Welcome to HSLC Guru! This page brings you a complete, exam-ready guide to Class 12 Physics Chapter 5 — Magnetism and Matter in English Medium, prepared strictly according to the latest ASSEB (Assam State School Education Board) syllabus. Every formula is rendered with KaTeX so that you can read the equations cleanly on mobile or desktop, and every concept is supported by clear diagrams drawn directly in SVG.

Magnetism is one of the most fascinating chapters in Class 12 Physics. From the humble bar magnet sitting on a refrigerator door to the vast magnetic field that surrounds our planet and protects life from solar radiation, the same fundamental principles operate at every scale. In this chapter we will travel from the microscopic origin of magnetism inside atoms, through the bulk magnetic properties of materials, all the way to Earth’s magnetic field and the practical magnetism of permanent magnets and electromagnets.

Chapter Summary

A bar magnet behaves like a magnetic dipole with two poles, North and South, of equal strength $q_m$ separated by an effective magnetic length $2l$. The product $m = q_m \cdot 2l$ defines the magnetic dipole moment, a vector quantity directed from the south pole to the north pole. Unlike electric charges, magnetic monopoles have never been observed in nature; cutting a bar magnet only produces two smaller magnets, each with its own pair of poles.

The magnetic field produced by a bar magnet shows a striking parallel with the electric field of an electric dipole. On the axial line (end-on position) the field is twice as strong as on the equatorial line (broadside position) at the same distance, and the directions of the two fields are mutually opposite. When such a dipole is placed in an external magnetic field, it experiences a torque $\vec{\tau} = \vec{m}\times\vec{B}$ that tends to align $\vec{m}$ with $\vec{B}$, and it stores potential energy $U = -\vec{m}\cdot\vec{B}$.

The Earth itself acts like a giant bar magnet whose magnetic axis is tilted by about $11.3^\circ$ from the geographic axis. Three quantities, called the magnetic elements of Earth, completely specify Earth’s field at a place: the angle of declination (between geographic and magnetic meridians), the angle of dip $\delta$ (between Earth’s field and the horizontal), and the horizontal component $B_H$. Inside matter, the response of a material to an applied field $\vec{H}$ is described by the magnetisation $\vec{M}$ and characterised by the magnetic susceptibility $\chi_m$. Materials are accordingly classified as diamagnetic, paramagnetic or ferromagnetic, the last category producing the spectacular phenomenon of hysteresis.

Key Formulas at a Glance

Quantity	Symbol & Formula	SI Unit
Magnetic dipole moment	$m = q_m \cdot 2l$	$\text{A m}^2$
Field on axis of bar magnet	$B_{axial}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$	tesla (T)
Field on equator	$B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$	tesla (T)
Torque on dipole	$\vec{\tau}=\vec{m}\times\vec{B}$	N m
Potential energy	$U=-\vec{m}\cdot\vec{B}$	joule (J)
Magnetising field	$H = \dfrac{B}{\mu_0}-M$	A/m
Susceptibility	$\chi_m = M/H$	dimensionless
Permeability	$\mu=\mu_0(1+\chi_m)=\mu_0\mu_r$	T m / A
Curie’s law	$\chi = C/T$	—
Earth’s components	$B_H=B\cos\delta$, $B_V=B\sin\delta$	T
Angle of dip	$\tan\delta = B_V/B_H$	—

The displayed forms of the most-used results are:

$$B_{axial}=\dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^3},\qquad B_{eq}=\dfrac{\mu_0}{4\pi}\cdot\dfrac{m}{r^3}$$

$$\vec{\tau}=\vec{m}\times\vec{B}\ \Rightarrow\ \tau=mB\sin\theta$$

$$U(\theta)=-\vec{m}\cdot\vec{B}=-mB\cos\theta$$

$$B=\sqrt{B_H^{2}+B_V^{2}},\qquad \tan\delta=\dfrac{B_V}{B_H}$$

$$\mu_r=1+\chi_m,\qquad \chi(T)=\dfrac{C}{T}\ \text{(Curie’s law)}$$

Diagram 1: Bar Magnet and Its Field Lines

Diagram 2: Earth’s Magnetic Field — Dip and Declination

Diagram 3: Hysteresis Loop of a Ferromagnet

Exercise — Short Answer Questions

Q1. Define magnetic dipole moment of a bar magnet. Write its SI unit and direction.

Answer: The magnetic dipole moment of a bar magnet is the product of the strength of either pole and the magnetic length (the distance between the two poles). If $q_m$ is the pole strength and $2l$ is the magnetic length, then $m = q_m \cdot 2l$. Its SI unit is ampere metre squared ($\text{A m}^2$). It is a vector quantity whose direction is taken from the south pole to the north pole inside the magnet.

Q2. Why do magnetic field lines form closed loops?

Answer: Magnetic field lines form closed continuous loops because magnetic monopoles do not exist. Field lines have neither a beginning nor an end. They emerge from the north pole, travel through the surrounding space, enter the south pole, and continue from south to north inside the magnet, completing a closed loop. This is mathematically expressed by Gauss’s law for magnetism, $\oint \vec{B}\cdot d\vec{A}=0$.

Q3. State and explain Gauss’s law for magnetism.

Answer: Gauss’s law for magnetism states that the net magnetic flux through any closed surface is always zero:

$$\oint_{S}\vec{B}\cdot d\vec{A}=0$$

This is a direct consequence of the absence of isolated magnetic charges (monopoles). For every field line that enters a closed surface, an equal field line leaves it. Hence inward flux equals outward flux, and the total flux vanishes.

Q4. Define angle of declination and angle of dip.

Answer: The angle of declination at a place is the angle between the geographic meridian and the magnetic meridian at that place. The angle of dip $\delta$ is the angle between the direction of the Earth’s total magnetic field $\vec{B}$ and the horizontal at that place. At the magnetic equator $\delta=0^\circ$ and at the magnetic poles $\delta=90^\circ$.

Q5. What are the magnetic elements of Earth?

Answer: The three independent quantities that completely specify the Earth’s magnetic field at a given place are called the magnetic elements of Earth: (i) angle of declination, (ii) angle of dip $\delta$, and (iii) horizontal component of Earth’s field $B_H$. Together they relate to the total field through $B_H = B\cos\delta$ and $B_V = B\sin\delta$, with $\tan\delta = B_V/B_H$.

Q6. Define magnetic susceptibility and magnetic permeability.

Answer: Magnetic susceptibility $\chi_m$ measures how strongly a material gets magnetised when placed in a magnetising field. It is defined by $\chi_m = M/H$, where $M$ is magnetisation and $H$ is the magnetising field. Magnetic permeability $\mu$ measures the ability of a medium to allow magnetic field lines to pass through it: $\mu = \mu_0(1+\chi_m)=\mu_0\mu_r$. The dimensionless ratio $\mu_r=\mu/\mu_0$ is the relative permeability.

Q7. State Curie’s law for paramagnetic materials.

Answer: Curie’s law states that the magnetic susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature:

$$\chi = \dfrac{C}{T}$$

where $C$ is the Curie constant (a positive material-dependent constant) and $T$ is the absolute temperature in kelvin. As temperature rises, thermal agitation disrupts the alignment of atomic dipoles and susceptibility falls.

Q8. What is hysteresis? Define retentivity and coercivity.

Answer: Hysteresis is the phenomenon in which the magnetic flux density $B$ of a ferromagnetic material lags behind the applied magnetising field $H$ during cycles of magnetisation and demagnetisation. The plot of $B$ versus $H$ traces a closed loop called the hysteresis loop. Retentivity (residual magnetism, $B_r$) is the value of $B$ that remains in the material when $H$ is reduced to zero. Coercivity ($H_c$) is the magnitude of reverse field required to reduce $B$ to zero.

Q9. Distinguish between diamagnetic, paramagnetic and ferromagnetic substances.

Property	Diamagnetic	Paramagnetic	Ferromagnetic
Susceptibility $\chi_m$	Small, negative	Small, positive	Very large, positive
Relative permeability $\mu_r$	$\mu_r<1$	$\mu_r>1$ (slightly)	$\mu_r\gg 1$
Behaviour in field	Repelled, weakly	Attracted, weakly	Attracted, strongly
Effect of temperature	Independent	Obeys $\chi=C/T$	$\chi=C/(T-T_c)$ above Curie point
Examples	Bi, Cu, water, gold	Al, Pt, $\text{O}_2$	Fe, Co, Ni, gadolinium

Q10. Why is soft iron used as the core of an electromagnet?

Answer: Soft iron has high permeability, high retentivity but very low coercivity, and a thin hysteresis loop. Therefore it can be magnetised easily to a high value of $B$, it loses its magnetisation almost completely when the current is switched off, and the energy lost per cycle of magnetisation (= area of the loop) is small. These properties are exactly what is required of an electromagnet, and are the reason soft iron — not steel — is chosen for transformer cores, motor cores and electromagnet cores.

Q11. Why is steel preferred for permanent magnets?

Answer: Steel has high retentivity and high coercivity. Its hysteresis loop is broad and tall, so once magnetised it retains a strong magnetisation and a large reverse field is required to demagnetise it. Hence steel (and modern alloys such as Alnico, Ferrites and rare-earth NdFeB) is suitable for permanent magnets, while soft iron is suitable for temporary (electro-) magnets.

Q12. What is the SI unit of magnetisation $M$? Show that $M$ has the same unit as the magnetising field $H$.

Answer: Magnetisation is defined as the magnetic dipole moment per unit volume, $M = m_{net}/V$. Its SI unit is therefore $\text{A m}^2/\text{m}^3 = \text{A m}^{-1}$. The magnetising field $H$ also has the unit $\text{A m}^{-1}$ (it equals the free-current per unit length producing the field in vacuum). Hence both $M$ and $H$ are measured in ampere per metre.

Exercise — Long Answer Questions

Q13. Derive an expression for the magnetic field on the axial line of a short bar magnet.

Answer: Consider a bar magnet of magnetic length $2l$ and pole strength $q_m$, so that $m = q_m\cdot 2l$. Let $P$ be a point on the axial line at a distance $r$ from the centre $O$ of the magnet.

Distance of $P$ from N-pole $= r-l$, distance from S-pole $= r+l$. The field due to N-pole (along NP) and the field due to S-pole (along PS, hence subtractive) are:

$$B_N=\dfrac{\mu_0}{4\pi}\dfrac{q_m}{(r-l)^2},\qquad B_S=\dfrac{\mu_0}{4\pi}\dfrac{q_m}{(r+l)^2}$$

Net axial field along the dipole direction:

$$B_{axial}=B_N-B_S=\dfrac{\mu_0 q_m}{4\pi}\left[\dfrac{1}{(r-l)^2}-\dfrac{1}{(r+l)^2}\right]=\dfrac{\mu_0 q_m}{4\pi}\cdot\dfrac{4rl}{(r^2-l^2)^2}$$

Substituting $m = q_m\cdot 2l$:

$$B_{axial}=\dfrac{\mu_0}{4\pi}\cdot\dfrac{2mr}{(r^2-l^2)^2}$$

For a short magnet $r\gg l$, so $r^2-l^2\approx r^2$ and

$$\boxed{\,B_{axial}=\dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^3}\,}$$

The direction of $\vec{B}_{axial}$ is along $\vec{m}$ (from S to N).

Q14. Derive the magnetic field on the equatorial line of a short bar magnet.

Answer: Let $P$ be on the equatorial line, distance $r$ from centre $O$. Distance from each pole $=\sqrt{r^2+l^2}$. The fields due to N and S have equal magnitudes

$$B_N=B_S=\dfrac{\mu_0}{4\pi}\dfrac{q_m}{r^2+l^2}$$

Their components perpendicular to the axis cancel; the components parallel to the dipole axis add and point opposite to $\vec{m}$:

$$B_{eq}=2B_N\cos\theta=\dfrac{\mu_0}{4\pi}\dfrac{2q_m}{r^2+l^2}\cdot\dfrac{l}{\sqrt{r^2+l^2}}=\dfrac{\mu_0}{4\pi}\dfrac{m}{(r^2+l^2)^{3/2}}$$

For a short dipole ($r\gg l$):

$$\boxed{\,B_{eq}=\dfrac{\mu_0}{4\pi}\cdot\dfrac{m}{r^3}\,}$$

Comparing with the axial result, $B_{axial}=2B_{eq}$ (for the same $r$). The directions of the two fields are mutually opposite — exactly as for an electric dipole.

Q15. Obtain the expression for torque and potential energy of a magnetic dipole in a uniform external field.

Answer: Let a bar magnet of moment $\vec{m}$ make an angle $\theta$ with a uniform external field $\vec{B}$. The N-pole experiences a force $q_m\vec{B}$ and the S-pole an equal and opposite force $-q_m\vec{B}$. These two forces form a couple of magnitude

$$\tau=q_m B\cdot 2l\sin\theta=mB\sin\theta\quad\Rightarrow\quad \vec{\tau}=\vec{m}\times\vec{B}$$

To find the potential energy, integrate the work done against the torque from $\theta_0=90^\circ$ (taken as zero of potential energy) to angle $\theta$:

$$U(\theta)=\int_{90^\circ}^{\theta}\tau\, d\theta=\int_{\pi/2}^{\theta}mB\sin\theta\, d\theta=-mB\cos\theta=-\vec{m}\cdot\vec{B}$$

Thus $U_{min}=-mB$ (when $\vec{m}\parallel\vec{B}$, stable equilibrium) and $U_{max}=+mB$ (when $\vec{m}\uparrow\downarrow\vec{B}$, unstable equilibrium).

Q16. State the properties of magnetic field lines.

Answer: (i) Magnetic field lines are continuous closed curves; outside a magnet they go from N to S and inside they continue from S to N. (ii) The tangent at any point gives the direction of $\vec{B}$ at that point. (iii) The number of lines per unit area (line density) is proportional to the magnitude of $\vec{B}$. (iv) Two field lines never intersect, because at the point of intersection the field would have two different directions, which is impossible. (v) They contract along their length and exert lateral pressure perpendicular to their length.

Exercise — Numerical Problems

N1. A short bar magnet placed with its axis at $30^\circ$ with an external field of $0.25$ T experiences a torque of $4.5\times 10^{-2}$ N m. What is the magnetic moment of the magnet?

Answer: $\tau = mB\sin\theta\Rightarrow m = \dfrac{\tau}{B\sin\theta}=\dfrac{4.5\times 10^{-2}}{0.25\times \sin 30^\circ}=\dfrac{4.5\times 10^{-2}}{0.25\times 0.5}=0.36\ \text{A m}^2$.

N2. A short bar magnet of moment $0.32\ \text{J/T}$ is placed in a uniform external field of $0.15$ T. If the magnet is free to rotate, find (a) the orientation of stable and unstable equilibrium, (b) the potential energy in each case.

Answer: (a) Stable equilibrium occurs when $\vec{m}\parallel\vec{B}$ ($\theta=0^\circ$); unstable when $\vec{m}\uparrow\downarrow\vec{B}$ ($\theta=180^\circ$).

$$U(0)=-mB=-0.32\times 0.15=-0.048\ \text{J}$$

$$U(180^\circ)=+mB=+0.048\ \text{J}$$

N3. A solenoid of $800$ turns and area $2.5\times 10^{-4}\,\text{m}^2$ carries a current of $3.0$ A. Find its magnetic moment.

Answer: $m = NIA = 800\times 3.0\times 2.5\times 10^{-4}=0.6\ \text{A m}^2$.

N4. A magnetic needle has magnetic moment $6.7\times 10^{-2}\,\text{A m}^2$ and moment of inertia $7.5\times 10^{-6}\,\text{kg m}^2$. It performs $10$ complete oscillations in $6.70$ s. What is the magnitude of the magnetic field?

Answer: Period $T=6.70/10=0.670$ s. From $T=2\pi\sqrt{I/(mB)}$,

$$B=\dfrac{4\pi^{2}I}{m T^{2}}=\dfrac{4\pi^{2}\times 7.5\times 10^{-6}}{6.7\times 10^{-2}\times (0.670)^{2}}\approx 0.01\ \text{T}$$

N5. The horizontal component of Earth’s field at a place is $0.30\ \text{G}$ and the angle of dip is $60^\circ$. Find the vertical component and the total field.

Answer: $B_V=B_H\tan\delta=0.30\times\tan 60^\circ=0.30\times\sqrt{3}=0.52\ \text{G}$.

$$B=\sqrt{B_H^{2}+B_V^{2}}=\sqrt{0.30^{2}+0.52^{2}}\approx 0.60\ \text{G}$$

N6. A solenoid has core of length $0.4$ m, area $4\times 10^{-4}\,\text{m}^2$, $1000$ turns, carries $2$ A and is filled with a magnetic material of relative permeability $400$. Find $B$ inside.

Answer: $H = nI = (1000/0.4)\times 2 = 5000\ \text{A/m}$. Therefore

$$B=\mu H=\mu_0\mu_r H=4\pi\times 10^{-7}\times 400\times 5000\approx 2.51\ \text{T}$$

N7. A bar magnet has a magnetic moment of $0.5\ \text{J/T}$ and is placed at $30^\circ$ to a uniform field $B=0.4$ T. Calculate the work done in rotating it from this orientation to (a) $90^\circ$, (b) $180^\circ$.

Answer: $W = U(\theta_2)-U(\theta_1)= -mB(\cos\theta_2-\cos\theta_1)$.

$$W_a=-mB(\cos 90^\circ-\cos 30^\circ)=-0.5\times 0.4\times(0-0.866)=0.173\ \text{J}$$

$$W_b=-mB(\cos 180^\circ-\cos 30^\circ)=-0.5\times 0.4\times(-1-0.866)=0.373\ \text{J}$$

Additional Conceptual Questions

A1. A bar magnet is cut into two equal pieces (a) along its axis and (b) perpendicular to its axis. How does the magnetic moment of each piece compare with the original?

Answer: Originally $m = q_m\cdot 2l$.

(a) Cut along the axis: length $2l$ stays the same, but each piece has half the cross-section, so pole strength becomes $q_m/2$. Hence $m’=(q_m/2)\cdot 2l=m/2$.

(b) Cut perpendicular to the axis: pole strength stays $q_m$ but the length of each piece becomes $l$. Hence $m”=q_m\cdot l = m/2$. In both cases each piece has half the original moment.

A2. Why do diamagnetic substances move from stronger to weaker regions of a magnetic field?

Answer: An external field induces atomic currents in a diamagnetic substance such that the induced moment is opposite to the applied field (Lenz’s law applied at atomic scale). The substance behaves like a tiny dipole anti-parallel to $\vec{B}$ and is therefore repelled from regions of strong field; it tends to move towards the weaker-field region.

A3. The susceptibility of a paramagnetic salt at $300$ K is $1.2\times 10^{-3}$. What will be its susceptibility at $200$ K?

Answer: By Curie’s law $\chi T=$ constant, hence

$$\chi_2=\chi_1\dfrac{T_1}{T_2}=1.2\times 10^{-3}\times\dfrac{300}{200}=1.8\times 10^{-3}$$

A4. Why is the area of the hysteresis loop important?

Answer: The area enclosed by the $B$–$H$ loop equals the energy dissipated as heat per unit volume of the specimen per cycle of magnetisation. Hence transformer cores must have thin loops (low loss) while permanent magnets need thick loops (high coercivity to resist demagnetisation).

A5. Show that the time period of a magnetic needle in a uniform field is $T=2\pi\sqrt{I/(mB)}$.

Answer: For small angular displacement $\theta$, restoring torque $=-mB\sin\theta\approx -mB\theta$. Equation of motion $I\ddot{\theta}=-mB\theta\Rightarrow \omega^{2}=mB/I$, giving

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

A6. At a place the angle of dip is $30^\circ$ and the horizontal component of the Earth’s field is $B_H=0.34\times 10^{-4}$ T. Find the resultant field.

Answer: $B = B_H/\cos\delta = 0.34\times 10^{-4}/\cos 30^\circ = 0.34\times 10^{-4}/0.866 \approx 3.93\times 10^{-5}\ \text{T}$.

A7. State two differences between the magnetic field of a bar magnet and that of a current-carrying solenoid.

Answer: Both produce nearly identical external dipole-type field patterns, so similarities dominate. Differences: (i) Inside the bar magnet, the field is weaker than at its surface, while inside an ideal long solenoid the field is uniform and equal to $\mu_0 nI$. (ii) The magnetic moment of a solenoid can be controlled by varying the current, but that of a permanent bar magnet is fixed.

A8. The horizontal component of Earth’s field is $B_H=0.36\ \text{G}$ at the magnetic equator. Find the magnetic moment of Earth treating it as a dipole. Take $R_E=6.4\times 10^{6}$ m.

Answer: At the equator, the surface field equals the equatorial dipole field:

$$B=\dfrac{\mu_0 m}{4\pi R_E^{3}}\Rightarrow m=\dfrac{4\pi R_E^{3} B}{\mu_0}$$

Substituting $B = 0.36\times 10^{-4}$ T, $\mu_0/4\pi = 10^{-7}$ T m/A and $R_E^{3}=(6.4)^{3}\times 10^{18}=2.62\times 10^{20}$ m³:

$$m=\dfrac{2.62\times 10^{20}\times 0.36\times 10^{-4}}{10^{-7}}\approx 9.43\times 10^{22}\ \text{A m}^2$$

Very Short Answer Questions (1 mark)

VS1. What does the dimensionless quantity $\chi_m$ represent?

Answer: Magnetic susceptibility — the ratio of magnetisation $M$ to magnetising field $H$ for a linear medium.

VS2. Why can’t we have an isolated north pole of a magnet?

Answer: Because magnetic monopoles have not been observed in nature. Cutting a magnet always produces dipoles; the elementary atomic currents responsible for magnetism are themselves loops, which have a north and a south face.

VS3. The magnetic moment of a current-carrying loop is $\vec{m}=NI\vec{A}$. What is the direction of $\vec{A}$?

Answer: Perpendicular to the plane of the loop, given by the right-hand rule with the curling fingers in the direction of conventional current.

VS4. Where on Earth is the angle of dip zero?

Answer: At the magnetic equator.

VS5. What happens to the susceptibility of a ferromagnet at the Curie temperature?

Answer: It falls sharply: above $T_c$ the substance behaves as a paramagnet obeying $\chi=C/(T-T_c)$ (Curie–Weiss law).

VS6. State the SI unit of pole strength.

Answer: Ampere metre (A m).

VS7. Two identical bar magnets each of moment $m$ are placed perpendicular to each other. What is the magnitude of the resultant moment?

Answer: $|\vec{m}_{net}|=\sqrt{m^{2}+m^{2}}=m\sqrt{2}$.

VS8. What is the dimensional formula of magnetic moment?

Answer: $[m]=[A\, L^{2}]=[A\,L^{2}]=[M^{0}L^{2}T^{0}A^{1}]$.

Multiple-Choice Questions

M1. The SI unit of magnetic moment is (a) A m, (b) $\text{A m}^2$, (c) $\text{A m}^{-1}$, (d) $\text{A}^2 \text{m}$.

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

M2. A magnet placed in a non-uniform field experiences (a) only torque, (b) only force, (c) both force and torque, (d) neither.

Answer: (c) both force and torque.

M3. The angle of dip at the magnetic poles is (a) 0°, (b) 30°, (c) 60°, (d) 90°.

Answer: (d) 90°.

M4. The relative permeability of a diamagnetic substance is (a) > 1, (b) < 1, (c) = 1, (d) zero.

Answer: (b) less than 1.

M5. The susceptibility of a paramagnetic substance varies with temperature as (a) $\chi\propto T$, (b) $\chi\propto 1/T$, (c) $\chi\propto T^{2}$, (d) independent of $T$.

Answer: (b) $\chi\propto 1/T$ (Curie’s law).

M6. Soft iron is preferred for the core of a transformer because of its (a) high retentivity, (b) low coercivity, (c) high coercivity, (d) low permeability.

Answer: (b) low coercivity (and a thin hysteresis loop).

M7. The unit of magnetisation $M$ is (a) tesla, (b) A/m, (c) A m, (d) Wb/m.

Answer: (b) ampere per metre.

M8. The relationship $\mu_r = 1+\chi_m$ is valid for (a) only diamagnetic, (b) only paramagnetic, (c) only ferromagnetic, (d) all magnetic substances.

Answer: (d) all magnetic substances.

M9. The substance that loses ferromagnetism above the Curie temperature becomes (a) diamagnetic, (b) paramagnetic, (c) non-magnetic, (d) anti-ferromagnetic.

Answer: (b) paramagnetic.

M10. Two unlike magnetic poles of strengths $0.4\ \text{A m}$ and $0.5\ \text{A m}$ are $10\ \text{cm}$ apart. The force between them is (a) $2\times 10^{-6}$ N attractive, (b) $2\times 10^{-5}$ N repulsive, (c) $2\times 10^{-5}$ N attractive, (d) zero.

Answer: $F=(\mu_0/4\pi)\,q_{m1}q_{m2}/r^{2}=10^{-7}\times(0.4\times 0.5)/(0.1)^{2}=2\times 10^{-6}$ N. Unlike poles attract — option (a).

M11. The retentivity of a substance is large for (a) soft iron, (b) steel, (c) copper, (d) bismuth.

Answer: (b) steel.

M12. The dimensional formula of magnetic susceptibility is (a) $[ML^{0}T^{0}]$, (b) $[M^{0}L^{0}T^{0}]$, (c) $[ML^{2}T^{-2}]$, (d) $[M^{0}L T^{0}]$.

Answer: (b) — it is dimensionless.

M13. The magnetic field at the centre of a current-carrying loop is along (a) the plane of the loop, (b) the radius, (c) the axis of the loop, (d) the tangent to the loop.

Answer: (c) along the axis (perpendicular to the plane of the loop).

M14. The unit of $\mu_0$ is (a) T m / A, (b) T A / m, (c) m / T A, (d) Wb / A.

Answer: (a) T m / A (equivalently H/m).

M15. A magnetic needle suspended in a uniform field oscillates with period $T$. If $B$ is doubled keeping $m$ and $I$ unchanged, the new period is (a) $T$, (b) $T/\sqrt{2}$, (c) $\sqrt{2}\,T$, (d) $2T$.

Answer: (b) $T/\sqrt{2}$, since $T\propto 1/\sqrt{B}$.

Long Answer — Earth’s Magnetism in Detail

Q17. Describe the Earth’s magnetic field. Explain its origin and the meaning of the magnetic meridian.

Answer: The Earth itself behaves like a giant dipole magnet with its magnetic axis tilted at about $11.3^\circ$ to the geographic axis. The geographic North pole lies near the magnetic south of this dipole and the geographic South pole near the magnetic north — the reason the N-pole of a freely suspended needle points (geographically) North. Modern theory attributes the origin of this field to electric currents flowing in molten iron and nickel of the Earth’s outer core; this is the so-called dynamo theory.

A vertical plane passing through the magnetic axis of a freely suspended magnetic needle at a place is called the magnetic meridian of that place. The angle between the magnetic meridian and the geographic meridian (vertical plane through the geographic axis) is the declination. Because the magnetic axis wanders slowly over centuries, declination at a given location also drifts with time — a phenomenon called secular variation.

Q18. Show that $\tan\delta = B_V/B_H$ and write down the relations connecting $B$, $B_H$, $B_V$ and $\delta$.

Answer: Resolve the total field $\vec{B}$ at a place into a horizontal component $B_H$ along the magnetic meridian and a vertical component $B_V$ perpendicular to it. From the right triangle formed:

$$B_H=B\cos\delta,\qquad B_V=B\sin\delta$$

$$\Rightarrow\ \tan\delta=\dfrac{B_V}{B_H},\qquad B=\sqrt{B_H^{2}+B_V^{2}}$$

At the magnetic equator $B_V=0$, so $\delta=0^\circ$ and $B=B_H$. At the magnetic poles $B_H=0$, so $\delta=90^\circ$ and $B=B_V$.

Long Answer — Magnetic Materials

Q19. Explain the molecular theory of magnetisation. How does it explain the behaviour of dia-, para- and ferromagnetic substances?

Answer: Each atom or molecule has electrons that move around the nucleus. The orbital motion and the spin together produce a tiny magnetic moment. In diamagnetic substances, the resultant atomic moment is zero (paired electrons). When an external field is applied, the orbits readjust and induce a small moment opposite to $\vec{B}$ — hence $\chi_m<0$.

In paramagnetic substances, atoms have a permanent dipole moment, but thermal agitation keeps them randomly oriented in zero field. An external field aligns them weakly along $\vec{B}$, giving a small positive $\chi_m$ that decreases with rising temperature (Curie’s law).

In ferromagnetic substances, strong exchange interaction creates regions of common alignment called domains. In an unmagnetised piece, domains are randomly oriented; an applied field grows favourable domains at the expense of the others, producing a very large $M$. Above the Curie temperature thermal energy destroys this alignment and the material becomes paramagnetic.

Q20. Derive $\vec{B}=\mu_0(\vec{H}+\vec{M})$ and obtain $\mu_r=1+\chi_m$.

Answer: Inside a magnetic material, the total field $\vec{B}$ has two contributions: one due to the free currents (giving $\mu_0\vec{H}$) and one due to the bound (Amperian) currents associated with the magnetisation $\vec{M}$ (giving $\mu_0\vec{M}$). Adding these contributions gives

$$\vec{B}=\mu_0(\vec{H}+\vec{M})$$

For a linear magnetic medium, $\vec{M}=\chi_m\vec{H}$, so

$$\vec{B}=\mu_0(1+\chi_m)\vec{H}=\mu_0\mu_r\vec{H}=\mu\vec{H}$$

Comparing, the relative permeability is $\mu_r = 1+\chi_m$ and the absolute permeability is $\mu = \mu_0\mu_r$. For free space $\chi_m=0$, $\mu_r=1$, $\mu=\mu_0$.

Q21. Define magnetic intensity $\vec{H}$ and show its relation to the free current per unit length.

Answer: The magnetic intensity (or magnetising field) $\vec{H}$ is defined by $\vec{H}=\vec{B}/\mu_0-\vec{M}$. Inside a long solenoid carrying free current $I_f$ with $n$ turns per metre, Ampère’s law for $\vec{H}$ gives

$$\oint\vec{H}\cdot d\vec{l}=I_{f,\,enc}\Rightarrow H=nI_f$$

Thus $H$ depends only on free currents and is independent of the magnetic material filling the solenoid. This makes $\vec{H}$ a particularly convenient field for engineering applications.

Worked-Out Numericals — Extended Set

N8. The earth’s magnetic field at a place has horizontal component $B_H = 0.36\ \text{G}$ and dip angle $\delta = 60^\circ$. Find $B_V$ and $B$.

Answer: $B_V = B_H\tan\delta = 0.36\times \sqrt{3}\approx 0.624\ \text{G}$.

$$B=B_H/\cos\delta=0.36/0.5=0.72\ \text{G}=0.72\times 10^{-4}\ \text{T}$$

N9. A magnetic dipole of moment $m=2.0\ \text{A m}^2$ is placed in a uniform field $B=0.5\ \text{T}$ such that $\vec{m}\parallel\vec{B}$. How much work is needed to rotate it through $180^\circ$?

Answer: $W = U(\pi)-U(0) = mB – (-mB) = 2mB = 2\times 2.0\times 0.5 = 2.0\ \text{J}$.

N10. A short bar magnet of moment $0.5\ \text{A m}^2$ is placed in space. Find the magnetic field on its axial line at $20\ \text{cm}$ and on its equatorial line at $20\ \text{cm}$.

Answer: Using $\mu_0/4\pi = 10^{-7}\ \text{T m/A}$, $r=0.20$ m, $r^3 = 8\times 10^{-3}$ m³.

$$B_{axial}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}=10^{-7}\times \dfrac{2\times 0.5}{8\times 10^{-3}}=1.25\times 10^{-5}\ \text{T}$$

$$B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}=10^{-7}\times \dfrac{0.5}{8\times 10^{-3}}=6.25\times 10^{-6}\ \text{T}$$

N11. A toroid with $3000$ turns has mean radius $25\ \text{cm}$ and carries $5\ \text{A}$. The core is filled with iron of $\mu_r = 800$. Find $B$ inside.

Answer: $H = NI/(2\pi r) = 3000\times 5/(2\pi\times 0.25)=9549\ \text{A/m}$.

$$B=\mu_0\mu_r H=4\pi\times 10^{-7}\times 800\times 9549\approx 9.6\ \text{T}$$

N12. A bar magnet has dimensions $5\ \text{cm}\times 1\ \text{cm}\times 1\ \text{cm}$, magnetic moment $m=10^{-1}\ \text{A m}^2$. Calculate its magnetisation $M$.

Answer: $V = 5\times 1\times 1\ \text{cm}^3=5\times 10^{-6}\ \text{m}^3$.

$$M=\dfrac{m}{V}=\dfrac{10^{-1}}{5\times 10^{-6}}=2\times 10^{4}\ \text{A/m}$$

N13. A magnetising field of $1.6\times 10^{3}\ \text{A/m}$ produces a magnetisation of $0.4\ \text{A/m}$ in a substance. Find $\chi_m$ and $\mu_r$. Identify the material.

Answer: $\chi_m = M/H = 0.4/(1.6\times 10^{3})=2.5\times 10^{-4}$. Hence $\mu_r=1+\chi_m\approx 1.00025$. Since $\chi_m$ is small and positive, the material is paramagnetic.

N14. The horizontal component of Earth’s field at a place is $0.4\ \text{G}$. A flat circular coil of $50$ turns and radius $10\ \text{cm}$ is placed in the magnetic meridian. What current must be passed so that a compass placed at the centre shows a deflection of $45^\circ$?

Answer: For a tangent-galvanometer-type set-up, $\tan\theta = B_{coil}/B_H$. The coil’s field is $B_{coil}=\mu_0 NI/(2R)$. With $\theta=45^\circ$, $B_{coil}=B_H$.

$$I=\dfrac{2RB_H}{\mu_0 N}=\dfrac{2\times 0.10\times 0.4\times 10^{-4}}{4\pi\times 10^{-7}\times 50}\approx 0.127\ \text{A}$$

Glossary of Key Terms

Term	Meaning
Pole strength ($q_m$)	Measure of how strongly a magnetic pole attracts/repels other poles; unit A m.
Magnetic dipole moment ($m$)	$m=q_m\cdot 2l$; vector from S to N; unit $\text{A m}^2$.
Magnetisation ($M$)	Net magnetic moment per unit volume; $M = m_{net}/V$; unit A/m.
Magnetising field ($H$)	The component of magnetic field due to free currents alone; unit A/m.
Susceptibility ($\chi_m$)	$\chi_m=M/H$; dimensionless; sign distinguishes dia/para/ferro materials.
Permeability ($\mu$)	$\mu=B/H$; unit T m/A; $\mu_r=\mu/\mu_0$.
Declination	Angle between geographic and magnetic meridians at a place.
Dip ($\delta$)	Angle between Earth’s $\vec{B}$ and the horizontal.
Horizontal component ($B_H$)	$B_H=B\cos\delta$; the part of Earth’s field along the horizontal.
Retentivity ($B_r$)	Magnetic flux density remaining after $H$ is reduced to zero.
Coercivity ($H_c$)	Reverse field needed to bring residual $B$ to zero.
Curie temperature ($T_c$)	Temperature above which a ferromagnet becomes paramagnetic.
Hysteresis	Lag of $B$ behind $H$ during cycles of magnetisation.

Common Pitfalls and Exam Tips

Below is a curated list of the mistakes students most often make in ASSEB Class 12 Physics Chapter 5, along with the correct way to think about each.

P1. Confusing geographic and magnetic poles. The magnetic north pole of a freely suspended needle points geographically north, but it actually points to the magnetic south of the Earth’s dipole. Always remember: like poles repel, so what attracts the needle’s N must itself be a magnetic S.

P2. Forgetting the factor 2 in the axial formula. $B_{axial}=2B_{eq}$ at the same distance for a short dipole. A common slip is to drop the factor 2 in $B_{axial}=(\mu_0/4\pi)(2m/r^{3})$, which leads to wrong numerical answers.

P3. Sign of potential energy. $U=-\vec{m}\cdot\vec{B}$ has a minus sign. At $\theta=0$, $U=-mB$ (minimum, stable). At $\theta=\pi$, $U=+mB$ (maximum, unstable). Forgetting the sign mis-identifies stable and unstable equilibrium.

P4. Using $B$ instead of $H$ inside a magnetic medium. The field due to free currents is described by $H = nI$ for a solenoid; the actual flux density inside the medium is $B=\mu H = \mu_0\mu_r H$, not $\mu_0 nI$. Always insert the relative permeability when a core is present.

P5. Confusing Gauss for $\vec{B}$ and Gauss for $\vec{E}$. $\oint\vec{E}\cdot d\vec{A} = q_{enc}/\varepsilon_0$ but $\oint\vec{B}\cdot d\vec{A}=0$. The right-hand side is zero for $\vec{B}$ because magnetic monopoles do not exist. Many students absent-mindedly write $\mu_0 q_{m,enc}$, which has no physical meaning.

P6. Not converting CGS units. ASSEB question papers sometimes give Earth’s magnetic field in gauss (G), where $1\ \text{G}=10^{-4}\ \text{T}$. Always convert before substituting into formulas that use SI units.

Quick Revision Notes

Use this as a last-night recap before the ASSEB Class 12 Physics examination. Read each line and try to recall the corresponding derivation or definition without looking back.

• A bar magnet is a magnetic dipole; magnetic monopoles do not exist; field lines are continuous closed loops; $\oint\vec{B}\cdot d\vec{A}=0$ everywhere.
• Magnetic moment of a bar magnet $m=q_m\cdot 2l$, vector from S to N; for a current loop $\vec{m}=NI\vec{A}$.
• Field of a short bar magnet: axial $B = (\mu_0/4\pi)(2m/r^{3})$; equatorial $B=(\mu_0/4\pi)(m/r^{3})$; ratio 2:1.
• Torque $\vec{\tau}=\vec{m}\times\vec{B}$ tends to align $\vec{m}$ with $\vec{B}$. Potential energy $U=-\vec{m}\cdot\vec{B}$.
• Stable equilibrium at $\theta=0$, unstable at $\theta=\pi$; period of oscillation $T=2\pi\sqrt{I/(mB)}$.
• Earth’s field is dipole-like with axis tilted $\sim 11.3^\circ$; described by declination, dip $\delta$ and horizontal $B_H$.
• $B_H=B\cos\delta$, $B_V=B\sin\delta$, $\tan\delta=B_V/B_H$, $B=\sqrt{B_H^{2}+B_V^{2}}$.
• Inside matter $\vec{B}=\mu_0(\vec{H}+\vec{M})$, $\vec{M}=\chi_m\vec{H}$, $\mu_r=1+\chi_m$, $\mu=\mu_0\mu_r$.
• Diamagnetic: small negative $\chi$, $\mu_r<1$, repelled by field, $T$-independent.
• Paramagnetic: small positive $\chi$, $\mu_r>1$ slightly, weakly attracted, $\chi=C/T$ (Curie’s law).
• Ferromagnetic: large positive $\chi$, $\mu_r\gg 1$, strongly attracted, exhibits hysteresis. Above Curie temperature it becomes paramagnetic.
• Hysteresis loop area = energy lost per unit volume per cycle. Soft iron: thin loop (electromagnets, transformers); steel: thick loop (permanent magnets).

Conclusion

This chapter ties together a remarkable variety of phenomena under one elegant theoretical umbrella. The magnetic dipole replaces the electric dipole, the magnetisation $\vec{M}$ replaces the polarisation $\vec{P}$, and the macroscopic field equations $\vec{B}=\mu_0(\vec{H}+\vec{M})$ and $\mu_r=1+\chi_m$ allow you to handle every type of material from a copper coin (diamagnetic) to a piece of iron (ferromagnetic) using a single framework.

Make sure you can derive the axial and equatorial fields, write the torque and PE expressions instantly, and reproduce the hysteresis loop diagram with retentivity and coercivity correctly labelled. These derivations and definitions are perennial favourites in ASSEB Class 12 Physics question papers. Best of luck — and keep visiting HSLC Guru for the remaining chapters of your Class 12 syllabus.