Bar Magnet and Magnetic Dipole and Magnetic Moment Questions in English

(A) The magnetic moment $(M)$ is defined as the product of pole strength $(m)$ and the magnetic length $(2l)$.
The $SI$ unit of pole strength is $Ampere \cdot meter$ $(A \cdot m)$.
The $SI$ unit of length is $meter$ $(m)$.
Therefore,the unit of magnetic moment is $(Ampere \cdot meter) \times meter = Ampere \cdot meter^2$ $(A \cdot m^2)$.

(A) magnetic dipole is characterized by its magnetic moment vector $\vec{M}$.
For a circular current loop,the magnetic moment $\vec{M}$ is directed along the axis of the loop.
Any point located on the line passing through the magnetic moment vector (the axis) is defined as the end-on position (also known as the axial position).
Therefore,a point on the axis of the loop is an end-on position.

(C) Neon $(Ne)$ is a noble gas with an atomic number of $10$.
Its electronic configuration is $1s^2 2s^2 2p^6$.
Since all the orbitals are completely filled,all electrons are paired.
Due to the pairing of electrons,the spin magnetic moments and orbital magnetic moments cancel each other out.
Therefore,the net resultant magnetic moment of a neon atom is $0$.

(B) The magnetic moment of a rod is given by $M = m \times L$,where $m$ is the pole strength and $L$ is the length of the rod.
When the rod 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,the radius $R = \frac{L}{\pi}$.
The new distance between the poles (the diameter of the semicircle) is $L' = 2R = \frac{2L}{\pi}$.
The pole strength $m$ remains unchanged.
Therefore,the new magnetic moment $M'$ is given by $M' = m \times L' = m \times \frac{2L}{\pi}$.
Since $M = m \times L$,we can substitute $M$ into the equation:
$M' = \frac{2M}{\pi}$.

(D) When a magnet is placed in iron powder and then removed,the maximum amount of iron powder adheres to the ends of the magnet. This occurs because the magnetic field strength is strongest at the poles (the ends) of the magnet,resulting in a greater magnetic force acting on the iron particles at these locations.

(B) The magnetic moment of a magnet is given by $M = m \times l$,where $m$ is the pole strength and $l$ is the magnetic length.
Case $1$: If the magnet is cut along its axis (longitudinally),the new pole strength becomes $m' = m/2$ and the length remains $l' = l$.
Therefore,the new magnetic moment $M' = m' \times l' = (m/2) \times l = M/2$.
Case $2$: If the magnet is cut perpendicular to its axis,the pole strength remains $m' = m$ and the new length becomes $l' = l/2$.
Therefore,the new magnetic moment $M' = m' \times l' = m \times (l/2) = M/2$.
In both cases,the magnetic moment of each part is $M/2$.

(B) The pole strength $m$ of a magnet is directly proportional to its cross-sectional area $A$ (i.e.,$m \propto A$).
When the magnet is cut into four parts such that the length and width of each part are halved,the cross-sectional area of each part becomes $A' = (width/2) \times (thickness/2) = A/4$ if we consider the cross-section as a rectangle. However,the question specifies the width is halved. If the magnet is cut along its length and width,the cross-sectional area $A$ is halved by the cut along the width. Thus,the new pole strength $m'$ is proportional to the new area $A' = A/2$.
Therefore,$m' = m/2$.

(A) The magnetism of a magnet is primarily due to the spin motion of electrons. Each electron in an atom revolves in an orbit around the nucleus, which is equivalent to a tiny current loop, creating an orbital magnetic dipole moment $\vec{M}_{l} = \text{current} \times \text{area}$.
In addition to orbital motion, every electron possesses an intrinsic spin motion around its own axis, which generates a spin magnetic dipole moment $\vec{M}_{s}$. The net magnetic dipole moment $\vec{M}$ of the atom is the vector sum of $\vec{M}_{l}$ and $\vec{M}_{s}$. Since $\vec{M}_{s}$ is significantly greater than $\vec{M}_{l}$, the magnetism of a magnet is predominantly attributed to the spin of electrons.

(D) The magnetic field on the axis of a short bar magnet at a distance $x$ is given by $B_{axis} = \frac{\mu_0}{4\pi} \frac{2M}{x^3}$.
The magnetic field on the equatorial line of a short bar magnet at a distance $y$ is given by $B_{equator} = \frac{\mu_0}{4\pi} \frac{M}{y^3}$.
Given that the magnetic fields are equal,$B_{axis} = B_{equator}$.
Therefore,$\frac{\mu_0}{4\pi} \frac{2M}{x^3} = \frac{\mu_0}{4\pi} \frac{M}{y^3}$.
Simplifying the equation,we get $\frac{2}{x^3} = \frac{1}{y^3}$.
Rearranging for the ratio,$\frac{x^3}{y^3} = 2$.
Taking the cube root on both sides,$\frac{x}{y} = 2^{1/3}$.

(B) For a short bar magnet of magnetic moment $M$,the magnetic intensity at an axial point (end-on position) at distance $d$ is given by $H_{axial} = \frac{2M}{4\pi d^3}$.
For the same magnet,the magnetic intensity at an equatorial point (broad side-on position) at the same distance $d$ is given by $H_{equatorial} = \frac{M}{4\pi d^3}$.
Given $OP_1 = OP_2 = d = 10 \ m$,where $P_1$ is at the equatorial position and $P_2$ is at the axial position.
Therefore,$H_1 = H_{equatorial} = \frac{M}{4\pi d^3}$ and $H_2 = H_{axial} = \frac{2M}{4\pi d^3}$.
The ratio is $\frac{H_1}{H_2} = \frac{M/4\pi d^3}{2M/4\pi d^3} = \frac{1}{2}$.
Thus,the ratio $H_1:H_2 = 1:2$.

(B) Magnetic lines of induction (or magnetic field lines) in a bar magnet are continuous and form closed loops.
Outside the magnet,the field lines emerge from the north pole and enter the south pole.
Inside the magnet,the field lines run from the south pole to the north pole to complete the closed loop.
Therefore,they run continuously through the bar and outside.

(A) The pole strength of a magnet is a property determined by the cross-sectional area and the intensity of magnetization of the material.
When a magnet is cut perpendicular to its length,the cross-sectional area remains unchanged.
Since the material properties and the cross-sectional area do not change,the pole strength of each piece remains the same as that of the original magnet.
Therefore,the ratio of the pole strengths of the two sections is $1 : 1$,which means they are equal.

(B) The magnetic field intensity $(H)$ at a point is defined as the force experienced by a unit north pole placed at that point.
Mathematically,it is given by $H = F/m$,where $F$ is the magnetic force and $m$ is the pole strength.
Therefore,it represents the magnetic induction force acting on a unit magnetic pole.
Thus,option $(b)$ is the correct definition.

(A) magnetic needle acts as a magnetic dipole.
In a non-uniform magnetic field,the magnetic force acting on the two poles of the needle will be different in both magnitude and direction.
Because the magnetic field is non-uniform,the net force on the dipole is non-zero,resulting in a translational force.
Additionally,because the forces on the two poles are not collinear and have different magnitudes,they create a net torque,causing the needle to rotate.
Therefore,the magnetic needle experiences both a force and a torque.

(B) The magnetic field $B$ at a distance $d$ from an isolated magnetic pole of strength $m$ is given by the formula $B = \frac{\mu_0}{4\pi} \frac{m}{d^2}$ in the $SI$ system.
In the $C.G.S.$ system,the constant $\frac{\mu_0}{4\pi}$ is taken as $1$. Therefore,the magnetic induction is $B = \frac{m}{d^2}$.

(C) When a magnetic needle is cut perpendicular to its length into two equal pieces,the pole strength $m$ of each new piece remains the same as the original,because the cross-sectional area of the magnet does not change.
The length of each new piece becomes $L' = \frac{2L}{2} = L$.
The magnetic moment $M'$ of each new piece is given by the product of its pole strength and its new length:
$M' = m \times L' = m \times L$.
Since the original magnetic moment $M = m \times (2L) = 2mL$,we can write:
$M' = \frac{M}{2}$.
Therefore,the magnetic moment of each piece is $\frac{M}{2}$ and the pole strength is $m$.

(B) The force $F$ between two magnetic poles of strengths $m_1$ and $m_2$ separated by a distance $r$ is given by Coulomb's law for magnetism: $F = k \frac{m_1 m_2}{r^2}$.
Given that the new pole strengths are $m_1' = 2m_1$ and $m_2' = 2m_2$,and the new distance is $r' = 2r$.
The new force $F'$ is given by: $F' = k \frac{(2m_1)(2m_2)}{(2r)^2}$.
$F' = k \frac{4 m_1 m_2}{4 r^2} = k \frac{m_1 m_2}{r^2} = F$.
Therefore,the force remains unchanged.

(C) The force $F$ between two magnetic poles of strengths $m_1$ and $m_2$ separated by a distance $r$ is given by Coulomb's law for magnetism:
$F = \frac{\mu_0}{4\pi} \cdot \frac{m_1 m_2}{r^2}$
Given that the pole strengths are unit pole strengths,$m_1 = 1 \, Am$ and $m_2 = 1 \, Am$.
The distance $r = 1 \, m$.
The value of $\frac{\mu_0}{4\pi} = 10^{-7} \, N/A^2$.
Substituting these values into the formula:
$F = 10^{-7} \times \frac{1 \times 1}{1^2} = 10^{-7} \, N$.

(C) The magnetic moment $M$ of a bar magnet is defined as the product of its pole strength $m$ and the magnetic length $2l$,given by $M = m \times 2l$.
When a hole is made at the centre of the bar magnet,the material of the magnet is removed from the central region.
However,the pole strength $m$ depends on the magnetic material at the ends of the magnet,which remains unaffected by a hole at the centre.
The magnetic length $2l$ (the distance between the poles) also remains unchanged.
Since both the pole strength and the magnetic length remain constant,the magnetic moment $M$ of the bar magnet does not change.

(D) Magnetic field lines are continuous closed loops.
Outside the bar magnet,the magnetic field lines originate from the north pole and terminate at the south pole.
Inside the bar magnet,the magnetic field lines travel from the south pole to the north pole to complete the closed loop.
Therefore,the correct description is that they move from the south pole to the north pole inside the magnet and from the north pole to the south pole outside the magnet.

(D) For two short bar magnets placed coaxially,the force of interaction $F$ is inversely proportional to the fourth power of the distance $r$ between their centers,i.e.,$F \propto \frac{1}{r^4}$.
Given the initial force $F_1 = 4.8 \ N$ at distance $r_1 = r$.
When the distance is increased to $r_2 = 2r$,the new force $F_2$ is given by:
$\frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^4$
$\frac{F_2}{4.8} = \left( \frac{r}{2r} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$
$F_2 = \frac{4.8}{16} = 0.3 \ N$.

(B) permanent magnet is a material that produces a magnetic field. This magnetic field exerts a force on magnetic materials (such as iron,nickel,and cobalt),causing them to be attracted to the magnet. Non-magnetic substances do not interact with the magnetic field in this way,meaning they are neither attracted nor repelled by the magnet. Therefore,a permanent magnet attracts only magnetic substances.

(C) Magnetic field lines are continuous closed loops. Outside a magnet,they travel from the north pole to the south pole. However,inside a magnet,they travel from the south pole to the north pole to complete the closed loop. Therefore,the statement in option $C$ is incorrect.

(B) The correct answer is $B$. Attraction can occur between a magnet and a magnetic material (like iron or steel) or between opposite poles of two magnets. However,repulsion only occurs when like poles of two magnets face each other. Therefore,repulsion is the only conclusive test for magnetism.

(A) The torque $\tau$ experienced by a magnetic dipole in a magnetic field is given by $\tau = M B \sin \theta$.
For maximum torque,$\theta = 90^\circ$,so $\tau_{\max} = M B$.
Given: $\tau_{\max} = 4 \times 10^{-5} \ N \cdot m$ and $B = 10^{-4} \ Wb/m^2$.
Substituting the values: $4 \times 10^{-5} = M \times 10^{-4}$.
Solving for $M$: $M = \frac{4 \times 10^{-5}}{10^{-4}} = 4 \times 10^{-1} = 0.4 \ A \cdot m^2$.

(B) The magnetic moment is a vector quantity. When two magnets of equal magnetic moment $M$ are placed at right angles to each other,their resultant magnetic moment $M_{net}$ is given by the vector sum of the individual magnetic moments.
Since the angle between the two magnetic moments is $90^{\circ}$,the magnitude of the resultant magnetic moment is calculated as:
$M_{net} = \sqrt{M^2 + M^2 + 2MM \cos(90^{\circ})}$
Since $\cos(90^{\circ}) = 0$,this simplifies to:
$M_{net} = \sqrt{M^2 + M^2} = \sqrt{2M^2} = \sqrt{2} \,M$

(B) Let the magnetic field be zero at point $P$,which lies at a distance $x$ from the $10$ unit pole.
Since the poles are like,the null point lies between them.
The distance of $P$ from the $40$ unit pole is $(30 - x)$.
At point $P$,the magnetic field intensities due to both poles must be equal in magnitude and opposite in direction:
$\frac{\mu_0}{4\pi} \cdot \frac{10}{x^2} = \frac{\mu_0}{4\pi} \cdot \frac{40}{(30 - x)^2}$
$\frac{1}{x^2} = \frac{4}{(30 - x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{30 - x}$
$30 - x = 2x$
$3x = 30 \Rightarrow x = 10 \, cm$.
Thus,the point is at a distance of $10 \, cm$ from the $10$ unit pole and $(30 - 10) = 20 \, cm$ from the stronger ($40$ unit) pole.
Therefore,the correct option is $(b)$.

(C) Let the original pole strength be $m$,length be $L$,and magnetic moment be $M = m \times L$.
Case $1$: Magnet $P$ is cut along its axial line (longitudinally).
Each piece has length $L' = L$ and pole strength $m' = \frac{m}{2}$.
The new magnetic moment is $M' = m' \times L' = \frac{m}{2} \times L = \frac{M}{2}$.
Case $2$: Magnet $Q$ is cut along its equatorial line (transversely).
Each piece has length $L' = \frac{L}{2}$ and pole strength $m' = m$.
The new magnetic moment is $M' = m' \times L' = m \times \frac{L}{2} = \frac{M}{2}$.
Thus,all four pieces obtained have a magnetic moment of $\frac{M}{2}$.

(D) couple is defined as two equal and opposite forces acting along different lines of action,which produces a torque.
For two bar magnets placed coaxially,the magnetic force between them acts along the axis connecting their centers.
Since the forces on the poles of the magnets act along the same line (the common axis),the perpendicular distance between the lines of action of the forces is zero.
Therefore,the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$.
Thus,no couple acts on the magnets.

(D) The magnetic moment $M$ of a bar magnet is given by $M = m \times L$,where $m$ is the pole strength and $L$ is the length of the magnet.
The force $F$ experienced by each pole in a uniform magnetic field $B$ is given by $F = m \times B$.
From this,the pole strength is $m = \frac{F}{B}$.
Substituting $m$ into the magnetic moment formula: $M = \left( \frac{F}{B} \right) \times L$.
Rearranging for $L$: $L = \frac{M \times B}{F}$.
Given: $M = 3.0 \, A-m^2$,$B = 2 \times 10^{-5} \, T$,and $F = 6 \times 10^{-4} \, N$.
$L = \frac{3.0 \times 2 \times 10^{-5}}{6 \times 10^{-4}} = \frac{6 \times 10^{-5}}{6 \times 10^{-4}} = 0.1 \, m$.
Therefore,the length of the magnet is $0.1 \, m$.

(D) Magnetic field lines represent the direction of the magnetic field at any given point. If two magnetic field lines were to intersect,it would imply that there are two different directions for the magnetic field at the point of intersection,which is physically impossible. Therefore,magnetic field lines due to a bar magnet never intersect each other.

(C) In magnetism,isolated magnetic monopoles do not exist. Even at the atomic level,magnetism arises from current loops or electron spin,which inherently form a magnetic dipole. Therefore,the smallest fundamental unit of magnetism is a magnetic dipole.

(D) is the correct option.
Magnetic field lines form continuous closed loops.
Outside the magnet,the field lines are directed from the north pole to the south pole.
Inside the magnet,the field lines are directed from the south pole to the north pole,ensuring the continuity of the closed loops.

(A) Magnetic field lines form continuous closed loops. Outside the magnet,the magnetic field lines emerge from the north pole and enter the south pole. Inside the magnet,to complete the closed loop,the magnetic field lines travel from the south pole to the north pole. Therefore,the correct option is $A$.

(A) The torque $\tau$ acting on a bar magnet of magnetic moment $M$ placed in an external magnetic field $B$ is given by the formula $\tau = MB \sin \theta$,where $\theta$ is the angle between the magnetic axis and the magnetic field.
For the earth's magnetic field,the torque is $\tau = M B_H \sin \theta$,where $B_H$ is the horizontal component of the earth's magnetic field.
The torque $\tau$ is maximum when $\sin \theta$ is maximum,which occurs at $\theta = 90^{\circ}$.
Therefore,the torque is maximum when the axis of the magnet is perpendicular to the earth's magnetic field.

(B) The magnetic dipole moment,denoted by $\vec{M}$ or $\vec{m}$,is defined as the product of the pole strength $(q_m)$ and the magnetic length $(2\vec{l})$ of the magnet.
Since it has both magnitude and a specific direction (from the South pole to the North pole),it is a vector quantity.
Therefore,the correct option is $B$.

(A) The magnetic moment $M$ of a bar magnet is given by the product of its pole strength $m$ and its magnetic length $L$.
Given:
Pole strength $m = 4.0\, Am$
Length $L = 10\, cm = 0.1\, m$
Formula: $M = m \times L$
Calculation: $M = 4.0\, Am \times 0.1\, m = 0.4\, A\, m^2$.
Therefore,the correct option is $A$.

(A) The initial magnetic moment $M$ is given by $M = m \times L$,where $m = 0.5 \, Am$ and $L = 31.4 \, cm = 0.314 \, m$.
So,$M = 0.5 \times 0.314 = 0.157 \, Am^2$.
When the magnet is bent into a semicircle,the new effective length (distance between poles) $L'$ becomes the diameter of the semicircle.
If $R$ is the radius,then the arc length $L = \pi R$,so $R = L / \pi$.
The new distance between poles is $L' = 2R = 2L / \pi$.
The new magnetic moment $M'$ is $M' = m \times L' = m \times (2L / \pi) = (2 / \pi) \times M$.
Substituting the values: $M' = (2 / 3.14) \times (0.5 \times 0.314) = (2 / 3.14) \times 0.157 = 2 \times 0.05 = 0.1 \, Am^2$.

(B) The time period of a freely suspended magnet is given by $T = 2\pi \sqrt{\frac{I}{MB_H}} = 4 \ s$.
When the magnet is cut into two equal halves along its length,the new magnetic moment becomes $M' = \frac{M}{2}$.
The new moment of inertia $I'$ for one half is calculated as $I' = \frac{m'l'^2}{12}$,where $m' = \frac{m}{2}$ and $l' = \frac{l}{2}$.
Thus,$I' = \frac{(m/2)(l/2)^2}{12} = \frac{1}{8} \cdot \frac{ml^2}{12} = \frac{I}{8}$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B_H}} = 2\pi \sqrt{\frac{I/8}{(M/2)B_H}}$.
Simplifying this,$T' = 2\pi \sqrt{\frac{1}{4} \cdot \frac{I}{MB_H}} = \frac{1}{2} \cdot 2\pi \sqrt{\frac{I}{MB_H}} = \frac{T}{2}$.
Substituting $T = 4 \ s$,we get $T' = \frac{4}{2} = 2 \ s$.

(D) The magnetic moment of an atom arises from the spin and orbital angular momentum of its electrons.
Different elements have different electronic configurations,leading to varying magnetic properties.
Some atoms have a permanent magnetic moment due to unpaired electrons,while others do not.
Furthermore,the interaction between atoms in a material determines the net magnetic behavior.
Therefore,it is not universally true that all atoms have a magnetic moment,nor is it true that none have one,nor do all atoms align in the same direction under an external field.
Thus,none of the provided statements are accurate.

(A) The electronic configuration of neon $(Ne)$ is $1s^2 2s^2 2p^6$.
Since all the orbitals are completely filled,all electrons are paired.
Due to the pairing of electrons,the spin magnetic moments and orbital magnetic moments cancel each other out.
Therefore,the net magnetic moment of a neon atom is zero.

(B) The magnetic dipole moment $M$ is given by the product of magnetization $I$ and the volume $V$ of the magnet,i.e.,$M = I \times V$.
Given: Magnetization $I = 5.30 \times 10^3 \, A/m$,length $l = 5 \, cm = 0.05 \, m$,and diameter $d = 1 \, cm$,so radius $r = 0.5 \, cm = 0.005 \, m$.
The volume of the cylindrical rod is $V = \pi r^2 l$.
Substituting the values:
$V = \pi \times (0.005)^2 \times 0.05 = \pi \times (2.5 \times 10^{-5}) \times 0.05 = 3.927 \times 10^{-6} \, m^3$.
Now,calculating the dipole moment:
$M = (5.30 \times 10^3) \times (3.927 \times 10^{-6}) \approx 2.08 \times 10^{-2} \, J/T$.

(C) $1$. Calculate the number of atoms per unit volume $(n)$:
$n = \frac{\rho N_A}{M} = \frac{7.78 \times 10^3 \times 6.02 \times 10^{23}}{56 \times 10^{-3}} \approx 8.36 \times 10^{28} \, atoms/m^3$.
$2$. Calculate the volume of the bar $(V)$:
$V = 5 \, cm \times 1 \, cm \times 1 \, cm = 5 \times 10^{-6} \, m^3$.
$3$. Calculate the total number of atoms $(N)$:
$N = n \times V = 8.36 \times 10^{28} \times 5 \times 10^{-6} = 4.18 \times 10^{23} \, atoms$.
$4$. Calculate the total magnetic moment $(M_{total})$:
$M_{total} = N \times \mu_{atom} = 4.18 \times 10^{23} \times 1.8 \times 10^{-23} \approx 7.52 \, A \cdot m^2$.
Rounding to the nearest provided option,the result is $7.54 \, A \cdot m^2$.

(B) The initial magnetic dipole moment is $M = m \cdot L$,where $m$ is the pole strength.
When the rod is bent into a semicircle of radius $R$,its length $L$ becomes the arc length of the semicircle,so $L = \pi R$,which gives $R = \frac{L}{\pi}$.
The new magnetic dipole moment $M'$ is the product of the pole strength $m$ and the straight-line distance between the poles (the diameter $2R$).
$M' = m \cdot (2R) = m \cdot \left( \frac{2L}{\pi} \right)$.
Since $M = m \cdot L$,we substitute $M$ into the equation:
$M' = \frac{2M}{\pi}$.

(D) The magnetic moment $M$ of a bar magnet is given by $M = m \cdot L$,where $m$ is the pole strength and $L$ is the length of the magnet.
The force $F$ acting on a pole in a magnetic field $B$ is given by $F = m \cdot B$.
From the first equation,we have $m = \frac{M}{L}$.
Substituting this into the force equation: $F = \frac{M}{L} \cdot B$.
Given $M = 3.0 \, A \cdot m^2$,$B = 2 \times 10^{-5} \, T$,and $F = 6 \times 10^{-4} \, N$.
Substituting the values: $6 \times 10^{-4} = \frac{3.0}{L} \times 2 \times 10^{-5}$.
$L = \frac{3.0 \times 2 \times 10^{-5}}{6 \times 10^{-4}} = \frac{6 \times 10^{-5}}{6 \times 10^{-4}} = 10^{-1} = 0.1 \, m$.

(C) The pole strength $m = 0.01 \, A \cdot m$. The distance between the poles is $d = 0.1 \, m$. The midpoint is at a distance $r = d/2 = 0.05 \, m$ from each pole.
At the midpoint,the magnetic field due to the North pole $(B_N)$ is directed away from it,and the magnetic field due to the South pole $(B_S)$ is directed towards it. Both fields point in the same direction.
The magnitude of the magnetic field due to a single pole is given by $B = \frac{\mu_0}{4\pi} \frac{m}{r^2}$.
$B_N = B_S = 10^{-7} \times \frac{0.01}{(0.05)^2} = 10^{-7} \times \frac{0.01}{0.0025} = 10^{-7} \times 4 = 4 \times 10^{-7} \, T$.
The net magnetic field $B_{net} = B_N + B_S = 4 \times 10^{-7} + 4 \times 10^{-7} = 8 \times 10^{-7} \, T$.

(B) The magnetic moment $M$ is given by the product of magnetization $I$ and volume $V$ of the rod.
$M = I \times V$
Given:
Magnetization $I = 5.30 \times 10^3 \, A/m$
Length $l = 5 \, cm = 5 \times 10^{-2} \, m$
Radius $r = \text{diameter} / 2 = 0.5 \, cm = 0.5 \times 10^{-2} \, m$
Volume $V = \pi r^2 l = \pi \times (0.5 \times 10^{-2})^2 \times (5 \times 10^{-2}) \, m^3$
$V = 3.14159 \times 0.25 \times 10^{-4} \times 5 \times 10^{-2} \approx 3.927 \times 10^{-6} \, m^3$
Now,calculate the magnetic moment:
$M = (5.30 \times 10^3) \times (3.927 \times 10^{-6})$
$M \approx 20.813 \times 10^{-3} \, J/T = 2.08 \times 10^{-2} \, J/T$.

(A) The magnetic moment $M$ of a bar magnet is given by $M = m \times L$,where $m$ is the pole strength and $L$ is the magnetic length of the magnet.
When a magnet is placed in a magnetic field $B$,the force experienced by each pole is given by $F = mB$.
From this,the pole strength is $m = F/B$.
Substituting the value of $m$ into the magnetic moment equation: $M = (F/B) \times L$.
Rearranging for the length $L$,we get: $L = MB/F$.

(B) Let $m$ be the pole strength of each pole of the bar magnet of length $l$. Then,the initial magnetic dipole moment is given by:
$M = m \times l$ ......... $(i)$
When the bar magnet is bent into an arc of radius $r$ subtending an angle $\theta = 60^{\circ} = \frac{\pi}{3} \text{ radians}$ at the center,the arc length $l$ is:
$l = r \theta = r \times \frac{\pi}{3}$
$r = \frac{3l}{\pi}$
The new magnetic dipole moment $M^{\prime}$ is the product of pole strength $m$ and the straight-line distance (chord length) between the two poles.
The chord length $d$ is given by $2r \sin(\frac{\theta}{2}) = 2r \sin(30^{\circ}) = 2r \times \frac{1}{2} = r$.
Thus,$M^{\prime} = m \times d = m \times r$.
Substituting $r = \frac{3l}{\pi}$:
$M^{\prime} = m \times \frac{3l}{\pi} = \frac{3}{\pi} (m \times l) = \frac{3}{\pi} M$ (using equation $(i)$).