part will be \(\frac{m}{2}\).

Hint

(b) The magnet with the hole will have a smaller magnetic moment.

Explanation:
The magnetic moment is a measure of the strength of a magnetic dipole (a magnet). It is directly proportional to the pole strength and the effective length of the magnet. A hole at the center of the magnet reduces the effective length of the magnetic material, thus decreasing the magnetic moment.

Hint

(c) For longitudinal position, magnetic field will be \(B_1 \propto \frac{2 M}{d^3}\)
For transverse position, magnetic field will be \(B_2 \propto \frac{M}{d^3}\)
\(\therefore \quad \frac{B_1}{B_2}=2: 1\)

Hint

(d) Magnetic field lines are closed loops. Inside a magnet, the field lines are directed from the south pole to the north pole.

Hint

(b) A diamagnetic substance is repelled by both the north and south poles of a bar magnet.

Explanation:
Diamagnetic materials create an induced magnetic field that is opposite to the applied external magnetic field. This opposition causes a repulsion between the diamagnetic substance and both the north and south poles of the magnet.

Hint

(b) ferromagnetic substance.
Explanation:
Ferromagnetic substances:
These materials have a very high positive magnetic susceptibility, meaning they are strongly attracted to an external magnetic field and can retain their magnetization even after the field is removed.

Paramagnetic substances:
While they have a positive susceptibility, it is much smaller than that of ferromagnetic materials.

Non-magnetic substances:
These have a susceptibility of zero, meaning they are not affected by an external magnetic field.

Diamagnetic substances:
These have a negative susceptibility, meaning they are slightly repelled by an external magnetic field.

Hint

\(
\begin{array}{lr}
\text { (b) Torque, } \tau =M B \sin \theta \\
\therefore \quad \tau_1 =M B_1 \sin 90^{\circ}=M B_1
\end{array}
\)
\(
\begin{array}{ll}
\text { and } & \tau_2=M B_2 \sin 90^{\circ}=M B_2 \text { or } \frac{M B_1}{M B_2}=\frac{\tau_1}{\tau_2} \\
\therefore & \frac{B_1}{B_2}=\frac{\tau_1}{\tau_2}
\end{array}
\)

Hint

(c)
Explanation:
A deflection magnetometer uses a small magnetic needle because it needs to be in a uniform magnetic field. A small needle experiences a more uniform field around it, allowing for a more accurate measurement of the magnetic field strength. A long pointer is used for better visibility and to allow for more precise readings on the circular scale.

Why other options are incorrect:
(a) needle cannot be made long:

While a needle can be made longer, it would not be suitable for a deflection magnetometer as it would experience a non-uniform magnetic field, leading to inaccurate measurements.
(b) circular scale cannot be made short:

The circular scale can be made shorter, but the main reason for the long pointer is to improve readability.
(d) pointer must be in a non-uniform field:

The pointer is used to indicate the direction of the magnetic field on the circular scale. It is designed to be in the same field as the needle, and a non-uniform field would make accurate readings impossible.

Hint

(c) Magnetism of a magnet falls with rise of temperature and becomes practically zero above curie temperature as shown in Fig. (c).

Hint

(b) The tangent deflection at position \(A\) is given by:
\(
\tan A=\frac{\mu_0 \cdot 2 M \cdot D}{4 \pi\left(D^2-L^2\right)^2}
\)
And the tangent deflection at position \(B\) is given by:
\(
\tan B=\frac{\mu_0 \cdot M}{4 \pi\left(D^2+L^2\right)^{3 / 2}}
\)
Where:
\(\mu_0\) is the permeability of free space,
\(M\) is the magnetic moment,
\(D\) is the distance between the two poles,
\(L\) is the distance from one pole to the center of the magnet.
\(
\frac{\tan A}{\tan B}=\frac{\frac{\mu_0 \cdot 2 M \cdot D}{4 \pi\left(D^2-L^2\right)^2}}{\frac{\mu_0 \cdot M}{4 \pi\left(D^2+L^2\right)^{3 / 2}}}=\frac{2 D}{\left(D^2-L^2\right)^2} \cdot\left(D^2+L^2\right)^{3 / 2}
\)
Assuming \(L\) is much smaller than \(D\) (i.e., \(\frac{L}{D} \ll 1\) ), we can neglect higher order terms:
\(
\frac{\tan A}{\tan B} \approx \frac{2 D \cdot D^3}{D^4}=2
\)
Thus, the ratio of the tangents of deflection at positions \(A\) and \(B\) is:
\(
\frac{\tan A}{\tan B}=2: 1
\)

Hint

(d) For a paramagnetic substance, the relative permeability \(\left(\boldsymbol{\mu}_{\boldsymbol{r}}\right)\) and magnetic susceptibility \((\chi)\) have the following characteristics:
Relative Permeability \(\left(\mu_r\right)\) : Paramagnetic materials have a relative magnetic permeability slightly greater than 1. This means they enhance the magnetic field within them when placed in an external magnetic field.
Magnetic Susceptibility ( \(\boldsymbol{x}\) ): Paramagnetic materials have a small, positive magnetic susceptibility. This indicates that they are slightly attracted by magnetic fields.

The relationship between relative permeability ( \(\boldsymbol{\mu}_{\boldsymbol{r}}\) ) and magnetic susceptibility \((\chi)\) is given by the formula: \(\mu_r=1+\chi\).
Since \(\chi\) is positive for paramagnetic materials, \(\mu_r\) will be greater than 1.
Therefore, for a paramagnetic substance, the correct relation is \(\mu_r>1\) and \(\chi>0\).

Based on the options provided:
(a) \(\mu_r<1, \chi<0\) – Incorrect (This describes diamagnetic materials)
(b) \(\mu_r<1, x>0\) – Incorrect
(c) \(\mu_r>1, x<0\) – Incorrect
(d) \(\mu_r>1, \chi>0\) – Correct

Hint

\(
\begin{aligned}
&\text { (d) Given, } B=4 B_0\\
&\therefore \quad \frac{B}{B_0}=4 \Rightarrow \mu_r=4 \quad\left[\because \mu_r=\frac{B}{B_0}\right]
\end{aligned}
\)

Hint

(a) rise. When a paramagnetic liquid is placed in a U-tube and one limb is subjected to a magnetic field, the liquid in that limb will rise. This is because paramagnetic materials are attracted to magnetic fields, causing the liquid level to increase in the limb within the field.

Explanation:
Paramagnetic Materials: Paramagnetic substances are weakly attracted to magnetic fields.
U-tube Experiment: In a U-tube, when one limb is placed in a magnetic field, the paramagnetic liquid within that limb is drawn towards the magnetic field.
Liquid Level Change: This attraction causes the liquid level to rise in the limb within the magnetic field, while the level in the other limb will fall.

Hint

(a) The magnetic moment or magnetic dipole moment is a measure of the strength of a magnetic source. The magnetic moment is defined as pole strength multiplied by the distance between the poles,
\(
M=m \times 2 l
\)
Given, \(2 l=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}, m=4 A-m\)
\(
\therefore M=0.10 \times 4=0.4 A-m^2
\)
Note: Magnetic moment is a vector quantity pointing along the axis of the magnet from \(S\) to \(N\).

Hint

(d) All magnetic materials lose their magnetic properties when strongly heated.

Explanation: When a magnet is heated to a high enough temperature, called the Curie temperature, the magnetic domains within the material become disordered, causing it to lose its magnetic properties.

Why the other options are incorrect:
(a) dipped in water:

Water does not affect the magnetic properties of a magnet. While prolonged exposure to saltwater can cause corrosion, it does not directly impact magnetism.
(b) dipped in oil:

Similar to water, dipping a magnet in oil does not affect its magnetic properties.
(c) brought near a piece of iron:

When a magnet is brought near a piece of iron, the magnetic force between them increases, actually strengthening the magnet’s magnetism due to the attraction of the iron filings.

Hint

(b) When heated above its Curie temperature, a ferromagnetic material becomes paramagnetic, meaning its ferromagnetic domains become random.

Explanation:
Ferromagnetic domains:
In a ferromagnetic material below its Curie temperature, the magnetic domains are aligned in a coordinated manner, creating a strong overall magnetization.

Above Curie temperature:
When heated above the Curie temperature, the thermal energy disrupts the alignment of these domains, causing them to become randomly oriented. This transition results in the loss of ferromagnetic properties and the material becoming paramagnetic.

Why other options are incorrect:
(a) Ferromagnetic domains are perfectly arranged:

This is only true below the Curie temperature. Above the Curie temperature, the domains become random.
(c) Ferromagnetic domains are not influenced:

This is incorrect. The temperature definitely influences the ferromagnetic domains, causing them to become disordered above the Curie temperature.
(d) Ferromagnetic material changes into diamagnetic material:

While a ferromagnetic material loses its ferromagnetic properties above the Curie temperature, it does not become diamagnetic. Diamagnetic materials have a very weak negative magnetic susceptibility, whereas a paramagnetic material (which a ferromagnetic material becomes above the Curie temperature) has a positive susceptibility.

Hint

(b) Above the Curie temperature, ferromagnetic substance become paramagnetic in nature. So, its susceptibility varies inversely as the absolute temperature.

Explanation:
Curie Temperature:
This is the temperature above which a ferromagnetic material loses its ferromagnetic properties and becomes paramagnetic.

Magnetic Susceptibility:
This is a measure of how easily a material can be magnetized in the presence of an external magnetic field.

Curie’s Law:
Above the Curie temperature, ferromagnetic materials follow Curie’s law, which states that the magnetic susceptibility ( \(\chi\) ) is inversely proportional to the absolute temperature ( \(\chi\) ), expressed as \(\chi \propto 1 / T\).

Paramagnetic Behavior:
Above the Curie temperature, the material’s magnetic domains become disordered, and the material exhibits paramagnetic behavior, where magnetization is directly proportional to the applied magnetic field and inversely proportional to temperature.

Hint

(b) In the figure depicts a material that is weakly repelled by a magnetic field, it is most likely diamagnetic. Diamagnetic materials are characterized by their negative magnetic susceptibility and are repelled by magnetic fields.

Note: If the figure shows a material that is strongly attracted to a magnetic field, then it could be ferromagnetic or ferrimagnetic. If the figure shows a material that is weakly attracted to a magnetic field, then it could be paramagnetic.

Hint

(b) If magnet is cut along the axis of magnet of length \(2 l\), then new pole strength becomes half, i.e. \(m^{\prime}=\frac{m}{2}\) and new length remains same, i.e. \(2 l^{\prime}=2 l\).
\(\therefore\) New magnetic moment,
\(
M^{\prime}=\frac{m}{2} \times 2 l=\frac{m 2 l}{2}=\frac{M}{2}
\)

Hint

(c) Net magnetic moment of the system is given by

\(
M=\sqrt{M^2+M^2}=\sqrt{2} M=\sqrt{2} \mathrm{ml}
\)

Hint

\(
\begin{aligned}
&\text { (b) Magnetic moment of the bar magnet is given by }\\
&M=\frac{\tau}{B \sin \theta}\\
&(\because \tau=M B \sin \theta)
\end{aligned}
\)
\(
=\frac{0.032}{0.16 \times \sin 30^{\circ}}=\frac{0.032 \times 2}{0.16}=0.4 \mathrm{~J} \mathrm{~T}^{-1}
\)

Hint

\(
\begin{aligned}
\text { (a) As, } \tau & =M B \sin \theta \\
\tau & =m(2 L) B \sin \theta \\
\therefore \quad m & =\frac{\tau}{(2 l) B \sin \theta}=\frac{25 \times 10^{-6}}{(0.05)\left(5 \times 10^{-2}\right) \sin 30^{\circ}} \\
& =2 \times 10^{-2} \mathrm{~A}-\mathrm{m}
\end{aligned}
\)

Hint

(d) Force experienced by each pole of bar magnet is given by
\(
F=m B
\)
where, \(m\) is pole strength of bar magnet.
\(
\begin{array}{ll}
\text { But } & M=m L \\
\therefore & F=\frac{M}{L} \times B \Rightarrow 6 \times 10^{-4}=\frac{3}{L} \times 2 \times 10^{-5} \\
\therefore & L=0.1 \mathrm{~m}
\end{array}
\)

Hint

(d) Work done in rotating the magnet from angle \(\theta_1\) to angle \(\theta_2\) is given by
\(
W=M B\left(\cos \theta_1-\cos \theta_2\right)
\)
Here, \(\theta_1=0^{\circ}\) and \(\theta_2=180^{\circ}\)
\(
W=M B\left(\cos 0^{\circ}-\cos 180^{\circ}\right)=2 M B
\)

Hint

(b) Work done in rotating the magnet in uniform magnetic field is given by
\(
W=M B\left(\cos \theta_1-\cos \theta_2\right)
\)
Here,
\(
\theta_1=0^{\circ}, \theta_2=90^{\circ}
\)
\(
\begin{array}{r}
\therefore \quad W=2 \times 0.1\left(\cos 0^{\circ}-\cos 90^{\circ}\right) \\
=2 \times 0.1(1-0)=0.2 \mathrm{~J}
\end{array}
\)

Hint

(c) Here, \(\mathbf{B}=0.5 \mathrm{~T} ; I=20 \mathrm{~A} ; N=15 ; \mathbf{A}=-0.04 \hat{\mathbf{i}} \mathrm{~m}^2\)
Magnetic moment of the coil,
\(
\begin{aligned}
\mathbf{M} & =\text { NIA } \\
& =(15)(20)(-0.04 \hat{\mathbf{i}})=-12 \hat{\mathbf{i}} \mathrm{Am}^2
\end{aligned}
\)
The potential energy of the coil in the given orientation,
\(
U=-\mathbf{M} \cdot \mathbf{B}=-(-12 \hat{\mathbf{i}}) \cdot(0.5 \hat{\mathbf{i}})=+6 \mathrm{~J}
\)

Hint

\(
\begin{aligned}
& \text { (d) } \therefore \frac{B}{0.2}=\frac{M /(5)^3}{2 M /(10)^3}=\frac{(10 / 5)^3}{2}=4 \mathrm{G} \\
& \Rightarrow \quad B=4 \times 0.2=0.8 \text { oersted }
\end{aligned}
\)

Note: The magnetic field along the axis of a short bar magnet is given by:
\(
B_{\mathrm{axis}}=\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r^3}
\)
The magnetic field at a point on the perpendicular bisector (equatorial line) is given by:
\(
B_{\text {perpendicular }}=\frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3}
\)
Set Up the Ratio of Fields:
We can set up the ratio of the magnetic fields at the two points:
\(
\frac{B_2}{B_1}=\frac{M / r_2^3}{2 M / r_1^3}
\)

Hint

(b) \(T \propto \sqrt{I}\)
By increasing mass to four times, the moment of inertia will also increase four times. Then \(T^{\prime}=\sqrt{4 I}=2 \sqrt{I}=2 T\)

Hint

(d) For a tangent galvanometer, if \(I\) ampere current flows through coil, then this current is proportional to tangent of angle of deflection (of the needle), i.e.
\(
\begin{array}{ll}
& I \propto \tan \theta \\
\therefore & \frac{I_1}{I_2}=\frac{\tan \theta_1}{\tan \theta_2} \Rightarrow \frac{2}{I_2}=\frac{\tan 30^{\circ}}{\tan 60^{\circ}} \\
\therefore & I_2=6 \mathrm{~A}
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (d) In series, current is same. }\\
&\Rightarrow \quad \frac{M_1}{M_2}=\frac{\tan \theta_1}{\tan \theta_2} \text { or } \frac{N_1}{N_2}=\frac{\tan 60^{\circ}}{\tan 45^{\circ}}=\sqrt{3}
\end{aligned}
\)

Hint

(a) Magnetic field due to a bar magnet at a distance \(r\) from the centre of magnet on axial position is
\(
\begin{aligned}
B & =\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r^3} \\
\frac{B_1}{B_2} & =\left(\frac{r_2}{r_1}\right)^3=\left(\frac{48}{24}\right)^3=\frac{8}{1}=8
\end{aligned}
\)

Hint

(b) Intensity of magnetisation, \(I=\frac{M}{V}=-\frac{M}{\text { mass } / \text { density }}\)
Given, mass \(=1 \mathrm{~g}=10^{-3} \mathrm{~kg}\)
and density \(=5 \mathrm{~g} / \mathrm{cm}^3=\frac{5 \times 10^{-3} \mathrm{~kg}}{\left(10^{-2}\right)^3}=5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\)
Hence, \(\quad I=\frac{6 \times 10^{-7} \times 5 \times 10^3}{10^{-3}}=3 \mathrm{~A} / \mathrm{m}\)

Hint

\(
\text { (a) } \because H=\frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3} \Rightarrow 4 \times 10^{-5}=10^{-7} \times \frac{M}{(20)^3 \times 10^{-6}}
\)
\(
\begin{aligned}
M & =\frac{(20)^3 \times 4 \times 10^{-5} \times 10^{-6}}{10^{-7}} \\
M & =3.2 \mathrm{~A}-\mathrm{m}^2
\end{aligned}
\)

Hint

(c) The original length is \(2 L\).
The new length is half of the original length: \(\frac{2 L}{2}=L\).
The original magnetic moment is \(M=m \cdot 2 L\).
The new magnetic moment is \(\boldsymbol{M}^{\prime}=\boldsymbol{m} \cdot \boldsymbol{L}\).
\(
M^{\prime}=\frac{M}{2}
\)
The pole strength remains the same when a magnet is broken. The new pole strength is \(m\).
The magnetic moment of each piece is \(\frac{M}{2}\) and the pole strength is \(m\).

Hint

(c) In CGS,
\(
\begin{aligned}
B_{\text {axial }} & =9=\frac{\mu_0}{4 \pi} \frac{2 M}{x^3} \dots(i) \\
B_{\text {equatorial }} & =\frac{\mu_0}{4 \pi} \frac{M}{\left(\frac{x}{2}\right)^2}=\frac{\mu_0}{4 \pi} \frac{8 M}{x^3} \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\text { From Eqs. (i) and (ii), we get }\\
&B_{\text {equatorial }}=4 B_{\text {axial }}=36 \mathrm{G}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (c) As, } & T \propto \frac{1}{\sqrt{H}} \text { and } H=B \cos \theta \\
\Rightarrow & \frac{T_1}{T_2}=\sqrt{\frac{B_2 \cos \theta_2}{B_1 \cos \theta_1}} \\
\Rightarrow & \frac{2}{3}=\sqrt{\frac{B_2}{B_1} \cdot \frac{\cos 60^{\circ}}{\cos 30^{\circ}}}=\frac{B_2}{\sqrt{3} B_1} \\
\Rightarrow & \frac{4}{9}=\frac{B_2}{\sqrt{3} B_1} \Rightarrow \frac{B_1}{B_2}=\frac{9}{4 \sqrt{3}}
\end{array}
\)

Hint

\(
\text { (b) } \quad T \propto \frac{1}{\sqrt{M}}
\)
\(
\begin{aligned}
M_1 & =M+2 M=3 M \text { and } M_2=2 M-M=M \\
T_2 & =\sqrt{3} T_1=3 \sqrt{3} \mathrm{~s}
\end{aligned}
\)

Hint

(b) As, we know \(T=2 \pi \sqrt{\frac{I}{M B}}\)
When magnet is broken into two equal parts, \(M^{\prime}=\frac{M}{2}[latex] and [latex]l^{\prime}=\frac{l}{2}\)
\(
\begin{array}{ll}
\therefore & \frac{T^{\prime}}{T}=\sqrt{\frac{I^{\prime}}{M^{\prime} B}} \times \sqrt{\frac{M B}{I}}=\sqrt{\frac{I}{8} \times \frac{2}{M}}=\sqrt{\frac{M}{I}}=\frac{1}{2} \\
\Rightarrow & T^{\prime}=\frac{T}{2}=\frac{4}{2}=2 \mathrm{~s}
\end{array}
\)

Hint

(d) \(\mu_r=1+\frac{I}{H} ;\) as we know \(I\) is dependent on \(H\), initially value of \(\frac{I}{H}\) is smaller so value of \(\mu_r\) increases with \(H\) but slowly but with further increases of \(H\) value of \(\frac{I}{H}\) also increases i.e, \(\mu_r\) increases rapidly. When material is fully magnetised \(I\) becomes constant, then with the increase of \(H\) ( \(\frac{I}{H}\) decreases) \(\mu_r\) decreases.

Hint

\(
\begin{aligned}
&\text { (a) Time period of vibrating magnet is given by }\\
&\begin{aligned}
& & T & =2 \pi \sqrt{\left(\frac{I}{M H}\right)} \\
& \therefore & \frac{T_1}{T_2} & =\sqrt{\left(\frac{M_2}{M_1}\right)} \\
& \therefore & \frac{M_1}{M_2} & =\frac{T_2^2}{T_1^2}=\frac{(60 / 15)^2}{(60 / 10)^2}=\frac{4}{9}
\end{aligned}
\end{aligned}
\)

Hint

(b) Suppose magnetic field is zero at point \(P\) which lies at a distance \(x\) from 10 unit pole.

Hence, at \(P\)
\(
\frac{\mu_0}{4 \pi} \cdot \frac{10}{x^2}=\frac{\mu_0}{4 \pi} \cdot \frac{40}{(30-x)^2} \Rightarrow x=10 \mathrm{~cm}
\)
So, from stronger pole distance is 20 cm.

Hint

\(
\begin{aligned}
&\text { (b) Time period, } T=2 \pi \sqrt{\frac{I}{M B}}\\
&\begin{aligned}
& \therefore & T & \propto \frac{1}{\sqrt{B}} \\
& \therefore & \frac{T_1}{T_2} & =\sqrt{\frac{B_2}{B_1}} \\
& \Rightarrow & \frac{60 / 40}{2.5} & =\sqrt{\frac{B_2}{0.1 \times 10^{-5}}} \\
& \therefore & B_2 & =0.36 \times 10^{-6} \mathrm{~T}
\end{aligned}
\end{aligned}
\)

Hint

(a) The needle is deflected through \(45^{\circ}\) means magnetic field at the centre of circular coil is equal (in magnitude) to the horizontal component of earth’s magnetic field.
\(
\begin{aligned}
\frac{\mu_0 N I}{2 r} & =H \\
I & =\frac{2 r H}{\mu_0 N}=\frac{2 \times 0.2 \times 0.37 \times 10^{-4}}{\left(4 \pi \times 10^{-7}\right)(20)} \approx 0.6 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\text { (b) As, } \mu=\mu_0(1+\chi) \text { or } \mu=4 \pi \times 10^{-7}(1+599)
\)
\(
\begin{aligned}
\mu & =7.536 \times 10^{-4} \mathrm{TmA}^{-1} \\
B & =\mu H=\left(7.536 \times 10^{-4} \times 1200\right) \mathrm{T} \\
\phi & =B A=\left(7.536 \times 10^{-4} \times 1200 \times 0.2 \times 10^{-4}\right) \mathrm{Wb} \\
& =1.81 \times 10^{-5} \mathrm{~Wb}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) As potential energy is given as }\\
&U=-M B(1-\cos \theta) \Rightarrow U=-M B\left[\text { when } \theta=0^{\circ}, U=\text { minimum potential energy.}\right]
\end{aligned}
\)
Here, \(P Q_6\) configuration provides the condition of lowest PE.

Hint

(a) Point \(P\) lies on equatorial line of magnet (1) and axial line of magnet (2) as shown

\(
\begin{aligned}
B_1 & =\frac{\mu_0}{4 \pi} \cdot \frac{M}{d^3} \\
& =10^{-7} \times \frac{1000}{(0.1)^3}=0.1 \mathrm{~T} \\
B_2 & =\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{d^3} \\
& =10^{-7} \times \frac{2 \times 1000}{(0.1)^3}=0.2 \mathrm{~T} \\
B_{\text {net }} & =B_2-B_1=0.1 \mathrm{~T}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (d) } \because & \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{12}{4}=\frac{3}{1} \\
\therefore & \frac{T_1}{T_2}=\frac{1}{3}
\end{array}
\)
Further, \(\quad T \propto \frac{1}{\sqrt{M}}\)
\(
\therefore \quad \sqrt{\frac{M_1-M_2}{M_1+M_2}}=\frac{1}{3}
\)
This equation gives, \(\frac{M_1}{M_2}=\frac{5}{4}\)

Hint

\(
\text { (d) At point } P \text {, net magnetic field, } B_{\text {net }}=\sqrt{B_1^2+B_2^2}
\)

\(
\begin{array}{ll}
\text { where, } & B_1=\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{d^3} \\
\text { and } & B_2=\frac{\mu_0}{4 \pi} \cdot \frac{M}{d^3} \\
\Rightarrow & B_{\text {net }}=\frac{\mu_0}{4 \pi} \cdot \frac{\sqrt{5} M}{d^3}
\end{array}
\)

Hint

(d) Frequency, \(v=\frac{1}{2 \pi} \sqrt{\frac{M H}{I}}\)
\(
\therefore \quad v^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{M(B+H)}{I}}
\)
where, \(B=\) magnetic field due to downward conductor.

\(
\begin{array}{ll}
\text { Then, } & B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{a} \\
\Rightarrow & B=10^{-7} \times \frac{2 \times 15}{20 \times 10^{-2}} \\
\text { So, } & B=15 \mu \mathrm{~T}
\end{array}
\)
\(
\begin{aligned}
\frac{v^{\prime}}{v} & =\sqrt{\frac{B+B H}{B}} \\
\frac{v^{\prime}}{10} & =\sqrt{\frac{15+12}{12}}=1.5 \\
v^{\prime} & =15 \mathrm{~Hz}
\end{aligned}
\)

Hint

(a) Using the formula for time period for magnetic system,
\(
T=2 \pi \sqrt{\left(\frac{I}{M H}\right)} \Rightarrow T \propto \frac{1}{\sqrt{M}} \dots(i)
\)
When similar poles placed at same side, then
\(
M_1=M+2 M=3 M
\)
So, from Eq. (i), \(\quad T_1 \propto \frac{1}{\sqrt{3 M}} \dots(ii)\)
When the polarity of a magnet is reversed, then
\(
M_2=2 M-M=M
\)
So, from Eq . (i)
\(
T_2 \propto \frac{1}{\sqrt{M}} \dots(iii)
\)
Now, dividing Eq. (ii) by Eq. (iii), we get
\(
\frac{T_1}{T_2}=\frac{\sqrt{M}}{\sqrt{3 M}}=\frac{1}{\sqrt{3}} \Rightarrow T_2=\sqrt{3} T_1
\)
Hence,
\(
T_1<T_2
\)

Hint

(b) Initially, the time period of the magnet,
\(
T=2 \pi \sqrt{\left(\frac{I}{M B}\right)}
\)
For each part of magnet, its moment of inertia is \(\frac{I}{27}\) and its magnetic moment \(=M / 3[latex].
When magnet is cut along its length into three equal parts and placed on each other, then new moment of inertia of system,
[latex]
I^{\prime}=\frac{I}{27} \times 3=\frac{I}{9}
\)
and magnetic moment of system, \(M^{\prime}=\frac{M}{3} \times 3=M\)
\(\therefore\) Time period of system,
\(
T^{\prime}=2 \pi \sqrt{\frac{I^{\prime}}{M^{\prime} B}}=\frac{1}{3} \times 2 \pi \sqrt{\frac{I}{M B}}=\frac{T}{3}=\frac{2}{3} \mathrm{~s}
\)

Hint

(a) When paramagnetic substances are kept in an external magnetic field, then it develops feeble magnetisation in the direction of external applied magnetic field, i.e. magnetic dipoles are aligned along external magnetic field weakly.
So, they gets poorly attracted in magnetic field.

Hint

(c) Coercivity and retentivity of soft iron is low, because its magnetisation disappears easily on the removal of external magnetising field.

Hint

(c) We know that, the magnetism of the magnet is achieved due to the spin motion of electrons. Also, the spinning electrons possess magnetic dipole moment. This dipole moment is much greater than that due to orbital moment of electrons around the nucleus.

Hint

(b) The magnetic susceptibility of a ferromagnetic substance decreases (in hyperbolic manner) with increase in temperature and above Curie’s temperature, the substance behaves like paramagnetic one.

Hint

(a) Magnetic permeability of substance,
\(
\mu=\frac{B}{H} \Rightarrow \mu \propto \frac{1}{H}
\)
where, \(H\) is the magnetising or magnetic field strength. As, the magnetising field of a ferromagnetic material is increasing thus its, permeability decreases.

Hint

(c) As, we know that, the magnetic field at a point due to the dipole is given by \(B=\frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3} \sqrt{1+3 \cos ^2 \theta}\) where, \(M\) is magnetic dipole moment.
\(
\Rightarrow \quad B \propto \frac{1}{r^3}
\)

Hint

(b) The coercivity of bar magnet \(=4 \times 10^3 \mathrm{Am}^{-1}\),
Length of solenoid \(=12 \mathrm{~cm}=12 \times 10^{-2} \mathrm{~m}\)
and number of turns \(=60\)
The coercivity of 12 cm length of solenoid
\(
=4 \times 10^3 \times 12 \times 10^{-2}=480 \mathrm{Am}^{-1}
\)
Now, current through each turn to demagnetise \(=\frac{480}{60}=8 \mathrm{~A}\)

Hint

(c) The effective length of magnet, \(l=31.4 \mathrm{~cm}=31.4 \times 10^{-2} \mathrm{~m}\)
Pole strength, \(m=0.8 \mathrm{Am}\)
Length of semi-circle \(=\pi \frac{D}{2}=l\)
where, \(D=\) diameter of circle
\(
\begin{aligned}
&\begin{array}{ll}
\Rightarrow & D=\frac{2 l}{\pi}=\frac{2 \times 31.4 \times 10^{-2}}{3.14} \\
\Rightarrow & D=20 \times 10^{-2} \mathrm{~m}
\end{array}\\
&\text { Now, the magnetic moment }\\
&\begin{aligned}
& =m D=0.8 \times 20 \times 10^{-2} \\
& =16 \times 10^{-2}=0.16 \mathrm{~A}-\mathrm{m}^2
\end{aligned}
\end{aligned}
\)

Hint

(c) Given, \(B_H=3 \times 10^{-5} \mathrm{~T}, \theta=45^{\circ}, n=50\)
\(
r=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}
\)
As, current
\(
\begin{aligned}
I & =\frac{2 r B_H}{\mu_0 n} \tan \theta \\
& =\frac{2 \times 20 \times 10^{-2} \times 3 \times 10^{-5} \times \tan 45^{\circ}}{4 \pi \times 10^{-7} \times 50} \\
& =\frac{120 \times 10^{-7}}{4 \pi \times 10^{-7} \times 50}=\frac{3}{5 \pi}=\frac{3}{5 \times 3.14}=\frac{3}{15.7}=0.19 A
\end{aligned}
\)

Hint

(b) The intensity of magnetisation increases for a ferromagnetic substance with the applied magnetic field and become constant, when the domains get perfectly aligned with respect to the external magnetic fields.

Hint

\(
\begin{aligned}
&\text { (c) The work done (in rotating) by external agent }\\
&\begin{aligned}
& =\text { change in potential energy } \\
& =-M B \cos \theta_2-\left(-M B \cos \theta_1\right) \\
& =-M B\left(\cos 45^{\circ}-\cos 0^{\circ}\right) \\
& =\left(1-\frac{1}{\sqrt{2}}\right) M B=\frac{0.41}{\sqrt{2}} \times \frac{1}{10} \approx 0.03 \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

(b) The core of electromagnets are made of ferromagnetic materials which have high permeability and low retentivity.

Explanation:
High permeability:
This means the material easily allows magnetic fields to pass through it, which is crucial for enhancing the strength of the magnetic field produced by the electromagnet.

Low retentivity:
This means the material loses its magnetization quickly once the external magnetic field is removed, which is desirable for electromagnets as they need to be able to quickly demagnetize when the current is switched off.

Why other options are incorrect:
(a) low permeability and high retentivity:

A low permeability would weaken the magnetic field, and high retentivity would mean the electromagnet would retain its magnetization even after the current is stopped, making it difficult to control the magnetic field.
(c) low permeability and low retentivity:

Low permeability would significantly decrease the magnetic field strength, and while low retentivity is desirable, a very low permeability would defeat the purpose of using a ferromagnetic material in the electromagnet’s core.
(d) high permeability and high retentivity:

High permeability is desirable, but high retentivity would make the electromagnet difficult to demagnetize, leading to slow switching and undesirable residual magnetism.

Hint

(c) Ferromagnetic.
Explanation:
Ferromagnetic materials
have a very high, positive magnetic susceptibility. A susceptibility value of 400 is indicative of this class of material.

Diamagnetic materials
have a small, negative susceptibility.
Paramagnetic materials
have a small, positive susceptibility, but much smaller than ferromagnetic materials.

Ferroelectric materials
refer to the electrical properties of a material, not its magnetic properties.

Hint

\(
\begin{aligned}
&\text { (d) For paramagnetic sample (Curie’s law), } I \propto B / T\\
&\begin{aligned}
& \text { where, } \quad I_1=0.8 \mathrm{~A} / \mathrm{m}, B_1=0.8 \mathrm{~T} \text {, } \\
& T_1=5 \mathrm{~K}, B_2=0.4 \mathrm{~T} \\
& \text { and } \quad T_2=20 \mathrm{~K} \\
& \Rightarrow \quad \frac{I_1}{I_2}=\frac{B_1 / T_1}{B_2 / T_2} \\
& \Rightarrow \quad \frac{I_1}{I_2}=\frac{B_1 \times T_2}{B_2 \times T_1} \\
& \Rightarrow \quad \frac{0.8}{I_2}=\frac{0.8 \times 20}{0.4 \times 5} \\
& \Rightarrow \quad I_2=\frac{0.4 \times 5}{20}=0.1 \mathrm{Am}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(a) When temperature increases beyond Curie temperature, the disorderness of the magnetic moments of the material increases, hence alignment of magnetic moments are disturbed, which caused material to behave as paramagnetic.

Hint

(a) We know that, intensity of magnetisation, \(I=\frac{M}{V}\) where, \(M=\) magnetic moment and \(V=\) volume.
So,
\(
\begin{aligned}
M & =I V=5 \times 10^4 \times \frac{12}{100} \times \frac{1}{(100)^2} \\
& =60 \times 10^4 \times 10^{-6} \\
& =0.6 \mathrm{~A}-\mathrm{m}^2 \text { (SI unit) }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a) Given, } \chi_m=299 \\
& \text { and } \quad \mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1} \\
& \text { We know that, } \mu=\mu_0\left(1+\chi_m\right) \\
& \therefore \quad \mu=4 \pi \times 10^{-7}(1+299)
\end{aligned}
\)
\(
\begin{aligned}
& =4 \times \frac{22}{7} \times 10^{-7} \times 300 \\
& =\frac{26400}{7} \times 10^{-7}=3771.4 \times 10^{-7} \mathrm{Hm}^{-1} \\
\mu & \cong 3771 \times 10^{-7} \mathrm{Hm}^{-1}
\end{aligned}
\)

Hint

\(
\text { (d) Given, } r=0.05 \mathrm{~nm}=0.05 \times 10^{-9} \mathrm{~m} \quad\left[\because 1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right]
\)
\(
\begin{aligned}
& n=10^{16} \text { revolutions/s } \\
& e=1.6 \times 10^{-19} \mathrm{C}
\end{aligned}
\)
\(
\begin{aligned}
&\text { We know that the magnetic moment, } M=A I\\
&\begin{aligned}
M & =\pi r^2 \times n e \\
& =3.14 \times\left(0.05 \times 10^{-9}\right)^2 \times 10^{16} \times 1.6 \times 10^{-19} \\
\text { or } \quad M & =1.26 \times 10^{-23} \mathrm{~A}-\mathrm{m}^2
\end{aligned}
\end{aligned}
\)

Hint

(a) When magnet is cut into equal parts, area of cross-section becomes half. Hence, pole strength becomes half.

Hint

(a) We know that, the magnetic field induction at a point,
\(
B=\frac{\mu_0}{4 \pi} \frac{M}{r^3}
\)
According to question, \(B_1: B_2: B_3=\frac{1}{(0.3)^3}: \frac{1}{(0.6)^3}: \frac{1}{(0.9)^3}\)
\(
=27: 3.37: 1
\)

Hint

(a) Given, \(M=120 \mathrm{Am}^2, T=12 \mathrm{~s}\) and \(B=0.4 \times 10^{-4} \mathrm{~T}\)
We know that, \(T=2 \pi \sqrt{\frac{I}{M B}}\) (squaring both sides)
So, \(\quad I=\frac{T^2 M B}{4 \pi^2}\)
or \(\quad I=\frac{(12)^2 \times(120) \times\left(0.4 \times 10^{-4}\right)}{4 \times 3.14 \times 3.14}\)
\(I=175.2 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2\)
In this question, the option (a) is approximately equivalent to \(175.2 \times 10^{-4}\), so the answer is \(I=172.8 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2\).

Hint

(a) When heated above the Curie temperature, a ferromagnetic substance becomes paramagnetic.

Explanation:
Ferromagnetic:
Below the Curie temperature, the magnetic domains in a ferromagnetic material are aligned, causing strong magnetization.

Paramagnetic:
Above the Curie temperature, these domains become disordered due to increased thermal energy, resulting in weaker magnetization and paramagnetic behavior.

Hint

(c) From the relation, susceptibility of the material,
\(
\chi_m=\frac{I}{H} \Rightarrow \chi_m \propto I
\)
Thus, it is obvious that greater the value of susceptibility of a material greater will be the value of intensity of magnetisation, i.e. more easily it can be magnetised.

Hint

\(
\begin{aligned}
&\text { (b) The magnetic field at a point on the axis of a bar magnet, }\\
&B=\frac{\mu_0}{4 \pi} \frac{2 M}{r^3}=10^{-7} \times \frac{2 \times 1.25}{(0.5)^3}=2 \times 10^{-6} \mathrm{NA}^{-1} \mathrm{~m}^{-1}
\end{aligned}
\)

Hint

(d) Rod is moving towards east, so induced emf across its end will be
\(
\begin{aligned}
B & =B_V v l=\left(B_H \tan \theta\right) v l \quad \quad\left[\therefore \tan \theta=B_V / B_H\right] \\
& =3 \times 10^{-4} \times \frac{4}{3} \times 10 \times 10^{-2} \times 0.25 \\
& =10^{-5} \mathrm{~V}=10 \mu \mathrm{~V}
\end{aligned}
\)

Hint

(c) When wire is bent in the form of semicircular arc, then
\(
l=\pi r
\)
\(\therefore\) The radius of semicircular arc, \(r=l / \pi\)
Distance between two end points of semicircular wire
\(
=2 r=\frac{2 l}{\pi}
\)
\(\therefore\) Magnetic moment of semicircular wire
\(
=m \times 2 r=m \times \frac{2 l}{\pi}=\frac{2}{\pi} m l
\)
But \(m l\) is the magnetic moment of wire, i.e. \(m l=M\)
\(
\therefore \quad \text { New magnetic moment }=\frac{2}{\pi} M
\)

Hint

(b) We know that, \(B=\mu_0 H+\mu_0 I\)
\(
\begin{aligned}
\therefore \quad & I=\frac{B-\mu_0 H}{\mu_0}=\frac{\mu H-\mu_0 H}{\mu_0}=\left(\frac{\mu}{\mu_0}-1\right) H \\
& I=\left(\mu_r-1\right) H
\end{aligned}
\)
For a solenoid having \(n\) turns/length and current \(I^{\prime}\),
\(
\begin{aligned}
H & =n I^{\prime} \\
I & =\left(\mu_r-1\right) n I^{\prime}=(1000-1) 500 \times 0.5 \\
& =2.5 \times 10^5 \mathrm{Am}^{-1}
\end{aligned}
\)
\(\therefore\) Magnetic moment, \(M=I V\)
\(
=2.5 \times 10^5 \times 10^{-4}=25 \mathrm{~A}-\mathrm{m}^2
\)

Hint

\(
\begin{aligned}
&\text { (b) Current in tangent galvanometer, }\\
&I=\frac{2 r H}{\mu_0 N} \tan \theta \dots(i)
\end{aligned}
\)

Here, \(R_1\) and \(R_2\) are in parallel
\(
\begin{aligned}
\therefore \quad & \frac{1}{R_{\text {net }}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_2+R_1}{R_1 R_2}=\frac{8+8}{8 \times 8} \\
& R_{\text {net }}=4 \Omega
\end{aligned}
\)
Hence, \(I=\frac{V}{R}=\frac{4}{4}=1 \mathrm{~A}\)
From Eq. (i), we get
\(
\begin{array}{ll}
& \frac{r \tan \theta}{N}=\frac{\mu_0 I}{2 H} \\
\therefore & \frac{r_A \tan \theta_A}{N_A}=\frac{r_B \tan \theta_B}{N_B} \quad\left(\because \frac{\mu_0 I}{2 H}=\text { constant }\right) \\
\Rightarrow & \frac{8 \times 1}{\sqrt{3} \times 2}=\frac{16 \times \sqrt{3}}{N_B}\left(\because \theta_A=30^{\circ} \text { and } \theta_B=60^{\circ}\right) \\
\therefore & N_B=12 \text { turns }
\end{array}
\)

Hint

(d) Paramagnetic materials will be feebly attracted, diamagnetic material will be feebly repelled and ferromagnetic material will be strongly attracted towards a magnet.

Hint

(b) If length of wire is \(2 l\), then magnetic moment,
\(
M=m \times 2 l=4 \pi \mathrm{~A}-\mathrm{m}^2 \text { (given) }
\)
As wire is bent in the form of semicircle, effective distance between the ends is \(2 r\).
So, new dipole moment, \(M^{\prime}=m \times 2 r\)
\(
\begin{aligned}
\text { As, } & & \pi r & =2 l \\
\Rightarrow & & r & =2 l / \pi
\end{aligned}
\)
\(
\begin{aligned}
M^{\prime} & =m \times 2 \times \frac{2 l}{\pi}=\frac{2}{\pi}(m \times 2 l) \\
& =\frac{2}{\pi} M=\frac{2}{\pi} 4 \pi=8 \mathrm{~A}-\mathrm{m}^2
\end{aligned}
\)

Hint

(b) Consider a point \(P\) located on the axial line of a short bar magnet of magnetic length \(2 l\) and strength \(m\). Let us find \(\mathbf{B}\) at a point \(P\) which is at a distance \(z\) from the centre of magnet. Magnetic field at point \(P\) due to \(N\)-pole,

\(
\begin{aligned}
&B_1=\frac{\mu_0}{4 \pi}\left[\frac{m}{(z-l)^2}\right] \text { along } N P\\
&\text { Similarly, magnetic field at point } P \text { due to } S \text {-pole, }\\
&B_2=\frac{\mu_0}{4 \pi}\left[\frac{m}{(z+l)^2}\right] \text { along } P S
\end{aligned}
\)
\(
\begin{aligned}
&\text { Net magnetic field at } P \text {, }\\
&\begin{aligned}
B & =B_1-B_2=\frac{\mu_0}{4 \pi}\left[\frac{m}{(z-l)^2}-\frac{m}{(z+l)^2}\right] \\
& =\frac{\mu_0}{4 \pi}\left[\frac{4 m l z}{\left(z^2-l^2\right)^2}\right] \\
& =\frac{\mu_0}{4 \pi}\left[\frac{m \times 2 l}{\left(z^2-l^2\right)^2} 2 z\right]
\end{aligned}
\end{aligned}
\)
\(
B=\frac{\mu_0}{4 \pi} \frac{2 M z}{\left(z^2-l^2\right)^2} \quad[\therefore M=m(2 l)]
\)
\(
B=\frac{\mu_0}{4 \pi} \cdot \frac{2 M z}{z^4} \quad(\because z \gg l)
\)
Hence, in vector form, magnetic field at point \(P\) due to \(N\)-pole can be written as
\(
\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{2 M}{z^3} \hat{\mathbf{M}}
\)
where, \(\hat{\mathbf{M}}=\) unit vector along magnetic dipole moment.

Hint

(d) The angle which the total magnetic field of earth makes with the surface of the earth is called inclination.