Magnetism and Matter: NEET and JEE questions with solutions

Answer: (B) $1280\pi^2$

Concept. A needle of moment $m$ in a field $B$ oscillates like a torsional pendulum with $T = 2\pi\sqrt{\dfrac{I}{mB}}$, so $m$ follows from the measured period.

Step 1. Find the period. $20$ oscillations take $5$ s:
$$T = \frac{5}{20} = 0.25\ \text{s}$$

Step 2. Solve for the moment. Squaring the period formula:
$$m = \frac{4\pi^2 I}{T^2 B} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{(0.25)^2 \times 0.049} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{3.0625 \times 10^{-3}} = 12.8\pi^2 \times 10^{-3}\ \text{A m}^2$$

Step 3. Compare with the given form. $12.8\pi^2 \times 10^{-3} = 1280\pi^2 \times 10^{-5}$, so $x = 1280\pi^2$.

Answer. $x = 1280\pi^2$ (Option B).

Solution by Sreeraj P, M.Sc Physics

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

Concept. Materials are classified by the sign and size of their susceptibility $\chi$.

Step 1. Match each class.

• Diamagnetic: small negative $\chi$, with $-1 \leq \chi < 0$ → II.
• Ferromagnetic: very large positive $\chi$, $\chi >> 1$ → III.
• Paramagnetic: small positive $\chi$ → IV.
• Non-magnetic: $\chi = 0$ → I.

Answer. A-II, B-III, C-IV, D-I (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{M}{2}$

Concept. Bending a magnet keeps the pole strength $m$; only the straight-line distance between the poles changes, and $M = m \times$ (pole separation).

Step 1. Original magnet. $M = mL$.

Step 2. Geometry after bending. Each arm is $\dfrac{L}{2}$ long and the arms meet at $60°$. The two arms and the line joining the ends form an equilateral triangle, so the poles are now $\dfrac{L}{2}$ apart.

Step 3. New moment.
$$M' = m \cdot \frac{L}{2} = \frac{M}{2}$$

Answer. $\dfrac{M}{2}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Zero

Concept. Gauss's law for magnetism: $\displaystyle\oint\vec{B}\cdot d\vec{S} = 0$.

Step 1. Reason. Magnetic monopoles do not exist and field lines form closed loops, so every line that enters a closed surface also leaves it. The inward and outward flux cancel exactly.

Answer. Zero (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$

Concept. Permeability depends only on the material: $\mu = \mu_0\mu_r = \mu_0(1 + \chi)$. The magnetising field is not needed.

Step 1. Relative permeability. $\mu_r = 1 + 599 = 600$.

Step 2. Permeability.
$$\mu = 4\pi \times 10^{-7} \times 600 = 2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$$

Answer. $2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\sqrt{2}$

Concept. Cutting a magnet along its axis halves the pole strength but keeps the length, so each piece has half the moment. Moments add as vectors.

Step 1. Moment of each piece. $\dfrac{M_1}{2}$.

Step 2. Combine at right angles.
$$M_2 = \sqrt{\left(\frac{M_1}{2}\right)^2 + \left(\frac{M_1}{2}\right)^2} = \frac{M_1}{\sqrt{2}}$$

Step 3. Ratio. $\dfrac{M_1}{M_2} = \sqrt{2}$.

Answer. $\sqrt{2}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $500\ \text{A m}^2$

Concept. For a magnet oscillating in a uniform field, $T = 2\pi\sqrt{\dfrac{I}{MB}}$, so $M = \dfrac{4\pi^2 I}{T^2 B}$.

Step 1. Substitute. With $4\pi^2 \approx 39.5$:
$$M = \frac{39.5 \times 0.49}{(8.8)^2 \times 0.5 \times 10^{-4}} = \frac{19.35}{77.44 \times 0.5 \times 10^{-4}} \approx 500\ \text{A m}^2$$

Answer. $500\ \text{A m}^2$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $25 : 7$

Concept. In a vibration magnetometer $f = \dfrac{1}{2\pi}\sqrt{\dfrac{MB}{I}}$. The combination has the same moment of inertia in both arrangements, so $f^2 \propto M$.

Step 1. Net moments. Similar poles on the same side: $M_1 + M_2$. Opposite poles on the same side: $M_1 - M_2$.

Step 2. Ratio of frequencies.
$$\frac{M_1 + M_2}{M_1 - M_2} = \frac{20^2}{15^2} = \frac{16}{9}$$

Step 3. Solve. $9M_1 + 9M_2 = 16M_1 - 16M_2 \Rightarrow 7M_1 = 25M_2$, so $\dfrac{M_1}{M_2} = \dfrac{25}{7}$.

Answer. $25 : 7$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $4$

Concept. $T = 2\pi\sqrt{\dfrac{I}{MB}}$. Cutting parallel to the length changes both $I$ and $M$.

Step 1. Moment of inertia of a piece. Each piece has $\dfrac{1}{4}$ of the mass and the same length; with small breadth $I \approx \dfrac{mL^2}{12}$, so $I' = \dfrac{I}{4}$.

Step 2. Magnetic moment of a piece. The pole strength is shared among the four pieces and the length is unchanged, so $M' = \dfrac{M}{4}$.

Step 3. New period. $\dfrac{I'}{M'} = \dfrac{I}{M}$, so the period is unchanged: $T' = 4$ s.

Answer. $4$ s (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0.1\ \text{A m}^2$

Concept. Cutting perpendicular to the length keeps the pole strength and halves the length; $M = m \times$ length.

Step 1. Length of each half. $\dfrac{10}{2} = 5$ cm $= 0.05$ m.

Step 2. Moment. $M = 2 \times 0.05 = 0.1\ \text{A m}^2$ (half of the original $0.2\ \text{A m}^2$).

Answer. $0.1\ \text{A m}^2$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.25 \times 10^{-5}$ T

Concept. For a short magnet, $B_{\text{axial}} = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ and $B_{\text{eq}} = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$.

Step 1. Find $\dfrac{\mu_0 M}{4\pi}$. At $r = 0.1$ m on the axis:
$$\frac{\mu_0}{4\pi}\frac{2M}{(0.1)^3} = 2 \times 10^{-4} \Rightarrow \frac{\mu_0 M}{4\pi} = 10^{-7}\ \text{T m}^3$$

Step 2. Equatorial field at $0.2$ m.
$$B = \frac{10^{-7}}{(0.2)^3} = \frac{10^{-7}}{8 \times 10^{-3}} = 1.25 \times 10^{-5}\ \text{T}$$

Answer. $1.25 \times 10^{-5}$ T (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.3$ N m

Concept. Torque on a magnetic dipole: $\tau = MB\sin\theta$.

Step 1. Substitute.
$$\tau = 2 \times 0.3 \times \sin 30° = 2 \times 0.3 \times 0.5 = 0.3\ \text{N m}$$

Answer. $0.3$ N m (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $2MB$

Concept. Work done in rotating a dipole from $\theta_1$ to $\theta_2$ equals the rise in potential energy: $W = MB(\cos\theta_1 - \cos\theta_2)$.

Step 1. Substitute. From $\theta_1 = 0°$ to $\theta_2 = 180°$:
$$W = MB(\cos 0° - \cos 180°) = MB(1 + 1) = 2MB$$

Answer. $2MB$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0.5\pi\ \text{A m}^2$

Concept. A current loop of $N$ turns has magnetic moment $M = NIA$.

Step 1. Area. $A = \pi(0.05)^2 = 2.5\pi \times 10^{-3}\ \text{m}^2$.

Step 2. Moment.
$$M = 100 \times 2 \times 2.5\pi \times 10^{-3} = 0.5\pi\ \text{A m}^2$$

Answer. $0.5\pi\ \text{A m}^2$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $6 \times 10^{-5}$ T

Concept. The horizontal component is $B_H = B\cos\delta$, where $\delta$ is the angle of dip.

Step 1. Solve for $B$.
$$B = \frac{B_H}{\cos\delta} = \frac{3 \times 10^{-5}}{\cos 60°} = \frac{3 \times 10^{-5}}{0.5} = 6 \times 10^{-5}\ \text{T}$$

Answer. $6 \times 10^{-5}$ T (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0°$

Concept. The angle of dip is the angle the Earth's total field makes with the horizontal.

Step 1. At the magnetic equator. The field there is horizontal ($B_V = 0$), so a dip needle rests horizontally: $\delta = 0°$. (At the magnetic poles it is $90°$.)

Answer. $0°$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\tan\delta = \dfrac{\sqrt{3}}{2}$

Concept. In a vertical plane at angle $\theta$ to the magnetic meridian, the dip needle feels only the component $B_H\cos\theta$ of the horizontal field, while the vertical component $B_V$ is unchanged.

Step 1. Apparent dip.
$$\tan\delta' = \frac{B_V}{B_H\cos\theta} = \frac{\tan\delta}{\cos\theta}$$

Step 2. Solve for the true dip.
$$\tan\delta = \tan\delta'\cos\theta = \tan 45° \times \cos 30° = \frac{\sqrt{3}}{2}$$

Answer. $\tan\delta = \dfrac{\sqrt{3}}{2}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) diamagnetic

Concept. The sign of $\chi = \mu_r - 1$ identifies the type of material.

Step 1. Susceptibility. $\chi = 0.9995 - 1 = -5 \times 10^{-4}$.

Step 2. Classify. A small negative susceptibility ($\mu_r$ slightly less than $1$) is diamagnetic. A superconductor would have $\chi = -1$.

Answer. Diamagnetic (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.5 \times 10^{-4}$

Concept. Curie's law for a paramagnet: $\chi \propto \dfrac{1}{T}$ with $T$ in kelvin.

Step 1. Temperatures. $27°$C $= 300$ K and $327°$C $= 600$ K.

Step 2. New susceptibility.
$$\chi_2 = \chi_1\frac{T_1}{T_2} = 3 \times 10^{-4} \times \frac{300}{600} = 1.5 \times 10^{-4}$$

Answer. $1.5 \times 10^{-4}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.32\pi$ T

Concept. The solenoid sets the magnetic intensity $H = nI$; the core multiplies the field: $B = \mu_0\mu_r H$.

Step 1. Magnetic intensity. $H = 1000 \times 2 = 2000$ A/m.

Step 2. Field with the core.
$$B = 4\pi \times 10^{-7} \times 400 \times 2000 = 0.32\pi\ \text{T} \approx 1.0\ \text{T}$$

Answer. $0.32\pi$ T (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $200$

Concept. Susceptibility is the magnetisation per unit magnetising field: $\chi = \dfrac{M}{H}$.

Step 1. Substitute.
$$\chi = \frac{3 \times 10^5}{1500} = 200$$
Such a large value means the sample is ferromagnetic.

Answer. $200$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) high retentivity and high coercivity

Concept. A permanent magnet must keep its magnetisation once the magnetising field is removed and must not lose it easily.

Step 1. Required properties.

• Keeping a large magnetisation at $H = 0$ needs high retentivity.
• Resisting demagnetisation by stray fields needs high coercivity.

Steel and alnico meet both. Electromagnet and transformer cores need the opposite (low coercivity, small hysteresis loss).

Answer. High retentivity and high coercivity (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) the energy lost as heat per unit volume per cycle

Concept. In one cycle of magnetisation the work done per unit volume on the material is $\oint H\,dB$, which is the area enclosed by the $B$-$H$ loop.

Step 1. Interpret. This energy is not recovered; it appears as heat. So the loop area measures the hysteresis loss per unit volume per cycle. Soft iron has a thin loop, which is why it is used in transformer cores.

Answer. The energy lost as heat per unit volume per cycle (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) moves from the stronger to the weaker region

Concept. A diamagnetic material acquires a weak magnetisation opposite to the applied field, so its energy is lower where the field is weaker.

Step 1. Direction of the force. The force pushes the sample from the stronger towards the weaker field region. (Paramagnets and ferromagnets are pulled the other way.)

Answer. It moves from the stronger to the weaker region (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 0.36

Concept. Torque on a magnetic dipole: $\tau = MB\sin\theta$.

Step 1. Solve for $M$.
$$M = \frac{\tau}{B\sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = 0.36\ \text{A m}^2$$

Answer. $0.36$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{1}{16\pi}$

Concept. $B=\mu_0\mu_rH\Rightarrow H=\dfrac{B}{\mu_0\mu_r}$.

Step 1. Apply. $\dfrac{1}{4\pi\times10^{-7}\times400}=\dfrac{10^5}{16\pi}$ A/m.

Answer. $\alpha=\dfrac{1}{16\pi}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 12

Concept. P is on the axis of the left magnet and on the equatorial line of the right one, $5$ cm from each. Axial field $\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ points along $\vec m$; equatorial field $\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$ points opposite to $\vec m$. The two are perpendicular, so add them as vectors.

Step 1. Axial field. $B_1=10^{-7}\times\dfrac{2\times3\sqrt5}{(0.05)^3}=10^{-7}\times6\sqrt5\times8000=4.8\sqrt5\times10^{-3}$ T, to the right.

Step 2. Equatorial field. Half as large: $B_2=2.4\sqrt5\times10^{-3}$ T, upward (the right magnet's moment points down, S to N).

Step 3. Resultant. $B=\sqrt{B_1^2+B_2^2}=2.4\sqrt5\times10^{-3}\sqrt{4+1}=2.4\times5\times10^{-3}=12\times10^{-3}$ T.

Answer. $\alpha=12$.

Solution by Sreeraj P, M.Sc Physics

Answer: 64

Concept. The torque gives the moment ($\tau=mB\sin\theta$); then the work from stable to unstable is $2mB$.

Step 1. Moment. $m=\dfrac{0.016}{0.08\times0.5}=0.4$ A m$^2$ (800 G $=0.08$ T).

Step 2. Work. $2\times0.4\times0.08=0.064$ J.

Answer. $\alpha=64$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4\%$

Concept. On the perpendicular bisector of a magnet of length $2l$ (pole strength $q_m$), $B=\dfrac{\mu_0}{4\pi}\dfrac{q_m(2l)}{(d^2+l^2)^{3/2}}\propto\dfrac{l}{(d^2+l^2)^{3/2}}$. Both $l$ and $d$ are lengths measured with $1\%$ uncertainty; maximum errors add.

Step 1. Error in $d^2+l^2$. $\Delta(d^2+l^2)=2d^2\dfrac{\Delta d}{d}+2l^2\dfrac{\Delta l}{l}=2(d^2+l^2)\times1\%$, so its relative error is $2\%$.

Step 2. Error in $B$. $\dfrac{\Delta B}{B}=\dfrac{\Delta l}{l}+\dfrac32\times2\%=1\%+3\%=4\%$.

Answer. $4\%$ (Option A).

Solution by Sreeraj P, M.Sc Physics

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

Concept. Recall the unit of each quantity.

Step 1. Units. Magnetic induction $B$: tesla or gauss (III). Magnetic intensity $H$: A/m (IV). Magnetic flux: weber (II). Magnetic moment: A m$^2$ (I).

Answer. (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $8\times10^{-2}\ \text{J}$

Concept. Energy of a dipole $U=-mB\cos\theta$. Most stable: $\theta=0$ ($U=-mB$); most unstable: $\theta=180^\circ$ ($U=+mB$). Work needed $=2mB$.

Step 1. Apply. $2\times0.5\times8\times10^{-2}=8\times10^{-2}$ J.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 6

Concept. Magnetic scalar potential on the axis of a short dipole: $V=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^2}$.

Step 1. Solve. $m=\dfrac{1.5\times10^{-5}\times0.04}{10^{-7}}=6$ A m$^2$.

Answer. $6$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A, C only

Concept. A paramagnet has a small positive susceptibility: it is magnetised weakly along the field and pulled weakly towards stronger field.

Step 1. Check. A: aligns along the field, true. B: "strongly" attracted is ferromagnetic behaviour, false. C: $\chi$ slightly above zero, true. D: moving from strong to weak field is diamagnetic behaviour, false.

Answer. A, C only (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 10

Concept. A magnet is demagnetised when the reverse $H=nI$ reaches its coercivity.

Step 1. Solve. $I=\dfrac{H}{n}=\dfrac{5\times10^3}{150/0.3}=10$ A.

Answer. $10$.

Solution by Sreeraj P, M.Sc Physics

Answer: 28

Concept. Magnetic moment $=$ pole strength $\times$ straight distance between the poles. Bending into a semicircle of arc length $l$: radius $\dfrac l\pi$, poles a diameter $\dfrac{2l}{\pi}$ apart.

Step 1. New moment. $m'=\dfrac{2}{\pi}m=\dfrac{2\times44\times7}{22}=28$ A m$^2$.

Answer. $28$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4\ \text{J}$

Concept. Work done in rotating a dipole: $W = mB(\cos\theta_1 - \cos\theta_2)$.

Step 1. Substitute. From parallel ($0°$) to antiparallel ($180°$):
$$W = mB(\cos 0° - \cos 180°) = 2mB = 2 \times 5 \times 0.4 = 4\ \text{J}$$

Answer. $4$ J (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.2\times10^3\ \text{A m}^{-1}$

Concept. Magnetic intensity inside a solenoid $H=nI$ (independent of the core).

Step 1. Apply. $\dfrac{1200}{2}\times2=1200$ A m$^{-1}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Both Statement I and Statement II are true

Concept. Diamagnets have small negative $\chi$ (down to $-1$ for a perfect diamagnet) and move towards weaker field.

Answer. Both statements true (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2\%$

Concept. Filling with a material multiplies $B$ by $\mu_r=1+\chi$; the increase is $\chi$ as a fraction.

Step 1. Apply. $2\times10^{-2}=2\%$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Statement I is incorrect but Statement II is true

Concept. Diamagnetism comes from the orbital motion of electrons reacting to the applied field (like Lenz's law); it does not depend on thermal alignment of permanent moments.

Step 1. Statement I. Diamagnetic susceptibility is (almost) independent of temperature: false.

Step 2. Statement II. The induced moment always opposes the magnetising field, which is why $\chi\lt0$: true.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 243

Concept. A compass needle swings in the horizontal plane, so $f\propto\sqrt{B_H}=\sqrt{B\cos\delta}$.

Step 1. Ratio. $\left(\dfrac{20}{30}\right)^2=\dfrac{B_1\cos30^\circ}{B_2\cos60^\circ}=\dfrac{B_1}{B_2}\sqrt3$.

Step 2. Solve. $\dfrac{B_1}{B_2}=\dfrac{4}{9\sqrt3}=\dfrac{4}{\sqrt{243}}$.

Answer. $x=243$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Inside a long solenoid the magnetic intensity is $H=nI$, with $n$ the turns per metre.

Step 1. Turns per metre. $8$ per cm $=800$ per m.

Step 2. Current. $I=\dfrac Hn=\dfrac{1.6\times10^3}{800}=2$ A.

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both A and R are correct and R is the correct explanation of A

Concept. An electromagnet must become strongly magnetised when current flows (high permeability) and lose its magnetism when the current is switched off (low retentivity, low coercivity).

Step 1. Assertion. Electromagnets use soft iron: true.

Step 2. Reason. Soft iron has high permeability and low retentivity: true, and these are exactly the properties that make it suitable.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 6

Concept. To demagnetise the magnet, the solenoid must supply a magnetic intensity $H = nI$ equal to the given (coercive) intensity.

Step 1. Turns per metre. $n = \dfrac{60}{0.15} = 400\ \text{m}^{-1}$.

Step 2. Current.
$$I = \frac{H}{n} = \frac{2.4 \times 10^3}{400} = 6\ \text{A}$$

Answer. $6$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) high permeability and low retentivity.

Concept. An electromagnet must magnetise strongly (high permeability) and lose its magnetism when the current stops (low retentivity, low coercivity).

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Both Statement I and Statement II are true.

Concept. Para- and ferromagnetic susceptibilities fall as temperature rises (Curie law, Curie–Weiss law). Diamagnetism arises from induced orbital moments opposing the field (Lenz's law).

Answer. Both true (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $14\ \text{J}$

Concept. $W=mB(\cos\theta_1-\cos\theta_2)$.

Step 1. Apply. $2\times10^5\times14\times10^{-5}\times(1-\tfrac12)=14$ J.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4\pi\times10^{-5}$

Concept. $\mu=\mu_0(1+\chi)$.

Step 1. Apply. $100\times4\pi\times10^{-7}=4\pi\times10^{-5}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\tan^{-1}\left(\sqrt{\dfrac32}\right)$

Concept. In a vertical plane at angle $\beta$ to the meridian, only $B_H\cos\beta$ acts horizontally, so the apparent dip satisfies $\tan\delta'=\dfrac{\tan\delta}{\cos\beta}$.

Step 1. True dip. $\tan\delta=\tan60^\circ\cos45^\circ=\dfrac{\sqrt3}{\sqrt2}$.

Answer. $\tan^{-1}\sqrt{\tfrac32}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $8 : 3$

Concept. A magnet oscillating in a field: $T=2\pi\sqrt{\dfrac{I}{mB}}$, so $m\propto\dfrac{I}{T^2}$ (same $B$).

Step 1. Ratio. $\dfrac{m_1}{m_2}=\dfrac32\times\dfrac{16}{9}=\dfrac83$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\sqrt3 : 4$

Concept. Same: $f^2\propto B\cos\delta$.

Step 1. Ratio. $4=\dfrac{B_P\cos30^\circ}{B_Q\cos60^\circ}=\sqrt3\dfrac{B_P}{B_Q}$.

Answer. $B_Q:B_P=\sqrt3:4$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 8

Concept. Same formula, solved for $B$: $B=\dfrac{4\pi^2I}{mT^2}$.

Step 1. Period. $T=\dfrac5{10}=0.5$ s.

Step 2. Field. $B=\dfrac{4\times9.85\times5\times10^{-6}}{9.85\times10^{-2}\times0.25}=8\times10^{-3}$ T.

Answer. $8$ mT.

Solution by Sreeraj P, M.Sc Physics

Answer: 250

Concept. Moment of the core $=\chi HV$, with $H$ and $V$ unchanged, so it scales with $\chi=\mu_r-1$.

Step 1. Change. $\dfrac{749-499}{499}=\dfrac{250}{499}$.

Answer. $x=250$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) may increase or decrease in size and change its orientation.

Concept. In a ferromagnet, an applied field makes the domains aligned with it grow at the expense of the others, and also turns the moments of domains towards the field.

Step 1. Apply. So some domains grow and others shrink, and orientations change.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\tan^{-1}\frac{\sqrt3}{2}$

Concept. Same relation, $\tan\delta=\tan\delta'\cos\beta$.

Step 1. Apply. $\tan\delta=1\times\dfrac{\sqrt3}{2}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2\pi\times10^{-4}\ \text{H m}^{-1}$

Concept. Same: $\mu=\mu_0(1+\chi)=500\mu_0$.

Step 1. Apply. $500\times4\pi\times10^{-7}=2\pi\times10^{-4}$ H m$^{-1}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Statement I is true but Statement II is false.

Concept. Thermal agitation destroys the alignment inside domains. Above the Curie temperature a ferromagnet becomes paramagnetic.

Step 1. Statement I. True.

Step 2. Statement II. At high temperature the domain structure breaks down (the domains disappear); the domain wall area does not grow. False.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) True dip is less than apparent dip.

Concept. $\tan\delta'=\dfrac{\tan\delta}{\cos\beta}\ge\tan\delta$ since $\cos\beta\le1$.

Answer. The true dip is less than (or equal to) the apparent dip (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Concept. Find the field of $m_1$ at the position of $m_2$, then $\tau=m_2B\sin\theta$.

Step 1. Field at P. P lies on the equatorial line of $m_1$ (which points up), at $r=1$ m: $B=\dfrac{\mu_0}{4\pi}\dfrac{m_1}{r^3}=10^{-7}$ T, pointing down (opposite to $\vec m_1$).

Step 2. Torque. $\vec m_2$ is horizontal, perpendicular to $\vec B$: $\tau=1\times10^{-7}\times\sin90^\circ=1\times10^{-7}$ N m.

Answer. $1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) (a) − (iii), (b) − (i), (c) − (iv), (d) − (ii)

Concept. Build each dimension from a defining equation.

Step 1. Magnetic induction. $F=qvB$: $B=\dfrac{F}{qv}=\dfrac{\mathrm{MLT^{-2}}}{\mathrm{AT\cdot LT^{-1}}}=\mathrm{MT^{-2}A^{-1}}$ (iii).

Step 2. Flux. $\phi=BA$: $\mathrm{ML^2T^{-2}A^{-1}}$ (i).

Step 3. Permeability. $B=\dfrac{\mu_0I}{2\pi r}$: $\mu=\dfrac{Br}{I}=\mathrm{MLT^{-2}A^{-2}}$ (iv).

Step 4. Magnetisation. Moment per volume: $\dfrac{\mathrm{AL^2}}{\mathrm{L^3}}=\mathrm{M^0L^{-1}A}$ (ii).

Answer. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $5\times10^2\ \text{A m}^2$

Concept. $M=\chi H$ with $\chi=\mu_r-1$, $H=nI$; moment $=MV$.

Step 1. Values. $H=1000\times0.5=500$ A/m; $\chi=999$; $M\approx5\times10^5$ A/m.

Step 2. Moment. $5\times10^5\times10^{-3}=5\times10^2$ A m$^2$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\phi_i = -\phi_0$

Concept. Magnetic field lines form closed loops: every line going up through the coil comes back down through the rest of the plane. Net flux through the whole infinite plane is zero.

Step 1. Balance. $\phi_i+\phi_0=0$.

Answer. $\phi_i=-\phi_0$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $P$: Large retentivity, large coercivity

Concept. A permanent magnet must keep its magnetism (large retentivity) and resist demagnetisation (large coercivity). A transformer core needs the opposite (small hysteresis loss).

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Zero

Concept. A perfect diamagnet ($\chi=-1$) expels the field completely (like a superconductor), so no field reaches the cavity, whatever fills it.

Answer. Zero (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $7.2\times10^{-2}\ \text{J}$

Concept. The torque reading gives $m$: $\tau=mB\sin\theta$. Turning from stable ($\theta=0$) to unstable ($\theta=180^\circ$) needs $W=mB(\cos0-\cos180^\circ)=2mB$.

Step 1. Moment. $m=\dfrac{0.018}{0.06\times\sin30^\circ}=0.6$ A m$^2$.

Step 2. Work. $W=2\times0.6\times0.06=7.2\times10^{-2}$ J.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $0.75\ \text{A/m}$

Concept. Curie's law: $M\propto\dfrac BT$.

Step 1. Apply. $6\times\dfrac{0.3}{0.4}\times\dfrac{4}{24}=0.75$ A/m.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.02\ \text{J}$

Concept. Reversing the moment (from along $\vec B$ to against it) needs $W=2mB$, with $B$ at that instant.

Step 1. Field at $t=1$ s. $B=\cos(0.125)=0.992$ T.

Step 2. Work. $2\times10^{-2}\times0.992\approx0.02$ J.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $T_h = T_c$

Concept. $T\propto\sqrt{\dfrac Im}$. The hoop has twice the moment of inertia of the cylinder ($MR^2$ vs $\tfrac12MR^2$) and twice the moment.

Step 1. Ratio. $\dfrac{T_h}{T_c}=\sqrt{\dfrac{2}{2}}=1$.

Answer. $T_h=T_c$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $6.5\times10^{-5}\ \text{N}$

Concept. Hanging freely, the needle lies along the total field, so dip $=45^\circ$ and $B_V=B_H$. Held horizontal, the vertical component exerts a torque $mB_V$, which the applied force must balance at a lever arm of half the length.

Step 1. Moment. $m=1.8\times0.12=0.216$ A m$^2$.

Step 2. Torque. $0.216\times18\times10^{-6}=3.89\times10^{-6}$ N m.

Step 3. Force. $\dfrac{3.89\times10^{-6}}{0.06}=6.5\times10^{-5}$ N.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2600\ \text{A/m}$

Concept. Same: coercivity $=nI$.

Step 1. Apply. $\dfrac{100}{0.2}\times5.2=2600$ A/m.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $3.3\times10^{-4}$

Concept. Magnetisation $M=\dfrac{\text{moment}}{\text{volume}}$; susceptibility $\chi=\dfrac MH$.

Step 1. Magnetisation. $\dfrac{20\times10^{-6}}{10^{-6}}=20$ A/m.

Step 2. Susceptibility. $\dfrac{20}{6\times10^4}=3.3\times10^{-4}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $3.267\times10^{-4}$

Concept. Curie's law for paramagnets: $\chi\propto\dfrac1T$.

Step 1. Apply. $2.8\times10^{-4}\times\dfrac{350}{300}=3.267\times10^{-4}$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $0.7$

Concept. Same: $f^2\propto B\cos\delta$.

Step 1. Ratio. $\dfrac{9}{16}=\dfrac{B_1\cos45^\circ}{B_2\cos30^\circ}$.

Step 2. Solve. $\dfrac{B_1}{B_2}=\dfrac{9}{16}\times\dfrac{0.866}{0.707}=0.69$.

Answer. About $0.7$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1\ \text{mA}$

Concept. To demagnetise the ferromagnet the solenoid must supply a reverse $H$ equal to the coercivity, the value of $H$ where the loop crosses $B=0$. In a solenoid $H=nI$.

Step 1. Coercivity. From the graph, $B=0$ at $H=\pm100$ A/m.

Step 2. Current. $n=1000$ per cm $=10^5$ per m: $I=\dfrac{100}{10^5}=10^{-3}$ A.

Answer. $1$ mA (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $6.65$ s

Concept. $T=2\pi\sqrt{\dfrac{I}{mB}}$.

Step 1. Period. $2\pi\sqrt{\dfrac{7.5\times10^{-6}}{6.7\times10^{-4}}}=2\pi\times0.1058=0.665$ s.

Step 2. Ten oscillations. $6.65$ s.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Zero potential energy when the torque is maximum

Concept. $U=-mB\cos\theta$ and $\tau=mB\sin\theta$. At $\theta=90^\circ$, $U=0$ and $\tau$ is maximum.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\chi = \dfrac{\mu}{\mu_0} - 1$

Concept. $B=\mu H=\mu_0(1+\chi)H$, so $\mu_r=\dfrac{\mu}{\mu_0}=1+\chi$.

Answer. $\chi=\dfrac{\mu}{\mu_0}-1$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $-80\ \text{J}$

Concept. $\tau=mB\sin\theta$ gives $mB$; then $U=-mB\cos\theta$.

Step 1. $mB$. $\dfrac{80\sqrt3}{\sqrt3/2}=160$.

Step 2. Energy. $-160\times\tfrac12=-80$ J.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac65\times10^{-3}\%$

Concept. Same: fractional increase $=\chi$.

Step 1. Apply. $1.2\times10^{-5}=1.2\times10^{-3}\%=\dfrac65\times10^{-3}\%$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Concept. A magnetic monopole would be a point where field lines begin or end.

Step 1. Reason. Magnetic field lines have no beginning or end: they form closed loops (Gauss's law for magnetism, net flux through any closed surface is zero). True.

Step 2. Assertion. Since no line starts or stops anywhere, there is no isolated pole: true, and it follows from the reason.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics