Question 4

   Question: The magnetic moment of an iron bar is $M$. It is now bent in such a way that it forms an arc section of a circle subtending an angle of $60^\circ$ at the centre. The magnetic moment of this arc section is

   Options: 

       A. $3M/\pi$

       B. $4M/\pi$

       C. $M/\pi$

       D. $2M/\pi$

   Correct Answer: A

   Year: NEET 2024 Re

   Solution (as Given in Source): $M = m \cdot L$;

$R\theta = L$; $\frac{R\pi}{3} = L \Rightarrow R = \frac{3L}{\pi}$. $M' = m(2R)\sin 30^\circ = m(2)\frac{3L}{\pi} \times \frac{1}{2} = \frac{3}{\pi}mL = \frac{3M}{\pi}$.

   Step Solution:

    1.  Define the initial magnetic moment of the straight bar as $M = m \times L$, where $m$ is the pole strength and $L$ is the length.

    2.  For the arc, the length $L$ remains the same. Use the arc length formula $L = R\theta$. Since $\theta = 60^\circ = \pi/3$ radians, then $L = R(\pi/3)$.

    3.  Calculate the radius of the arc: $R = 3L/\pi$.

    4.  Find the new effective length (the straight-line distance between the poles of the arc): $L' = 2R \sin(\theta/2) = 2R \sin(30^\circ)$.

    5.  Substitute $R$ into the formula: $M' = m \times (2 \times \frac{3L}{\pi} \times \frac{1}{2}) = m \times \frac{3L}{\pi} = \frac{3M}{\pi}$.

   Difficulty Level: Medium

   Concept Name: Magnetic Dipole Moment of a Bent Magnet

   Short cut solution: Use the general formula for a magnet bent into an arc: $M' = \frac{M \sin(\theta/2)}{\theta/2}$ (where $\theta$ is in radians). For $60^\circ$ ($\pi/3$ rad): $M' = \frac{M \sin(30^\circ)}{\pi/6} = \frac{M(1/2)}{\pi/6} = \frac{3M}{\pi}$.

Question 9

   Question: An iron bar of length $L$ has magnetic moment $M$. It is bent at the middle of its length such that the two arms make an angle $60^\circ$ with each other. The magnetic moment of this new magnet is:

   Options:

       A. $M$

       B. $M/2$

       C. $2M$

       D. $M/\sqrt{3}$

   Correct Answer: B

   Year: NEET 2024

   Solution (as Given in Source):$M = ml$.$\Delta l = 2 \frac{l}{2} \sin 30^\circ = \frac{l}{2}$. $M' = ml/2 = M/2$.

   Step Solution:

    1.  The initial magnetic moment is $M = m \times L$.

    2.  When bent at the middle, the magnet consists of two arms, each of length $L/2$.

    3.  The poles are now separated by a distance $L'$ which is the base of an isosceles triangle with sides $L/2$ and an included angle of $60^\circ$.

    4.  Since the triangle is equilateral (two sides $L/2$ and angle $60^\circ$), the new effective length (distance between poles) is $L' = L/2$.

    5.  The new magnetic moment is $M' = m \times L' = m \times (L/2) = M/2$.

   Difficulty Level: Easy

   Concept Name: Resultant Magnetic Moment

   Short cut solution: For two segments of a magnet bent at an angle, the effective length is the third side of the triangle formed. If a magnet is bent at $60^\circ$ at its center, it forms an equilateral triangle with the distance between the poles, making $L' = L/2$, thus $M' = M/2$.

Question 13

   Question: A uniform conducting wire of length $12a$ and resistance '$R$' is wound up as a current carrying coil in the shape of, (i) an equilateral triangle of side '$a$'. (ii) a square of side '$a$'. The magnetic dipole moments of the coil in each case respectively are

   Options:

       A. $3\sqrt{3} Ia^2$ and $3 Ia^2$

       B. $3 Ia^2$ and $Ia^2$

       C. $3 Ia^2$ and $4 Ia^2$

       D. $4 Ia^2$ and $3 Ia^2$

   Correct Answer: A

   Year: NEET 2021

   Solution (as Given in Source): Current in the loop will be $V/R = I$ which is same for both loops. Now magnetic moment of Triangle loop $= NIA_{M1} = (\frac{12a}{3a}) \cdot I \cdot \frac{\sqrt{3}}{4} a^2 = \sqrt{3} Ia^2$. Magnetic moment of square loop $= N'IA' = (\frac{12a}{4a}) \cdot I \cdot a^2 = 3 Ia^2$.

   Step Solution:

    1.  Calculate number of turns ($N$) for the triangle: Total length / Perimeter $= 12a / 3a = 4$ turns.

    2.  Calculate magnetic moment for triangle ($M_1$): $N \times I \times \text{Area} = 4 \times I \times (\frac{\sqrt{3}}{4} a^2) = \sqrt{3} Ia^2$.

    3.  Calculate number of turns ($N$) for the square: Total length / Perimeter $= 12a / 4a = 3$ turns.

    4.  Calculate magnetic moment for square ($M_2$): $N \times I \times \text{Area} = 3 \times I \times (a^2) = 3 Ia^2$.

    5.  The results are $\sqrt{3} Ia^2$ and $3 Ia^2$. (Note: The source text lists Option A as the answer despite the $3\sqrt{3}$ typo in the option text; the math in the solution confirms the $\sqrt{3}$ coefficient).

   Difficulty Level: Medium

   Concept Name: Magnetic Moment of a Coil ($M = NIA$)

   Short cut solution: Note that $M \propto N \times \text{Area}$. For a fixed length $L$, $N = L/\text{Perimeter}$. Thus, $M \propto \frac{\text{Area}}{\text{Perimeter}}$. Triangle: $\frac{(\sqrt{3}/4)a^2}{3a} \propto \frac{\sqrt{3}}{12}$. Square: $\frac{a^2}{4a} \propto \frac{1}{4}$. Multiply both by total length $12a \times I$ to get $\sqrt{3}Ia^2$ and $3Ia^2$.

Question 22

   Question: Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment $\vec{m}$. Which configuration has highest net magnetic dipole moment?

   Options: 

       A. (1)

       B. (2)

       C. (3)

       D. (4)

   Correct Answer: C

   Year: 2014

   Solution (as Given in the Source): The direction of magnetic dipole moment is from south to north pole of the magnet. In configuration (2), $m_{net} = m - m = 0$.

   Step Solution:

    1.  The magnetic dipole moment $\vec{m}$ is a vector quantity. The net moment $M$ for two magnets at an angle $\theta$ is $M = \sqrt{m^2 + m^2 + 2m^2\cos\theta}$.

    2.  In configuration (1), magnets are perpendicular ($90^\circ$): $M = \sqrt{m^2 + m^2} = \sqrt{2}m \approx 1.41m$.

    3.  In configuration (2), magnets are antiparallel ($180^\circ$): $M = m - m = 0$.

    4.  In configuration (3), magnets are parallel ($0^\circ$): $M = m + m = 2m$.

    5.  Comparing the results ($2m > 1.41m > 0$), configuration (3) yields the maximum resultant.

   Difficulty Level: Medium

   Concept Name: Vector Addition of Magnetic Dipole Moments

   Short cut solution: The resultant of two equal vectors is maximum when the angle between them is minimum. For $\theta = 0^\circ$ (parallel), the value is $2m$, which is the highest possible.

Question 24

   Question: A bar magnet of length '$l$' and magnetic dipole moment '$M$' is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

   Options:

       A. $2M/\pi$

       B. $M/2$

       C. $M$

       D. $3M/\pi$

   Correct Answer: D

   Year: 2013 NEET

   Solution (as Given in the Source): Let $m$ be strength of each pole of bar magnet of length $l$. Then $M = m \times l$. When the bar magnet is bent in the form of an arc, $l = \frac{\pi}{3} \times r \implies r = \frac{3l}{\pi}$. New magnetic dipole moment $M' = m \times 2r \sin 30^\circ = m \times 2 \times \frac{3l}{\pi} \times \frac{1}{2} = \frac{3M}{\pi}$.

   Step Solution:

    1.  Start with the original moment: $M = \text{pole strength} (m) \times \text{length} (l)$.

    2.  For an arc subtending $60^\circ$ ($\pi/3$ radians), the length $l$ is related to radius $r$ by $l = r(\pi/3)$.

    3.  Rearrange to find the radius: $r = 3l/\pi$.

    4.  The new effective length $l'$ is the straight-line distance between the poles (chord length): $l' = 2r \sin(\theta/2) = 2r \sin(30^\circ)$.

    5.  Substitute $r$ and $\sin 30^\circ$: $M' = m \times (2 \times \frac{3l}{\pi} \times \frac{1}{2}) = \frac{3ml}{\pi} = \frac{3M}{\pi}$.

   Difficulty Level: Medium

   Concept Name: Magnetic Dipole Moment of a Bent Magnet

   Short cut solution: Use the general formula for a magnet bent into an arc of angle $\theta$ (in radians): $M' = M \frac{\sin(\theta/2)}{\theta/2}$. For $60^\circ$ ($\pi/3$ rad): $M' = M \frac{1/2}{\pi/6} = \frac{3M}{\pi}$.

Question 38

   Question: A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by

   Options:

       A. $qvR/2$

       B. $qvR^2/2$

       Note: Source shows options A and D as identical ($qvR/2$)

   Correct Answer: D (or A)

   Year: 2007

   Solution (as Given in the Source): Magnetic moment $\mu = IA$. Since $T = 2\pi R/v$, then $I = q/T = qv/2\pi R$. Therefore, $\mu = (\frac{qv}{2\pi R})(\pi R^2) = \frac{qvR}{2}$.

   Step Solution:

    1.  The magnetic moment of a current loop is $\mu = I \times A$.

    2.  Current $I$ is defined as charge flow per unit time: $I = q/T$.

    3.  The time $T$ for one revolution at speed $v$ is $T = \text{circumference} / \text{speed} = 2\pi R/v$.

    4.  Substitute $T$ to get current: $I = \frac{q}{2\pi R/v} = \frac{qv}{2\pi R}$.

    5.  Calculate moment using area $A = \pi R^2$: $\mu = \frac{qv}{2\pi R} \times \pi R^2 = \frac{qvR}{2}$.

   Difficulty Level: Easy

   Concept Name: Magnetic Moment of a Circulating Charge

   Short cut solution: Use the gyromagnetic ratio relationship $\mu = \frac{q}{2m} L$. For a particle in a circle, angular momentum $L = m v R$. Thus, $\mu = \frac{q}{2m} (m v R) = \frac{qvR}{2}$.

Question 39

   Question: If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by $\mu_d$, $\mu_p$ and $\mu_f$ respectively, then.

   Options: 

       A. $\mu_d = 0$ and $\mu_p \neq 0$

       B. $\mu_d \neq 0$ and $\mu_p = 0$

       C. $\mu_d \neq 0$ and $\mu_f \neq 0$

       D. $\mu_p = 0$ and $\mu_f \neq 0$.

   Correct Answer: A.

   Year: 2005.

   Solution (as Given in the Source): Materials with no unpaired, or isolated electrons are considered diamagnetic. Diamagnetic substances do not have magnetic dipole moments and have negative susceptibilities. However, materials having unpaired electrons whose spins do not cancel each other are called paramagnetic. These substances have positive magnetic moments and susceptibilities. $\mu_d = 0, \mu_p \neq 0$.

   Step Solution:

    1.  Identify that diamagnetic materials consist of atoms where electron spins and orbital moments cancel out, resulting in no net magnetic moment.

    2.  Conclude that for diamagnetic atoms, $\mu_d = 0$.

    3.  Identify that paramagnetic and ferromagnetic materials have atoms with unpaired electrons that provide a permanent magnetic moment.

    4.  Conclude that for these materials, the atomic moments are non-zero: $\mu_p \neq 0$ and $\mu_f \neq 0$.

    5.  Select Option A as it correctly states $\mu_d = 0$ and $\mu_p \neq 0$.

   Difficulty Level: Easy

   Concept Name: Atomic Magnetic Moments of Materials

   Short cut solution: In diamagnetism, individual atoms have zero net magnetic moment due to paired electrons. Only Option A correctly identifies $\mu_d = 0$.

Question 49

   Question: A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of each part will be.

   Options:

       A. M

       B. 2M

       C. zero

       D. 0.5 M.

   Correct Answer: D.

   Year: 1997.

   Solution (as Given in the Source): Magnetic moment = pole strength $\times$ length. $\therefore M' = \frac{M}{2} = 0.5 M$.

   Step Solution:

    1.  Define the initial magnetic moment as $M = m \times l$, where $m$ is the pole strength and $l$ is the original length.

    2.  When the magnet is cut into two parts of equal length, the new length becomes $l' = l/2$.

    3.  Note that cutting a magnet perpendicular to its axis does not change the pole strength ($m$).

    4.  Calculate the new magnetic moment: $M' = m \times l' = m \times (l/2)$.

    5.  Since $M = ml$, substitute to find $M' = M/2 = 0.5M$.

   Difficulty Level: Easy

   Concept Name: Magnetic Moment of a Cut Magnet

   Short cut solution: The magnetic moment $M$ is directly proportional to the length of the magnet for a constant pole strength. If the length is halved, the magnetic moment is also halved, becoming $0.5M$.