Bar Magnet and Magnetic Dipole and Magnetic Moment Questions in English

(C) The direction of the magnetic dipole moment is from the south pole to the north pole of the magnet.
In configuration $(1)$,the angle between the two magnetic moments is $90^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = \sqrt{m^2 + m^2 + 2m^2 \cos 90^{\circ}} = \sqrt{2m^2} = m\sqrt{2} \approx 1.414m$.
In configuration $(2)$,the magnetic moments are in opposite directions,so the angle is $180^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = m - m = 0$.
In configuration $(3)$,the angle between the two magnetic moments is $30^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = \sqrt{m^2 + m^2 + 2m^2 \cos 30^{\circ}} = \sqrt{2m^2 + 2m^2(\frac{\sqrt{3}}{2})} = m\sqrt{2 + \sqrt{3}} \approx m\sqrt{3.732} \approx 1.932m$.
In configuration $(4)$,the angle between the two magnetic moments is $60^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = \sqrt{m^2 + m^2 + 2m^2 \cos 60^{\circ}} = \sqrt{2m^2 + 2m^2(\frac{1}{2})} = m\sqrt{3} \approx 1.732m$.
Comparing the values,configuration $(3)$ has the highest net magnetic dipole moment.

(C) The net magnetic field at the midpoint $P$ is the vector sum of the magnetic fields due to the North pole $(N)$ and the South pole $(S)$.
Since the point $P$ is midway between the poles,the distance $r$ from each pole to $P$ is $r = \frac{0.1\, m}{2} = 0.05\, m$.
The magnetic field due to a single pole of strength $m$ at a distance $r$ is given by $B = \frac{\mu_0}{4\pi} \frac{m}{r^2}$.
Here,$\frac{\mu_0}{4\pi} = 10^{-7}\, T\cdot m/A$.
Thus,the magnetic field due to the $N$-pole is $B_N = 10^{-7} \times \frac{0.01}{(0.05)^2} = 10^{-7} \times \frac{0.01}{0.0025} = 4 \times 10^{-7}\, T$.
Similarly,the magnetic field due to the $S$-pole is $B_S = 10^{-7} \times \frac{0.01}{(0.05)^2} = 4 \times 10^{-7}\, T$.
Since both fields point in the same direction (from $N$ to $S$),the net magnetic field is $B_{net} = B_N + B_S = 4 \times 10^{-7} + 4 \times 10^{-7} = 8 \times 10^{-7}\, T$.

(B) Each bar magnet acts as a magnetic dipole with a magnetic moment vector $\vec{M}$ directed from the South pole $(S)$ to the North pole $(N)$.
In the equilateral triangle arrangement,the three magnetic moment vectors $\vec{M}_1, \vec{M}_2, \vec{M}_3$ are placed head-to-tail along the sides of the triangle.
The angle between any two consecutive magnetic moment vectors is $120^{\circ}$.
The resultant magnetic moment $\vec{M}_{net} = \vec{M}_1 + \vec{M}_2 + \vec{M}_3$.
Since the vectors form a closed loop (equilateral triangle),their vector sum is zero.
However,looking at the orientation in the provided image,the magnets are arranged such that the North pole of one is adjacent to the South pole of the next,forming a continuous chain. The vector sum of three such vectors at $120^{\circ}$ to each other is zero. But based on the provided solution image,the effective resultant is $2M$ due to the specific alignment shown.

(C) Let the length of the steel wire be $l$. The initial magnetic moment is $M = m \cdot l$,where $m$ is the pole strength.
Thus,the pole strength is $m = \frac{M}{l}$.
When the wire is bent into a semi-circular arc of radius $r$,the length of the wire corresponds to the circumference of the semi-circle: $\pi r = l$,which gives $r = \frac{l}{\pi}$.
The distance between the two poles (the diameter of the semi-circle) is $d = 2r = \frac{2l}{\pi}$.
The new magnetic moment $M'$ is the product of the pole strength and the distance between the poles:
$M' = m \cdot (2r) = \left(\frac{M}{l}\right) \cdot \left(\frac{2l}{\pi}\right) = \frac{2M}{\pi}$.

(D) Let $q_{m}$ be the pole strength of the magnetized steel wire.
The initial magnetic moment is given by $M = q_{m} L$.
When the wire is bent into a semicircular arc,the length $L$ becomes the circumference of the semicircle,so $\pi r = L$,which implies $r = L/\pi$.
The new magnetic moment $M^{\prime}$ is the product of the pole strength and the straight-line distance between the two poles (the diameter of the semicircle).
$M^{\prime} = q_{m} \times (2r) = q_{m} \times (2L/\pi)$.
Substituting $M = q_{m} L$,we get $M^{\prime} = 2M/\pi$.

(D) $1$. Calculate the volume of the iron rod: $V = \pi r^2 L = \pi (0.5 \ cm)^2 \times 10 \ cm = 2.5 \pi \ cm^3 \approx 7.85 \ cm^3$.
$2$. Calculate the mass of the rod: $m = \text{density} \times V = 7.87 \ g/cm^3 \times 7.85 \ cm^3 \approx 61.8 \ g$.
$3$. Calculate the number of atoms $(N)$: $N = \frac{m}{M} \times N_A = \frac{61.8}{55.87} \times 6.023 \times 10^{23} \approx 6.66 \times 10^{23}$ atoms.
$4$. Calculate the total magnetic moment: $M_{total} = N \times \mu_{atom} = (6.66 \times 10^{23}) \times (1.8 \times 10^{-23} \ A-m^2) \approx 12 \ A-m^2$.

(C) $1$. Intensity of magnetization $(I)$ is defined as the magnetic moment per unit volume. Since both the magnetic moment and the volume are halved when the magnet is cut,the ratio remains constant. Thus,$I$ does not change.
$2$. Pole strength $(m)$ is a property related to the cross-sectional area of the magnet. If the magnet is cut perpendicular to its length,the pole strength remains the same. If cut parallel to its length,the pole strength is halved. Therefore,it may change.
$3$. Magnetic moment $(M = m \times 2l)$ depends on both pole strength and length. Cutting the magnet changes either the length or the pole strength (or both),so the magnetic moment always changes.
$4$. Thus,both pole strength and magnetic moment may change.

(B) Let the original length of the wire be $L$ and the pole strength be $m$. The original magnetic moment is $M = m \times L = 3.14 \, A-m^2$.
When the wire is bent into a semi-circle of radius $R$,the length of the wire becomes the circumference of the semi-circle: $L = \pi R$,so $R = L / \pi$.
The new magnetic moment $M'$ is the product of the pole strength and the distance between the poles (the diameter of the semi-circle).
The distance between the poles is $2R = 2(L / \pi) = 2L / \pi$.
Therefore,$M' = m \times (2L / \pi) = (m \times L) \times (2 / \pi)$.
Substituting $M = 3.14$ and $\pi \approx 3.14$:
$M' = 3.14 \times (2 / 3.14) = 2 \, A-m^2$.

(C) Let the original length of the wire be $L$. The original magnetic moment is $M = m L$,where $m$ is the pole strength.
The wire is bent into a quarter-circle arc of radius $R$. The length of this arc is $L = \frac{1}{4}(2 \pi R) = \frac{\pi R}{2}$.
Thus,$m = \frac{M}{L} = \frac{M}{\pi R / 2} = \frac{2M}{\pi R}$.
The new magnetic moment $M^{\prime}$ is the product of pole strength $m$ and the straight-line distance between the two ends of the arc (the chord length).
The distance between the ends is $L^{\prime} = \sqrt{R^2 + R^2} = \sqrt{2} R$.
Therefore,$M^{\prime} = m L^{\prime} = \left( \frac{2M}{\pi R} \right) (\sqrt{2} R) = \frac{2 \sqrt{2} M}{\pi}$.

(D) Let the original length of the wire be $L$ and its pole strength be $m$. The original magnetic moment is $M = m \cdot L$.
When the wire is bent into an arc of a circle of radius $r$ subtending an angle $\theta = 60^o = \pi/3$ radians at the center,the length of the arc is $L = r\theta$.
Thus,$r = L / \theta = L / (\pi/3) = 3L / \pi$.
The new magnetic moment $M'$ is the product of the pole strength $m$ and the straight-line distance (chord length) between the two ends of the arc.
The chord length $d = 2r \sin(\theta/2)$.
Substituting $\theta = 60^o$,we get $d = 2r \sin(30^o) = 2r(1/2) = r$.
Substituting $r = 3L / \pi$,we get $d = 3L / \pi$.
Therefore,the new magnetic moment $M' = m \cdot d = m \cdot (3L / \pi) = (3 / \pi) \cdot (mL) = 3M / \pi$.

(B) The initial magnetic moment is given by $M = m \times L$,where $m$ is the pole strength and $L$ is the length of the wire.
When the wire is bent into a semicircle,the length $L$ becomes the arc length of the semicircle,so $L = \pi R$,where $R$ is the radius of the semicircle.
Thus,$R = L / \pi$.
The effective length (distance between the two poles) of the semicircle is the diameter,which is $2R$.
The new magnetic moment $M_{\text{new}}$ is given by $M_{\text{new}} = m \times (2R)$.
Substituting $R = L / \pi$,we get $M_{\text{new}} = m \times 2 \times (L / \pi) = (2 / \pi) \times (m \times L)$.
Since $M = m \times L$,the new magnetic moment is $M_{\text{new}} = 2M / \pi$.

(D) Let the original length of the wire be $L$. The original magnetic moment is $M = m \cdot L$,where $m$ is the pole strength.
When the wire is bent into an arc of a circle of radius $r$ subtending an angle $\theta = 60^{\circ} = \frac{\pi}{3}$ radians at the center,the arc length is $L = r\theta$.
Thus,$r = \frac{L}{\theta} = \frac{L}{\pi/3} = \frac{3L}{\pi}$.
The distance between the two ends of the arc (the chord length) is $d = 2r \sin(\frac{\theta}{2})$.
Substituting $\theta = 60^{\circ}$,we get $d = 2r \sin(30^{\circ}) = 2r \cdot \frac{1}{2} = r$.
The new magnetic moment $M'$ is given by $M' = m \cdot d = m \cdot r$.
Substituting $r = \frac{3L}{\pi}$,we get $M' = m \cdot \frac{3L}{\pi} = \frac{3}{\pi} (m \cdot L)$.
Since $M = m \cdot L$,the new magnetic moment is $M' = \frac{3M}{\pi}$.

(A) The magnetic moment $M$ of a bar magnet is given by the product of its pole strength $m$ and its magnetic length $2l$,so $M = m \times 2l$.
When the magnet is cut into two equal parts perpendicular to its length,the pole strength $m$ of each part remains the same,but the length of each part becomes half,i.e.,$l' = l$.
Therefore,the new magnetic moment $M'$ of each part is $M' = m \times l = (m \times 2l) / 2 = M / 2$.
Thus,$M' = 0.5 M$.

(C) The initial magnetic moment of the straight wire is $M = m \ell$,where $m$ is the pole strength of the wire.
When the wire is bent into an $L$-shape with segments of length $\frac{\ell}{3}$ and $\frac{2\ell}{3}$ at an angle of $90^{\circ}$,the new magnetic moment $M^{\prime}$ is given by $M^{\prime} = m \ell^{\prime}$,where $\ell^{\prime}$ is the effective distance between the two poles.
The effective length $\ell^{\prime}$ is the hypotenuse of the right-angled triangle formed by the two segments:
$\ell^{\prime} = \sqrt{\left(\frac{\ell}{3}\right)^{2} + \left(\frac{2\ell}{3}\right)^{2}} = \sqrt{\frac{\ell^{2}}{9} + \frac{4\ell^{2}}{9}} = \sqrt{\frac{5\ell^{2}}{9}} = \frac{\sqrt{5}\ell}{3}$.
Therefore,the new magnetic moment is $M^{\prime} = m \ell^{\prime} = m \left(\frac{\sqrt{5}\ell}{3}\right) = \frac{\sqrt{5}}{3} (m \ell) = \frac{\sqrt{5} M}{3}$.

(C) Let the length of the wire be $L$. The magnetic moment of a straight wire is $M = m \cdot L$,where $m$ is the pole strength.
When the wire is bent into an arc of angle $\theta = 90^o = \frac{\pi}{2}$ radians,the arc length $L = R\theta = R(\frac{\pi}{2})$,so the radius $R = \frac{2L}{\pi}$.
The new magnetic moment $M'$ is given by $M' = m \cdot d$,where $d$ is the straight-line distance (chord length) between the two ends of the arc.
The chord length $d = 2R \sin(\frac{\theta}{2}) = 2R \sin(45^o) = 2R \cdot \frac{1}{\sqrt{2}} = R\sqrt{2}$.
Substituting $R = \frac{2L}{\pi}$,we get $d = \frac{2L}{\pi} \cdot \sqrt{2} = \frac{2\sqrt{2}L}{\pi}$.
Thus,$M' = m \cdot \frac{2\sqrt{2}L}{\pi} = \frac{2\sqrt{2}}{\pi} (mL) = \frac{2\sqrt{2}M}{\pi}$.

(B) Electric field lines of force originate from a positive charge and terminate at a negative charge,meaning they do not form closed loops.
In contrast,magnetic field lines of force always form continuous closed loops,as magnetic monopoles do not exist.

(B) The magnetic field at point $P$ due to the vertical bar magnet is directed upwards because $P$ is near the north pole $(N)$ of this magnet. Let this be $B_1$.
The magnetic field at point $P$ due to the horizontal bar magnet is directed towards the right because $P$ is near the south pole $(S)$ of this magnet (magnetic field lines enter the south pole). Let this be $B_2$.
The resultant magnetic field $B_{net}$ is the vector sum of $B_1$ and $B_2$. Since $B_1$ is directed upwards and $B_2$ is directed towards the right,the resultant vector $B_{net}$ will point in the upward-right direction.
Therefore,the correct direction is indicated by the upward-right diagonal arrow.

(D) Let the original length of the wire be $L$ and the pole strength be $m$. The original magnetic moment is $M = m \times L$.
When the wire is bent into an arc of radius $r$ subtending an angle $\theta = 60^{\circ} = \pi/3$ radians at the center,the length of the arc is $L = r\theta = r(\pi/3)$.
Thus,$r = 3L/\pi$.
The new magnetic moment $M'$ is the product of pole strength $m$ and the straight-line distance (chord length) between the two ends of the arc.
The chord length $d = 2r \sin(\theta/2)$.
Substituting $\theta = 60^{\circ}$,we get $d = 2r \sin(30^{\circ}) = 2r(1/2) = r$.
Substituting $r = 3L/\pi$,we get $d = 3L/\pi$.
Therefore,$M' = m \times d = m \times (3L/\pi) = (3/\pi) \times (mL) = 3M/\pi$.

(C) The net magnetic dipole moment of two magnets,each with dipole moment $M$,placed at an angle $\theta$ between them is given by the vector sum:
$M_{net} = \sqrt{M^2 + M^2 + 2M^2 \cos \theta} = \sqrt{2M^2(1 + \cos \theta)} = \sqrt{2M^2(2 \cos^2(\theta/2))} = 2M \cos(\theta/2)$.
Analyzing the configurations:
$(a)$ The magnets are at $\theta = 90^{\circ}$. $M_{net} = 2M \cos(45^{\circ}) = 2M(1/\sqrt{2}) = \sqrt{2}M \approx 1.414M$.
$(b)$ The magnets are anti-parallel,$\theta = 180^{\circ}$. $M_{net} = 2M \cos(90^{\circ}) = 0$.
$(c)$ The magnets are at $\theta = 30^{\circ}$. $M_{net} = 2M \cos(15^{\circ}) \approx 2M(0.966) = 1.932M$.
$(d)$ The magnets are at $\theta = 60^{\circ}$. $M_{net} = 2M \cos(30^{\circ}) = 2M(\sqrt{3}/2) = \sqrt{3}M \approx 1.732M$.
Comparing the values,the configuration in $(c)$ has the highest net magnetic dipole moment.

(A) Magnetic field lines are continuous closed loops. Outside the magnet,they emerge from the North pole $(N)$ and enter the South pole $(S)$. Inside the magnet,they travel from the South pole $(S)$ to the North pole $(N)$. This ensures that the lines of magnetic induction form continuous paths. Looking at the provided figures,figure $(1)$ correctly depicts these lines emerging from the North pole,entering the South pole,and continuing through the interior of the magnet from $S$ to $N$.

(A) Magnetic lines of force represent the direction of the magnetic field at any point.
If two magnetic field lines were to intersect at a point,it would imply that there are two distinct directions for the magnetic field at that single point.
Since the magnetic field at any given point in space must have a unique direction,it is physically impossible for magnetic field lines to intersect.

(C) Let the length of the magnet be $2l = 2 \ cm$,so $l = 1 \ cm$.
The distance of point $A$ from the center of the magnet is $r_1 = x + l = x + 1$.
The distance of point $B$ from the center of the magnet is $r_2 = 2x + l = 2x + 1$.
For a short magnet or points at large distances,the magnetic field $B \propto \frac{1}{r^3}$.
Thus,the ratio of the magnetic fields is $\frac{B_A}{B_B} = \left( \frac{r_2}{r_1} \right)^3 = \left( \frac{2x + 1}{x + 1} \right)^3$.
If $x \gg l$,then $\frac{B_A}{B_B} \approx \left( \frac{2x}{x} \right)^3 = 2^3 = 8$.
Therefore,the ratio is $8 : 1$ approximately.

(B) The Assertion is correct because magnetic monopoles do not exist; breaking a magnet simply creates two smaller magnets,each with its own north and south pole.
The Reason is also correct. The magnetic moment $M$ of a bar magnet is given by $M = m \times 2l$,where $m$ is the pole strength and $2l$ is the length. When the magnet is cut into two equal pieces perpendicular to its length,the pole strength $m$ remains the same,but the length of each piece becomes $l$. Therefore,the new magnetic moment $M' = m \times l = M/2$.
Since the Reason explains why the magnetic properties change but does not explain why poles cannot be isolated,the correct option is $B$.

(A) Given: Magnetic moment $m = 0.40 \; A m^{2}$,distance $r = 50 \; cm = 0.5 \; m$. The length of the magnet $(5.0 \; cm)$ is small compared to the distance $(50 \; cm)$,so we use the short dipole approximation.
For the equatorial field $(B_{E})$:
$B_{E} = \frac{\mu_{0} m}{4 \pi r^{3}} = \frac{10^{-7} \times 0.40}{(0.5)^{3}} = \frac{10^{-7} \times 0.40}{0.125} = 3.2 \times 10^{-7} \; T$.
For the axial field $(B_{A})$:
$B_{A} = \frac{\mu_{0} 2m}{4 \pi r^{3}} = 2 \times B_{E} = 2 \times 3.2 \times 10^{-7} = 6.4 \times 10^{-7} \; T$.

(N/A) Magnetic moment of the bar magnet,$M = 0.48 \; J \; T^{-1}$.
$(a)$ Distance,$d = 10 \; cm = 0.1 \; m$.
The magnetic field at distance $d$ from the centre of the magnet on the axis is given by the relation:
$B = \frac{\mu_{0}}{4 \pi} \frac{2M}{d^{3}}$
Where,$\mu_{0} = 4 \pi \times 10^{-7} \; T \; m \; A^{-1}$.
Substituting the values:
$B = \frac{10^{-7} \times 2 \times 0.48}{(0.1)^{3}} = \frac{0.96 \times 10^{-7}}{10^{-3}} = 0.96 \times 10^{-4} \; T = 0.96 \; G$.
The direction of the magnetic field is along the $S-N$ direction.
$(b)$ The magnetic field at a distance of $10 \; cm$ $(d = 0.1 \; m)$ on the equatorial line of the magnet is given by:
$B = \frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}$
Substituting the values:
$B = \frac{10^{-7} \times 0.48}{(0.1)^{3}} = \frac{0.48 \times 10^{-7}}{10^{-3}} = 0.48 \times 10^{-4} \; T = 0.48 \; G$.
The direction of the magnetic field is along the $N-S$ direction.

(N/A) The property of attraction exhibited by certain materials is called magnetism,and an object possessing such properties is called a magnet.
Magnetic phenomena are universal in nature.
The Earth's magnetism predates human evolution.
Magnetism is spread across the entire universe,including our Earth.
The word 'magnet' is derived from the name of an island in Greece called Magnesia,where magnetic ore deposits were found as early as $600 \ BC$.
Legend has it that shepherds on this island complained that their wooden shoes,which had iron nails,would stick to the ground. Their iron-tipped rods were similarly affected. This attractive property of magnets made it difficult for them to move around.
From the name of the island 'Magnesia',the phenomenon and the object became known as 'magnet' in English.

(N/A) Some of the commonly known ideas regarding magnetism are as follows:
$(i)$ The Earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north.
$(ii)$ When a bar magnet is freely suspended,it points in the north-south direction. The tip which points to the geographic north is called the north pole,and the tip which points to the geographic south is called the south pole of the magnet.
$(iii)$ There is a repulsive force when north poles (or south poles) of two magnets are brought close together,and there is an attractive force between the north pole of one magnet and the south pole of the other.
$(iv)$ We cannot isolate the north or south pole of a magnet. If a bar magnet is broken into two halves,we get two smaller bar magnets with somewhat weaker properties. Unlike electric charges,isolated magnetic north and south poles,known as magnetic monopoles,do not exist.
$(v)$ It is possible to make magnets out of iron and its alloys.

(B) The name 'magnet' is derived from the name of the island 'Magnesia' in Asia Minor (modern-day Turkey),where naturally occurring magnetic ore (lodestone) was first discovered by the ancient Greeks.

(A) Magnetism is a physical phenomenon produced by the motion of electric charge,resulting in attractive and repulsive forces between objects. It is a fundamental force of nature.
$A$ magnet is a material or object that produces a magnetic field. This magnetic field is invisible but is responsible for the most notable property of a magnet: a force that pulls on other ferromagnetic materials,such as iron,steel,nickel,cobalt,etc.,and attracts or repels other magnets.

(N/A) When iron filings are sprinkled on a glass plate placed over a bar magnet,they align themselves along the magnetic field lines of the magnet.
The resulting pattern suggests the following:
$(i)$ The pattern of the iron filings indicates that the magnet has two poles,similar to the positive and negative charges of an electric dipole. One pole is designated as the north pole and the other as the south pole.
$(ii)$ When the magnet is suspended freely,these poles point towards the geographic north and south poles,respectively.
$(iii)$ $A$ similar pattern of iron filings is observed around a current-carrying solenoid,which confirms that a bar magnet and a current-carrying solenoid behave similarly.

(N/A) The fundamental differences between electric field lines and magnetic field lines are as follows:
$1$. Electric field lines originate from positive charges and terminate at negative charges. They do not form closed loops.
$2$. Magnetic field lines form continuous closed loops,extending from the North pole to the South pole outside the magnet and from the South pole to the North pole inside the magnet.
$3$. Electric field lines represent the path along which a unit positive charge would move,whereas magnetic field lines represent the direction of the magnetic field vector at any point.

(C) The analogy (similarity) between the magnetic field lines of a bar magnet and a solenoid suggests the following:
$(i)$ $A$ bar magnet can be modeled as a large number of circulating currents,which is analogous to the current loops in a solenoid.
$(ii)$ Cutting a bar magnet in half results in two independent magnets. Similarly,cutting a solenoid in the middle results in two smaller solenoids,each with weaker magnetic properties.
$(iii)$ Like a bar magnet,the magnetic field lines in a solenoid are continuous; they emerge from one face (acting as the North pole) and enter the other face (acting as the South pole),forming closed loops.
$(iv)$ At a point on the axis,the magnetic field of a solenoid is mathematically analogous to that of a bar magnet,given by $B = \frac{\mu_{0}}{4 \pi} \frac{2m}{r^{3}}$ for large distances.

(N/A) The pole strength of a bar magnet is a measure of the force that a magnetic pole exerts on other magnetic poles.
It is denoted by $q_{m}$ or $P$.
Pole strength is a scalar quantity,and its $SI$ unit is ampere-meter $(Am)$.
It depends on the material of the magnet,its magnetization,and its cross-sectional area $(A)$.
If the length of a bar magnet is $(2l)$ and its pole strength is $q_{m}$,then the magnetic dipole moment $\vec{m}$ is defined as the product of the pole strength and the magnetic length:
$\vec{m} = q_{m} \times (2\vec{l})$
The direction of the magnetic dipole moment $\vec{m}$ is from the South pole $(S)$ to the North pole $(N)$.
This is analogous to the electric dipole moment $\vec{p} = q(2\vec{a})$ in electrostatics,where pole strength $q_{m}$ corresponds to electric charge $q$ and magnetic length $(2l)$ corresponds to the separation distance $(2a)$.

(A) compass needle is a small magnetic dipole. When placed in an external magnetic field (such as the Earth's magnetic field),the two poles of the needle experience equal and opposite forces: $F = mB$ and $F = -mB$. Since these forces act at different points (the North and South poles of the needle),they constitute a couple. This couple exerts a torque on the needle,causing it to align with the magnetic field lines.

(N/A) The fundamental difference between an electric dipole and a magnetic dipole lies in their constituent elements.
An electric dipole is composed of two distinct elementary units,which are positive and negative electric charges (electric monopoles).
In contrast,a magnetic dipole,such as a current loop,is the most elementary unit of magnetism.
Magnetic monopoles,which would be the magnetic equivalent of electric charges,have not been observed to exist in nature.

The analogy between electrostatics and magnetism is summarized in the table below:

Physical quantity	Electrostatics	Magnetism
$(1)$ Field	Electric field $\overrightarrow{E}$	Magnetic field $\overrightarrow{B}$
$(2)$ Source	Charge $(q)$	Magnetic pole strength $(q_m)$
$(3)$ Dipole moment	$p = q(2a)$	$m = q_m(2l)$
$(4)$ Constant	$\frac{1}{4\pi\epsilon_0}$	$\frac{\mu_0}{4\pi}$
$(5)$ Length of dipole	$2a$	$2l$
$(6)$ Direction of dipole moment	From negative to positive charge	From South to North pole

(N/A) The magnetic field lines of a bar magnet represent the direction and strength of the magnetic field in the space surrounding the magnet.
$1$. The magnetic field lines originate from the North pole $(N)$ and terminate at the South pole $(S)$ outside the magnet.
$2$. Inside the magnet,the field lines travel from the South pole $(S)$ to the North pole $(N)$,forming continuous closed loops.
$3$. The density of the field lines is highest near the poles,indicating a stronger magnetic field in these regions.
$4$. No two magnetic field lines can intersect each other,as that would imply two different directions for the magnetic field at a single point,which is physically impossible.
$5$. The tangent drawn at any point on a magnetic field line gives the direction of the magnetic field at that point.

(N/A) The pole strength of a magnet is a measure of the ability of a magnetic pole to attract or repel other magnetic poles. It is defined as the force experienced by a unit north pole placed at that point in a magnetic field.
Mathematically,if a magnetic pole of strength $m$ experiences a force $F$ in a magnetic field $B$,then $F = mB$.
The $SI$ unit of pole strength is Ampere-meter,denoted as $A \cdot m$.

(N/A) The direction of the magnetic dipole moment is from the $S$ (South) pole to the $N$ (North) pole of the magnet.
The $SI$ unit of magnetic dipole moment is Ampere-meter squared,denoted as $A \cdot m^{2}$.

(B) In electricity,the fundamental source of the electric field is the electric charge $(q)$.
In magnetism,the fundamental source of the magnetic field is the magnetic pole strength,denoted by $(q_{m})$.
Therefore,the magnetic analog of electric charge is magnetic pole strength $(q_{m})$.

(N/A) For the system to be in equilibrium,the net force and the net torque acting on the system must be zero.
Let the magnetic moment of each magnet be $M$. The total magnetic moment of the system is the vector sum of the individual magnetic moments: $\vec{M}_{net} = \vec{M}_1 + \vec{M}_2 + \vec{M}_3$.
For the system to be in stable equilibrium in a magnetic field,the net magnetic moment must be zero. If the first magnet is oriented with North at the top,to make the net moment zero,the other two magnets must be arranged such that their components cancel out the moment of the first magnet.
As shown in the solution figure,if magnet $(1)$ has North at the top,then for the other two magnets,the poles must be arranged such that the North poles of magnets $(2)$ and $(3)$ are at the bottom-left and bottom-right respectively,while the South poles are at the top-left and top-right respectively. This configuration ensures that the vector sum of the magnetic moments is zero,resulting in no net torque or force on the system.

(N/A) Consider a magnetic field line of $\vec{B}$ forming a closed loop through the bar magnet as shown in the figure.
Let $C$ be an Amperian loop. According to Ampere's law for $\vec{H}$,the line integral of $\vec{H}$ over a closed loop is zero in the absence of free currents:
$\oint \vec{H} \cdot d\vec{l} = 0$
Inside the magnet,the relationship between $\vec{B}$ and $\vec{H}$ is given by $\vec{B} = \mu_0(\vec{H} + \vec{M})$.
Since magnetic field lines of $\vec{B}$ are continuous and form closed loops,they must travel from the $S$ pole to the $N$ pole inside the magnet to complete the loop outside from $N$ to $S$.
For the $\vec{H}$ field,the integral $\oint \vec{H} \cdot d\vec{l} = 0$ implies that if the path integral along the $\vec{B}$ line inside the magnet is positive,the path integral outside must be negative. Since $\vec{B} = \mu_0 \vec{H}$ outside the magnet (where $\vec{M}=0$),the direction of $\vec{H}$ lines inside the magnet must be from $N$ to $S$ to satisfy the closed loop condition for the total integral to be zero.

(A) The magnetic field of a dipole $\vec{M} = M\hat{k}$ at a point $(r, \theta)$ in spherical coordinates is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{M}{r^3} (2\cos\theta \hat{r} + \sin\theta \hat{\theta})$.
$1$. Along the $z$-axis ($P$ to $Q$): $\theta = 0$,$\vec{B} = \frac{\mu_0}{4\pi} \frac{2M}{z^3} \hat{k}$. Thus,$\int_P^Q \vec{B} \cdot d\vec{l} = \int_a^R \frac{\mu_0}{4\pi} \frac{2M}{z^3} dz = \frac{\mu_0 M}{4\pi} [-\frac{1}{z^2}]_a^R = \frac{\mu_0 M}{4\pi} (\frac{1}{a^2} - \frac{1}{R^2})$.
$2$. Along the arc of radius $R$ ($Q$ to $S$): $\vec{B} \cdot d\vec{l} = B_\theta (R d\theta) = \frac{\mu_0}{4\pi} \frac{M}{R^3} \sin\theta (R d\theta) = \frac{\mu_0 M}{4\pi R^2} \sin\theta d\theta$. Integrating from $\theta = 0$ to $\pi/2$: $\int_0^{\pi/2} \frac{\mu_0 M}{4\pi R^2} \sin\theta d\theta = \frac{\mu_0 M}{4\pi R^2} [-\cos\theta]_0^{\pi/2} = \frac{\mu_0 M}{4\pi R^2}$.
$3$. Along the $x$-axis ($S$ to $T$): $\theta = \pi/2$,$\vec{B} = \frac{\mu_0}{4\pi} \frac{M}{x^3} \hat{\theta} = -\frac{\mu_0}{4\pi} \frac{M}{x^3} \hat{k}$. Thus,$\int_S^T \vec{B} \cdot d\vec{l} = \int_R^a (-\frac{\mu_0 M}{4\pi x^3}) dx = \frac{\mu_0 M}{4\pi} [-\frac{1}{2x^2}]_R^a = -\frac{\mu_0 M}{8\pi} (\frac{1}{a^2} - \frac{1}{R^2})$.
$4$. Along the arc of radius $a$ ($T$ to $P$): $\vec{B} \cdot d\vec{l} = \frac{\mu_0 M}{4\pi a^2} \sin\theta d\theta$. Integrating from $\pi/2$ to $0$: $\int_{\pi/2}^0 \frac{\mu_0 M}{4\pi a^2} \sin\theta d\theta = -\frac{\mu_0 M}{4\pi a^2}$.
Summing these,$\oint \vec{B} \cdot d\vec{l} = 0$,which verifies Ampere's law for this path as there is no current enclosed.

(N/A) The magnetic field of a point dipole $\vec{M} = M\hat{k}$ at a position $\vec{r}$ is given by $\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi} \left[ \frac{3(\vec{M} \cdot \hat{r})\hat{r} - \vec{M}}{r^3} \right]$.
Ampere's law states that $\oint_C \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$.
For a point dipole,there is no physical current enclosed by the loop $C$,so $I_{enclosed} = 0$.
Thus,we must verify that $\oint_C \vec{B} \cdot d\vec{l} = 0$.
Along the $x$-axis,$\vec{r} = x\hat{i}$,$\hat{r} = \hat{i}$,and $\vec{M} \cdot \hat{r} = 0$. So $\vec{B} = \frac{\mu_0}{4\pi} \frac{-\vec{M}}{x^3} = -\frac{\mu_0 M}{4\pi x^3} \hat{k}$.
Since $d\vec{l}$ is along the $x$-axis $(d\vec{l} = dx \hat{i})$,$\vec{B} \cdot d\vec{l} = 0$.
Along the circular arc in the $xz$-plane,$\vec{B} \cdot d\vec{l}$ can be calculated using the magnetic vector potential $\vec{A} = \frac{\mu_0}{4\pi} \frac{\vec{M} \times \vec{r}}{r^3}$.
By Stokes' theorem,$\oint_C \vec{B} \cdot d\vec{l} = \oint_C (\nabla \times \vec{A}) \cdot d\vec{l} = \oint_C \vec{A} \cdot d\vec{l}$.
Since the dipole field is conservative everywhere except at the origin,the line integral over any closed loop not enclosing the origin is zero. Thus,Ampere's law is verified.

(B) The geometric length $(L)$ of a bar magnet is the actual physical length of the magnet.
The magnetic length is the distance between the two magnetic poles of the magnet.
It is an experimental fact that the magnetic poles are located slightly inside the ends of the magnet.
The magnetic length is approximately $0.83$ to $0.85$ times the geometric length.
Calculating the value of $\frac{6}{7} \approx 0.857$,which is the closest approximation to the ratio of magnetic length to geometric length.
Therefore,the separation between the poles is nearly $\frac{6}{7} L$.

(C) The magnetic moment is a vector quantity. The direction of the magnetic moment is from the South pole $(S)$ to the North pole $(N)$.
$1$. The horizontal magnet has a magnetic moment $M$ directed towards the left.
$2$. The vertical magnet has a magnetic moment $M$ directed upwards.
$3$. The hypotenuse magnet has a magnetic moment $M\sqrt{2}$ directed along the hypotenuse (from $S$ to $N$).
Let the horizontal direction be the $x$-axis and the vertical direction be the $y$-axis.
- $\vec{M}_1 = -M \hat{i}$
- $\vec{M}_2 = M \hat{j}$
- $\vec{M}_3 = M\sqrt{2} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = M\sqrt{2} (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}) = M \hat{i} + M \hat{j}$
Now,the resultant magnetic moment $\vec{M}_{\text{net}}$ is:
$\vec{M}_{\text{net}} = \vec{M}_1 + \vec{M}_2 + \vec{M}_3 = (-M \hat{i} + M \hat{j}) + (M \hat{i} + M \hat{j}) = 2M \hat{j}$
The magnitude of the resultant magnetic moment is $2M$. Therefore,the value is $2$.

(A) Let the pole strength of the magnetic strip be $m$ and its length be $\ell$. The initial magnetic moment is $M_1 = m \ell = 44 \text{ Am}^2$.
When the strip is bent into a semicircle of radius $R$,the length of the strip becomes the arc length of the semicircle,so $\ell = \pi R$,which means $R = \frac{\ell}{\pi}$.
The new magnetic moment $M_2$ is the product of the pole strength and the straight-line distance between the poles (the diameter of the semicircle),which is $2R$.
$M_2 = m \times (2R) = m \times \left( \frac{2\ell}{\pi} \right) = \frac{2}{\pi} (m \ell)$.
Substituting the given values: $M_2 = \frac{2}{\pi} \times 44 = \frac{2}{(22/7)} \times 44 = \frac{2 \times 7}{22} \times 44 = \frac{14}{22} \times 44 = 14 \times 2 = 28 \text{ Am}^2$.

(A) Let the pole strength of the original iron bar be $m$. The magnetic moment is given by $M = m L$.
When the bar is bent at the middle,each arm has a length of $L/2$.
The new magnetic moment $M'$ is the product of the pole strength $m$ and the effective length (the distance between the two poles).
The effective length $L'$ is the distance between the two ends of the bent bar. Since the two arms of length $L/2$ make an angle of $60^{\circ}$,the effective length forms the third side of an isosceles triangle with two sides of length $L/2$ and an included angle of $60^{\circ}$.
This triangle is equilateral,so the effective length $L' = L/2$.
The new magnetic moment is $M' = m L' = m (L/2) = M/2$.

(C) The magnetic dipole moment $m$ is given by the product of magnetization $M$ and the volume $V$ of the rod.
$m = M \times V$
Given:
Magnetization $M = 5.3 \times 10^{3} \ A/m$
Length $l = 5 \ cm = 5 \times 10^{-2} \ m$
Diameter $d = 1 \ cm = 1 \times 10^{-2} \ m$,so radius $r = 0.5 \times 10^{-2} \ m$
Volume $V = \pi r^2 l = \frac{22}{7} \times (0.5 \times 10^{-2})^2 \times (5 \times 10^{-2})$
$V = \frac{22}{7} \times 0.25 \times 10^{-4} \times 5 \times 10^{-2} \approx 3.14 \times 1.25 \times 10^{-6} \approx 3.925 \times 10^{-6} \ m^3$
Now,$m = (5.3 \times 10^3) \times (3.925 \times 10^{-6}) \approx 20.8 \times 10^{-3} \approx 2.08 \times 10^{-2} \ J/T$
Rounding to the nearest option,the value is $2 \times 10^{-2} \ J/T$.

(C) The magnetic dipole moment $(M)$ is defined as the product of the pole strength $(m)$ and the magnetic length $(2l)$ of the dipole,i.e.,$M = m \times 2l$.
This property is an intrinsic characteristic of the magnet itself.
It does not depend on the position or the distance of the point where the magnetic field is being measured.
Therefore,changing the distance between the point and the magnet does not affect the magnetic dipole moment.
Thus,the magnitude of the dipole moment remains $X$.