NCERT Solutions Class 12 Physics Chapter 5 Magnetism and Matter

Vector picture. Pointing the right-hand fingers from m (along the magnet's axis from S to N) toward B, the thumb points along τ. That is the rotation axis around which the magnet tries to swing. The torque drives the magnet toward θ = 0 (stable equilibrium) — i.e., m aligns parallel to B.

Why two units for m? Magnetic moment has SI units A m² current × area, for a current loop. For a bar magnet, an equivalent unit is J T^-1 (energy per tesla), because the magnet's potential energy in a field is U = -mBcosθ. Both are identical: 1 A m² = 1 J T^-1. Verify: T = kg s^-2A^-1, so J T^-1 = kg m²s^-2-1s²A = m²A. Same thing.

Common mistake. Students sometimes plug the angle in degrees into a calculator set to radians. If sin 30 is computed in radians it gives sin 30 rad ≈ -0.988 — completely wrong. Always check your calculator mode, or remember the standard exact values: sin 30^∘ = 1/2, sin 45^∘ = 1/2, sin 60^∘ = 3/2.

Sanity check. A typical small bar magnet has m∼ 0.1 to 1 A m². Our 0.36 A m² sits comfortably in that range — answer looks physical.

Real-world connection. Compass needles use this exact principle. Earth's horizontal field ∼ 310^-5 T exerts a tiny torque on the needle whose magnetic moment is ∼ 0.01 A m², enough to swing it within seconds toward magnetic north.

Why stability comes from the minimum of U. Imagine a marble in a bowl (minimum of potential) — push it slightly, it rolls back. That's stable. A marble balanced on top of an inverted bowl (maximum of potential) — any tiny push and it rolls off forever. That's unstable. Same logic with the magnet's angular position: at θ = 0^∘, any small kick produces a restoring torque τ = -mBsinθ that pushes θ back to zero. At θ = 180^∘, any small kick produces a torque that swings θ further away from 180^∘, eventually flipping the magnet all the way to 0^∘.

Energy difference between extremes. Δ U = U_180^∘ - U_0^∘ = +0.048 - (-0.048) = 0.096 J = 2mB. This is the work an external agent must do to flip the magnet from parallel to antiparallel — important for things like NMR/MRI, where radio-frequency photons supply exactly this energy gap to flip nuclear spins.

Small-oscillation period. Near the stable position, for small angles sinθ ≈ θ, the restoring torque is τ ≈ -mBθ — a linear restoring "spring." If the magnet has moment of inertia I, it oscillates with period T = 2π√ImB. This is the principle behind vibration-magnetometer measurements of B in laboratory experiments.

Common pitfall. Don't confuse "stable equilibrium" (lowest energy) with "highest torque" maximum at θ = 90^∘. They're different concepts. At θ = 90^∘, torque is largest but U is neither min nor max — it's just zero (a particular value, not an extremum of U).

Why this works — the magnetic-equivalence theorem. Ampère first noticed that the far-field of any current loop is mathematically identical to the field of a tiny bar magnet (a "magnetic dipole") with moment m = IA, where A is the area vector. This isn't a coincidence — it's because the curl of the magnetic field is sourced by current, and a small current loop and a small magnetic dipole produce the same circulating-field signature when viewed from far away. So the solenoid here is, magnetically, indistinguishable from a bar magnet with m = 0.60 A m².

Right-hand grip rule — the practical recipe. Curl the fingers of your right hand in the direction the current flows around each loop. The thumb points along m, i.e., from the South face of the solenoid toward its North face. (Equivalently: when looking at the end of the solenoid, if the current looks anticlockwise, that end is the North pole; if clockwise, it's the South pole.)

Where the limit breaks down. The "equivalent bar magnet" description is excellent far from the solenoid. Inside the solenoid (or very close to one end), the actual field of the solenoid differs from a true bar magnet — it's nearly uniform inside B = ₀ n I, while a real bar magnet has a complicated, non-uniform internal field due to atomic dipoles. For exam problems about torque or field outside, treat as a bar magnet; for problems about field inside, use B = ₀ n I instead.

Common mistake. Forgetting the factor N. A single loop has moment IA; a solenoid with N loops has moment NIA. Many students drop the N and get an answer 800× too small here.

Why the solenoid behaves identically to a bar magnet. Earlier we computed the solenoid's equivalent magnetic moment as m = NIA. Once you have m, the field doesn't care whether that m came from a permanent magnet's atomic dipoles or from a coil's current — both produce the same external field and feel the same external torque τ = mBsinθ. This is a key idea: the magnetic moment fully summarises the dipolar behaviour of an object.

Direction of the torque (vector view). Using the right-hand rule on τ = m×B:

• Point fingers along m (solenoid's axis from S to N).
• Curl them toward B (horizontal direction).
• Thumb points along τ — vertical.

Since the solenoid is constrained to rotate about the vertical, this is exactly the axis along which the torque acts; the rotation will be in the horizontal plane.

Alternative method — using U = -m·B. The torque is related to potential energy via τ = -dU/dθ: U(θ) = -mBcosθ ⇒ τ = -dUdθ = -mBsinθ. The magnitude is mBsinθ — same answer.

Real-world parallel. Galvanometers work on exactly this principle. A small current-carrying coil sits between magnet poles. When current flows, the coil acquires moment m = NIA, and the external field exerts a torque proportional to the current. A spring resists, and the angle of deflection is read off as the meter reading.

Watch out. The problem prints J for torque units in Q 5.1 and Q 5.4. Strictly, torque is N m, not J, but they're dimensionally identical. The distinction is that J is normally reserved for energy. Don't let the unit confuse you.

Why work to 180^∘ is exactly twice work to 90^∘. Look at the formula W = mB1-cos_f. At 90^∘, 1-cos = 1-0 = 1. At 180^∘, 1-cos = 1-(-1) = 2. The ratio is exactly 2:1. This is a clean result that often appears in JEE/NEET problems — flipping a dipole completely takes 2mB of work, regardless of m or B.

Why τ = 0 at θ = 180^∘ even though it's unstable. Equilibrium just means zero net torque. There are TWO orientations with zero torque: the stable one θ = 0 and the unstable one θ = 180^∘. In both, the field and the moment are colinear, so m×B = 0. The difference is what happens when you perturb slightly: at θ = 0, the torque restores; at θ = 180^∘, the torque amplifies the deviation. (Think of a pencil balanced on its tip vs lying on the table — both equilibria, one stable, one not.)

Integration check. Work done against the field-torque, computed by integration: W = _0^f τ(θ) dθ = _0^f mBsinθ dθ = mB1 - cos_f. Same formula — comforting.

Where this matters in real physics. The energy difference Δ U = 2mB between aligned and antialigned dipoles in a field underlies NMR, MRI, and ESR (electron spin resonance) spectroscopies. In NMR, you flip nuclear spins between aligned and antialigned states using radio-frequency photons; the photon energy must equal 2_n B where _n is the nuclear magnetic moment.

Common mistake. Confusing "work done by the field" with "work done by the external agent." The work done by the field is W_field = -W_external — they're negatives of each other. Here we report the work the external agent does, which is positive (energy goes into the system as it's pushed up the potential hill).

Why the net force vanishes in a uniform field. The force on a dipole is F = ∇(m·B). In a uniform field, B is constant everywhere, so its gradient is zero — hence F = 0. This is one of the cleanest statements in dipole physics. A bar magnet only feels a net force when the field is NON-uniform (e.g., near another magnet's pole, near an iron core, etc.); a uniform field gives only torque.

Compare with a moving charge. A moving point charge in a uniform field feels a force F = qv×B. A dipole at rest feels zero force in the same field. This is a key distinction — currents and charges feel qv×B, but a complete current loop (dipole) integrates to zero net force in a uniform field.

Force on a dipole in a NON-uniform field. For comparison, near a bar magnet's pole, the field gradient is significant and the dipole feels F_z = m dB_zdz. This is why one magnet attracts another, and why iron filings get pulled toward poles. It's also the principle behind Stern–Gerlach experiments, where a non-uniform field is used to separate atoms by their spin orientation.

Alternative: torque from m×B. Here m lies in the horizontal plane and B lies in the horizontal plane at 30^∘. The cross product m×B points vertically — exactly the axis of suspension, so the solenoid is free to respond to it. Good engineering: the experimenter chose the suspension axis to coincide with the torque axis.

Common mistake. Students sometimes try to compute F = IL×B for the solenoid wires individually and add them up. While correct, it's the hard way. The dipole formula is much faster and gives the same answer.

The 2:1 ratio. For any short dipole, B_axial/B_eq = 2 at the same distance. This is a clean geometric fact: the axial position is "looking down the dipole" so both pole contributions add constructively; the equatorial position sees the two poles from equal but slightly oblique angles, and the components perpendicular to the axis cancel while the parallel components partially add, giving exactly half.

Vector form of the dipole field. The full expression at a general point r (with the dipole at origin) is B(r) = ₀4π 3(m·r)r - mr³. At the axial position, r is parallel to m, so m·r = m and 3mr - m = 2m, giving B = ₀/42m/r³ — matches our axial formula. On the equator, r⊥m, so m·r = 0 and B = -₀/4πm/r³ — the minus sign explains the antiparallel direction.

Why the equatorial field is antiparallel. Picture the dipole as N (top) and S (bottom) separated by a small distance. Stand to the side at the equator. The N-pole pulls a north-seeking probe toward itself; the S-pole pushes it away. Geometrically, the net field points from N back to S — opposite to m. (Field lines go from N to S in the OUTSIDE world; on the equator, they're going parallel to the dipole-axis but opposite.)

Sanity check with the units. ₀/4π· /r³= T m A^-1·A m²m³ = T.

Comparison to Earth's field. Earth's surface field is ∼ 5× 10^-5 T. Our axial value 9.6× 10^-5 T is about twice that — so a small bar magnet at 10 cm competes with the geomagnetic field. That's why a compass needle held very close to a small bar magnet wildly deflects.

Common mistake. Many students forget the factor of 2 in the axial formula or swap the two formulae. Remember the mnemonic: "Axial is twice equatorial." If you write B ∝ 1/r³ and confirm the 2:1 ratio, you're set.

Why 45^∘ means equal magnitudes. When two perpendicular vectors of magnitudes A and B add, the resultant makes an angle α with the B-vector where tanα = A/B. For α = 45^∘, tan 45^∘ = 1, so A = B. This is the crisp condition the problem hinges on — without it the algebra is messier.

The 2:1 ratio strikes again. Since the axial field is twice the equatorial field at the same distance, the distance at which the axial field equals B_E must be larger than the equatorial distance — exactly by a factor 2^1/3 ≈ 1.26. Verify: 6.30/5.00 = 1.26. Always a useful cross-check.

Direction of the resultant.

• Equatorial case (a): The magnet's field at the equatorial point is antiparallel to its moment m. Combined with B_E perpendicular to m, the resultant tilts 45^∘ away from B_E toward -m.
• Axial case (b): The magnet's field at the axial point is parallel to m. The resultant tilts 45^∘ toward +m.

So the two resultants are on opposite sides of Earth's field direction.

Real-world example — neutral point. When the magnet's field and Earth's field are antiparallel (instead of perpendicular as here), there's a special distance where they exactly cancel — the "neutral point" where a compass needle has no preferred direction. The location depends on the same dipole-field formula B ∝ m/r³, set equal to B_E but in the antiparallel geometry.

Common mistake. Confusing gauss with tesla. 1 gauss = 10^-4 tesla. Earth's surface field is conventionally written as 0.3 to 0.6 G, which is 3 to 6× 10^-5 T. Forgetting the 10^-4 factor throws the answer off by a factor of ∼ 50 in r (since r∝ B_E^-1/3, and 10^41/3 ≈ 21.5).

A simple mnemonic — "ANticlockwise = North." Capital letters of both words match, making it easy to remember. Equivalently: ClockWise = South — "C" pairs with "S" in the alphabet through their nearby positions, but the ANticlockwise–North link is what most students remember.

Why the moment direction matters. The torque on the solenoid in an external field is τ = m×B; the potential energy is U = -m·B; the force on it in a non-uniform field is F = ∇m·B. Every one of these requires knowing the direction of m, not just its magnitude. Get the direction wrong and you predict the solenoid swings the WRONG way in a field — a fatal error in any vector problem.

Picturing the field lines. Inside the solenoid, field lines run from S to N (a strong, nearly uniform field). Outside, they loop around from N to S — exactly like a bar magnet. The solenoid is the "honest" version of a bar magnet — it makes explicit that the source of permanent-magnet fields is microscopic current loops (electron spin and orbital motion in atoms).

Reversing the current. Flip the current direction (now clockwise from the same end), and the right-hand rule reverses: that end becomes the South pole instead. So the solenoid's "polarity" can be switched at will — this is the basis of electromagnets (and electromagnetic relays, doorbells, MRI scanners, particle accelerators).

Common mistake. Students sometimes use the LEFT hand by accident — typically because they're holding a pen in the right hand. Always set down what you're holding and use the right hand for these direction rules.

What "angle of dip" measures, geometrically. Stand at a point on Earth. The horizontal local plane is your reference. The magnetic field at that point points down (in the Northern hemisphere) into the ground at angle δ. At the magnetic equator, δ = 0^∘ (field is horizontal); at the magnetic poles, δ = 90^∘ (field is vertical, straight down or straight up). A δ = 22^∘ reading suggests a location at a relatively low magnetic latitude — perhaps somewhere in southern India.

The three "elements" of Earth's field. A complete description of geomagnetism at a point requires three numbers:

• Horizontal component B_H (magnitude in the horizontal plane).
• Declination D angle between true geographic north and the horizontal projection of B_E.
• Dip δ (angle below horizontal).

Given any two of these plus B_H or the total B_E, the full field vector is determined. The needle in our problem measures δ directly because the needle aligns with B_E when free to tilt vertically.

Alternative computation. One could equally compute B_V = B_Htanδ = 0.3522^∘ ≈ 0.35× 0.404 ≈ 0.141 G, then B_E = √B_H²+B_V² = √0.35²+0.141² = √0.1225+0.0199 = √0.1424 ≈ 0.377 G. Matches — within rounding error.

Why N-tip pointing DOWN means Northern hemisphere. Earth's geographic North pole is actually a magnetic South pole (field lines RE-enter the Earth there). So compass needles, whose N-poles are attracted to that magnetic-South, dip downward into the ground in the Northern hemisphere. In the Southern hemisphere, the N-tip points UP instead.

Common mistake. Students sometimes use B_E = B_Hsinδ — wrong gives B_V, not B_E. Or use B_E = B_H cosδ — wrong gives less than B_H, which is geometrically impossible since B_E is the hypotenuse. Always start from the right triangle: B_H is the adjacent side, B_V is opposite, B_E is the hypotenuse.

Why the N-tip points UPWARD in southern Africa. Earth's magnetic field lines form closed loops emerging from the magnetic-south (which is near the geographic North pole) and re-entering at the magnetic-north (near the geographic South pole). So in the Southern hemisphere, field lines come OUT of the ground and curve upward. A dip needle aligns along these lines — its N-tip thus tilts UP rather than down.

Computing the components in full.

• B_H = 0.16 G (given, horizontal).
• B_V = B_Esinδ = 0.3260^∘ = 0.32× 0.866 ≈ 0.277 G (vertical, pointing upward in Southern hemisphere).
• B_E = 0.32 G (total).

Cross-check: B_E = √B_H²+B_V² = √0.0256+0.0768 = √0.1024 = 0.32 G. ✓

Why declination matters. If you're navigating by compass, the needle points along the local magnetic meridian, not toward the geographic north pole. To get to true north, you must rotate D = 12^∘ east of the needle's reading. Maps usually print declination so navigators can correct. Declination varies slowly with time (decade scale) and significantly with location — in some places it's zero (the agonic line); in northern Canada it can exceed 20^∘; near the magnetic poles, it can be very large and unstable.

Geomagnetic-pole geometry. The dip is 60^∘ above horizontal, so the magnetic latitude (angular distance from the magnetic equator) is given by -lat = 12tanδ, giving mag-lat ≈ 40.9^∘ south of the magnetic equator. Combined with declination 12^∘ W, this points to a location somewhere in southern Africa — consistent with the problem statement.

Common mistake. Reading cos 60^∘ as 0.866 instead of 0.5. Remember: sin 30^∘ = cos 60^∘ = 0.5; sin 60^∘ = cos 30^∘ = 0.866. Memorise the complementary-angle pattern.

Sanity check. Earth's surface field varies from ∼ 0.25 G (near the equator) to ∼ 0.65 G (near the magnetic poles). Our 0.32 G for a mid-latitude location in Africa is right in the expected range.