A · Units, dimensions and conversions
10 questionsThe SI unit of magnetic moment is
Given
Magnetic moment of a current loop, m = NIA.
Asked
Its SI unit.
Concept applied
Build the unit from the defining formula rather than memorising it. m = NIA is a pure current times an area, so the unit follows immediately.
Formula to use
m = N I A ⇒ unit = (A)(m2) = A m2
Baby steps
- N is a pure number of turns and carries no unit.
- I is measured in amperes (A).
- A is an area, measured in m2.
- So the unit of m is A m2.
- Cross-check from the torque relation: τ = mB sin θ gives m = τ/B, whose unit is N m / T = J T−1.
- A m2 and J T−1 are the same unit written two ways — NCERT uses both freely.
Assumption
SI system throughout.
Shortcut trick
Two names, one unit: A m2 = J T−1. If an option list contains both, they are equally correct and the question is faulty — but usually only one appears, so recognise it in either dress.
Answer: (B) A m2
The dimensional formula of magnetic moment is
Given
m = NIA, with unit A m2.
Asked
The dimensional formula.
Concept applied
Convert the unit directly into dimensions: amperes give A, square metres give L2. Nothing else enters.
Formula to use
[m] = [current] × [area] = [A][L2] = [L2 A]
Baby steps
- The unit of m is A m2.
- Ampere is a base quantity with dimension [A].
- Square metre has dimension [L2].
- Combining: [m] = [L2 A].
- Sanity check against magnetisation: M = m/V has dimension [L2A]/[L3] = [L−1A], which is the printed dimension of M and matches its unit A m−1. ✔
- That consistency check confirms [L2A] is right.
Assumption
The dimensions table on NCERT page 151 prints the magnetic moment as [L−2 A]. That is a typographical error in the book — the correct dimension is [L2 A], as the unit A m2 printed in the very same row confirms.
Shortcut trick
When a printed table and a printed unit disagree, trust the unit. Units are checked against equations far more often than dimension columns, and here A m2 settles it beyond doubt.
Answer: (A) [L2 A]
Magnetisation M and magnetic intensity H are measured in
Given
M = mnet/V and H = B/μ0 − M.
Asked
Their units.
Concept applied
They must share a unit, because B = μ0(H + M) adds them together. Quantities can only be added if their units match — this is a structural requirement, not a coincidence.
Formula to use
B = μ0(H + M) requires [H] = [M]
Baby steps
- Derive M's unit from its definition: moment per unit volume = A m2 / m3 = A m−1.
- H appears added to M inside the bracket of B = μ0(H + M).
- Addition demands identical units, so H is also in A m−1.
- Independent confirmation: for a solenoid H = nI, with unit m−1 × A = A m−1. ✔
- Both therefore have dimension [L−1 A].
- Answer: the same unit, A m−1.
Assumption
SI units; H defined as in NCERT Eq. 5.11.
Shortcut trick
Use the addition rule as a free check throughout physics. Anything summed inside a bracket must share units — it instantly rules out options and catches formula misremembering.
Answer: (B) the same unit, A m−1
Which of the following pairs is dimensionless?
Given
The quantities χ, μr, μ, μ0, M, H, m, φB.
Asked
Which pair carries no dimensions.
Concept applied
Both χ and μr are defined as ratios of like quantities, so the units cancel. Everything else in the list is a physical quantity with genuine dimensions.
Formula to use
χ = M/H (both A m−1) μr = μ/μ0 (both T m A−1)
Baby steps
- χ = M/H. Numerator and denominator are both in A m−1, so the units cancel ⇒ dimensionless.
- μr = μ/μ0. Both are permeabilities in T m A−1, so they cancel ⇒ dimensionless.
- Confirm with μr = 1 + χ: you may only add a pure number to a pure number, so if one is dimensionless the other must be. ✔
- μ and μ0 both carry [MLT−2A−2] — not dimensionless.
- M and H carry [L−1A]; m carries [L2A]; φB carries [ML2T−2A−1].
- Answer: χ and μr.
Assumption
Linear isotropic materials, so χ is a scalar ratio.
Shortcut trick
Anything defined as a ratio of two like quantities is dimensionless. In this chapter that is exactly χ and μr — and NCERT's table marks both with a dash in the dimensions column.
Answer: (B) χ and μr
The SI unit of magnetic flux is the weber, which is equivalent to
Given
φB = B·ΔS.
Asked
The weber expressed in base-derived units.
Concept applied
Flux is field times area, so its unit is the unit of B times the unit of area.
Formula to use
φB = B A ⇒ weber = tesla × metre2
Baby steps
- φB = B·ΔS, a field multiplied by an area.
- B is measured in tesla (T).
- Area is measured in m2.
- So 1 weber = 1 T m2, exactly as printed in NCERT's table (W = T m2).
- Dimensions follow: [B][L2] = [MT−2A−1][L2] = [ML2T−2A−1].
- Option (A) inverts the area — that would be a field per unit area, which is not a physical quantity here.
Assumption
Uniform B perpendicular to the area, so the dot product reduces to a simple product.
Shortcut trick
Flux is always field × area, in every branch of physics. Electric flux is N m2 C−1; magnetic flux is T m2. Build the unit from the definition and you never need to memorise it.
Answer: (B) T m2
The dimensional formula of magnetic flux is
Given
φB = B × area; [B] = [MT−2A−1].
Asked
The dimensional formula of φB.
Concept applied
Multiply the dimensions of B by those of area. The dimension of B itself comes from F = BIL, giving [MT−2A−1].
Formula to use
[φB] = [B][L2]
Baby steps
- Establish [B] from F = BIL: B = F/(IL), so [B] = [MLT−2]/([A][L]) = [MT−2A−1].
- Area has dimension [L2].
- Multiply: [φB] = [MT−2A−1] × [L2].
- [φB] = [M L2 T−2 A−1].
- Cross-check via energy: this is [energy]/[A] = [ML2T−2][A−1]. ✔ Indeed weber = joule per ampere.
- Option (B) is the dimension of B alone — the planted intermediate.
Assumption
SI base dimensions with current A as a base quantity.
Shortcut trick
Anchor the whole chapter on one dimension: [B] = [MT−2A−1]. Flux is that times L2; permeability is that divided by [L−1A]. Everything else is one step away.
Answer: (A) [M L2 T−2 A−1]
The Earth's magnetic field is about 0.5 gauss. In tesla this is
Given
B = 0.5 gauss; 104 gauss = 1 tesla.
Asked
The field in tesla.
Concept applied
The gauss is the CGS unit of magnetic field. NCERT gives the conversion directly in its table of physical quantities.
Formula to use
1 T = 104 G ⇒ 1 G = 10−4 T
Baby steps
- From the conversion, 1 gauss = 10−4 tesla.
- Multiply by the given value: B = 0.5 × 10−4 T.
- Write in standard scientific form: 0.5 × 10−4 = 5 × 10−5 T.
- Sense check: the tesla is a very large unit, so an everyday field must be a small number of tesla. ✔
- For comparison, a strong laboratory electromagnet reaches a few tesla — about 105 times the Earth's field.
- Option (A) is the answer for 5 gauss, and option (C) inverts the conversion entirely.
Assumption
The standard conversion 104 G = 1 T as printed in NCERT.
Shortcut trick
Direction of conversion is the whole difficulty. Gauss is the small unit, so converting gauss → tesla must make the number smaller. If your answer grew, you multiplied when you should have divided.
Answer: (B) 5 × 10−5 T
The permeability of free space μ0 has units
Given
B = μ0nI for a solenoid.
Asked
The units of μ0.
Concept applied
Rearrange any equation containing μ0 and read the units off. The solenoid relation is the cleanest one available.
Formula to use
μ0 = B/(nI) ⇒ unit = T ÷ (m−1 · A)
Baby steps
- Start from B = μ0nI and rearrange: μ0 = B/(nI).
- Unit of B is T; unit of n is m−1; unit of I is A.
- So the unit of μ0 is T / (m−1 A) = T m A−1.
- Answer: T m A−1.
- NCERT's table lists this alongside the equivalent form N A−2, which follows from the force-between-currents formula.
- Numerically μ0 = 4π × 10−7 T m A−1, so μ0/4π = 10−7.
Assumption
Long solenoid so B = μ0nI applies exactly.
Shortcut trick
μ and μ0 share units, because μr = μ/μ0 is dimensionless. Establishing one gives the other free.
Answer: (A) T m A−1
Which of these magnetic quantities is a scalar?
Given
The quantities listed in NCERT's summary table on page 151.
Asked
Which one is a scalar.
Concept applied
Flux is defined by a dot product, B·ΔS, and a dot product of two vectors is always a scalar. The other three are genuine vectors with direction in space.
Formula to use
φB = B·ΔS — a dot product, hence scalar
Baby steps
- M has a direction — it points along the net alignment of the atomic moments. Vector.
- H has a direction — along the applied magnetising field. Vector.
- m has a direction — from S to N inside the magnet. Vector.
- φB is formed by taking B·ΔS, a scalar product of two vectors.
- A dot product yields a number, not a direction, so flux is a scalar.
- It can be positive or negative depending on orientation, but a signed number is still a scalar.
Assumption
Standard classification as printed in the NCERT table.
Shortcut trick
Flux carries a sign but no direction — that is what makes Gauss's law a sum of signed numbers rather than a vector sum. The other scalars in the table are χ, μr, μ and μ0.
Answer: (C) magnetic flux φB
A magnet has moment 0.5 J T−1. Expressed in A m2, this is
Given
m = 0.5 J T−1.
Asked
The same quantity in A m2.
Concept applied
J T−1 and A m2 are two names for one unit, so the numerical value is unchanged. No conversion factor exists because none is needed.
Formula to use
J T−1 = A m2 ⇒ conversion factor is exactly 1
Baby steps
- Break down J T−1: joule = N m; tesla = N A−1 m−1.
- So J T−1 = (N m) ÷ (N A−1 m−1).
- Cancel the newtons: = m × A m = A m2.
- The two unit names are identical, so the number does not change.
- m = 0.5 A m2.
- This is why NCERT quotes some magnets in J T−1 and others in A m2 without ever converting.
Assumption
SI units throughout.
Shortcut trick
Before hunting for a conversion factor, check whether the two units are secretly the same. In this chapter, J T−1 = A m2 and Wb = T m2 = J A−1 are both identities, not conversions.
Answer: (A) 0.5 A m2
B · Magnetic moment of a revolving charge
8 questionsAn electron of charge e revolves in a circular orbit of radius r with speed v. The magnitude of its orbital magnetic moment is
Given
Electron of charge e in a circular orbit of radius r at speed v.
Asked
The magnitude of its magnetic moment.
Concept applied
A revolving charge is a current loop. Convert the orbital motion into an equivalent current (charge past a point per second), then apply m = IA.
Formula to use
I = e/T = ev/2πr; m = IA = Iπr2
Baby steps
- Period of one revolution: T = 2πr/v.
- Equivalent current: I = e/T = ev/(2πr).
- Area enclosed by the orbit: A = πr2.
- Magnetic moment: m = IA = [ev/(2πr)] × πr2.
- Cancel π and one r: m = evr/2.
- Direction: perpendicular to the orbital plane; because the electron is negative, m points opposite to the direction of its angular momentum.
Assumption
Uniform circular motion; classical (Bohr-style) picture of the orbit.
Shortcut trick
The chain is always orbit → current → moment, and the factor 1/2 survives because πr2/(2πr) = r/2. Option (A) is what you get by forgetting the 2π in the period.
Answer: (B) evr/2
For an orbiting electron, the ratio of magnetic moment to orbital angular momentum, m/L, equals
Given
Electron of mass me, charge e, in a circular orbit of radius r at speed v.
Asked
The ratio m/L, called the gyromagnetic ratio.
Concept applied
Both m and L are proportional to v and r, so forming their ratio cancels the orbit details entirely. What survives depends only on the particle's charge and mass.
Formula to use
m = evr/2; L = mevr; m/L = e/2me
Baby steps
- Magnetic moment of the orbit: m = evr/2.
- Orbital angular momentum: L = mevr.
- Form the ratio: m/L = (evr/2) ÷ (mevr).
- Cancel v and r from numerator and denominator.
- m/L = e/2me.
- Numerically: 1.6 × 10−19 / (2 × 9.1 × 10−31) ≈ 8.8 × 1010 C kg−1.
Assumption
Classical circular orbit; orbital motion only, ignoring electron spin, which has a different gyromagnetic ratio.
Shortcut trick
The cancellation of v and r is the whole point: every orbit of a given particle has the same m/L, regardless of size or speed. That universality is what makes it a named constant.
Answer: (B) e/2me
For an orbiting electron, the vector relation between magnetic moment and angular momentum is
Given
An electron, charge −e, in a circular orbit.
Asked
The vector relation including sign.
Concept applied
The magnitudes are linked by e/2me, but the directions are opposite because the electron's charge is negative — conventional current runs opposite to the electron's motion.
Formula to use
m = −(e/2me)L
Baby steps
- The electron physically travels in one sense around the orbit.
- Conventional current is defined as the flow of positive charge, so it runs in the opposite sense.
- m is fixed by the conventional current via the right-hand rule.
- L = r × p is fixed by the electron's actual motion.
- The two therefore point in opposite directions: m = −(e/2me)L.
- Reject option (D): m and L are antiparallel, not perpendicular — both are normal to the orbital plane.
Assumption
e denotes the magnitude of the electronic charge, so the sign is carried explicitly by the minus.
Shortcut trick
For any negative charge, moment and angular momentum are antiparallel. For a positive charge they are parallel. The magnitude relation is the same in both cases; only the sign flips.
Answer: (B) m = −(e/2me)L
An electron moves at 2.2 × 106 m s−1 in an orbit of radius 0.53 Å. Its orbital magnetic moment is about (e = 1.6 × 10−19 C)
Given
v = 2.2 × 106 m s−1; r = 0.53 Å = 0.53 × 10−10 m; e = 1.6 × 10−19 C
Asked
The orbital magnetic moment.
Concept applied
Direct substitution into m = evr/2. The data are those of the ground-state hydrogen orbit, so the answer should come out close to one Bohr magneton.
Formula to use
m = evr/2
Baby steps
- Convert the radius: 0.53 Å = 0.53 × 10−10 m.
- Multiply e × v: 1.6 × 10−19 × 2.2 × 106 = 3.52 × 10−13.
- Multiply by r: 3.52 × 10−13 × 0.53 × 10−10 = 1.866 × 10−23.
- Divide by 2: m = 9.33 × 10−24.
- m ≈ 9.3 × 10−24 A m2.
- This is essentially the Bohr magneton, 9.27 × 10−24 A m2 — a useful check that the arithmetic is right.
Assumption
Classical circular orbit; these are the standard Bohr first-orbit values for hydrogen.
Shortcut trick
Option (B) is the answer with the ÷2 omitted — the single most common slip in this calculation. Recognising that the result should land near 9.3 × 10−24 catches it immediately.
Answer: (A) 9.3 × 10−24 A m2
A charge q revolves f times per second in a circle of radius r. The equivalent current and magnetic moment are
Given
Charge q; frequency of revolution f; orbit radius r.
Asked
The equivalent current and the magnetic moment.
Concept applied
Current is charge past a point per second. If the charge completes f circuits per second, then q passes any point f times per second, so I = qf directly — no period conversion needed.
Formula to use
I = qf; m = IA = qfπr2
Baby steps
- The charge passes any fixed point on the orbit f times each second.
- Each pass carries charge q, so the charge per second is I = qf.
- Area enclosed: A = πr2.
- Moment: m = IA = qfπr2.
- Consistency with the earlier form: f = v/2πr, so m = q(v/2πr)πr2 = qvr/2. ✔ Same result.
- Option (B) uses the period instead of the frequency — the classic inversion error.
Assumption
Uniform circular motion at constant frequency; single revolving charge.
Shortcut trick
When a question gives frequency, use I = qf and skip the period entirely. When it gives speed, use I = qv/2πr. Choosing the matching form avoids an unnecessary inversion where sign and factor errors creep in.
Answer: (A) qf and qfπr2
The smallest value of the magnetic moment associated with an orbiting electron, called the Bohr magneton, is approximately
Given
The natural unit of atomic magnetic moment.
Asked
Its approximate value.
Concept applied
Combining m = (e/2me)L with the quantised angular momentum L = h/2π gives a fixed smallest moment — the Bohr magneton.
Formula to use
μB = eh / 4πme ≈ 9.27 × 10−24 A m2
Baby steps
- Start from m = (e/2me)L.
- Insert the smallest allowed orbital angular momentum, L = h/2π.
- m = (e/2me)(h/2π) = eh/(4πme).
- Substituting e = 1.6 × 10−19 C, h = 6.63 × 10−34 J s, me = 9.1 × 10−31 kg gives ≈ 9.27 × 10−24 A m2.
- The distractors are the electronic charge, the electron mass and Planck's constant — all quoted with their own units, none of them a magnetic moment.
- Check the previous question: the hydrogen ground-state orbit gave 9.3 × 10−24 A m2, matching this. ✔
Assumption
Bohr quantisation L = nh/2π with n = 1. The rationalised NCERT Chapter 5 does not derive μB, though 'magnetic dipole moment of a revolving electron' appears in the NEET syllabus — treat the derivation as gap content.
Shortcut trick
Spot the answer by units alone. Only one option is plausibly a magnetic moment; 10−19 C, 10−31 kg and 10−34 J s are recognisable as charge, mass and action. Elimination is faster than recall here.
Answer: (A) 9.27 × 10−24 A m2
Two electrons orbit at different radii and different speeds. The quantity that is necessarily the same for both is
Given
Two electrons in different circular orbits.
Asked
Which quantity must be identical.
Concept applied
m and L each depend on v and r, so both differ between the orbits. But their ratio is e/2me, built only from the particle's charge and mass — identical for every electron.
Formula to use
m/L = e/2me, independent of v and r
Baby steps
- m = evr/2 depends on both v and r ⇒ differs between the orbits.
- L = mevr also depends on both ⇒ differs between the orbits.
- I = ev/2πr depends on v and r as well ⇒ differs.
- But m/L = (evr/2)/(mevr) = e/2me, in which v and r have cancelled.
- This ratio is built only from e and me, which are the same for every electron.
- Answer: the ratio of magnetic moment to angular momentum.
Assumption
Both particles are electrons; classical orbital motion only.
Shortcut trick
When a question asks 'what is necessarily the same', look for the combination in which the variable quantities cancel. Forming ratios is usually where that happens.
Answer: (C) the ratio of magnetic moment to angular momentum
Can a system whose net electric charge is zero possess a magnetic moment?
Given
A system with zero net electric charge, e.g. an atom.
Asked
Whether it can have a magnetic moment.
Concept applied
Magnetic moment arises from charges in motion, not from a net charge. An atom has equal positive and negative charge but its electrons still circulate, and those loops need not cancel.
Formula to use
net charge = 0 does not imply ∑mloops = 0
Baby steps
- Net charge is the algebraic sum of all charges present.
- Magnetic moment comes from current loops, which depend on how charges move, not on the total.
- In an atom the electrons orbit and spin, forming loops each with its own moment.
- These vector moments may or may not cancel — there is no requirement that they do.
- In paramagnetic atoms they do not cancel, giving a net permanent moment despite zero net charge.
- Answer: Yes. NCERT's Example 5.4(d): the average of the charge may be zero, yet the mean of the magnetic moments due to various current loops may not be zero.
- Reject option (C): paramagnetic atoms show this too, not only ferromagnetic ones.
Assumption
Ordinary matter composed of atoms with orbiting electrons.
Shortcut trick
Charge and moment answer different questions. Charge asks how much; moment asks how it moves. A neutral atom can be a perfectly good little magnet — this is the entire microscopic basis of paramagnetism.
Answer: (B) Yes — the moments of internal current loops need not cancel
C · Two magnets combined at an angle
10 questionsTwo identical bar magnets, each of moment m, are placed with their axes perpendicular to each other. The resultant magnetic moment is
Given
m1 = m2 = m; angle between them θ = 90°.
Asked
The resultant magnetic moment.
Concept applied
Magnetic moment is a vector, so moments combine by the parallelogram law, never by simple addition. At 90° the cross term vanishes and it reduces to Pythagoras.
Formula to use
mR = √(m12 + m22 + 2m1m2 cos θ)
Baby steps
- Substitute θ = 90°, so cos θ = 0 and the cross term drops out.
- mR = √(m2 + m2).
- = √(2m2).
- mR = m√2 ≈ 1.41 m.
- Direction: at 45° to each magnet, by symmetry, since the two moments are equal.
- Sense check: the answer must lie between 0 (antiparallel) and 2m (parallel). 1.41m does. ✔
Assumption
The magnets are joined rigidly at their centres so both moments act at one point.
Shortcut trick
Bracket every answer before computing: the resultant of two moments must lie between |m1 − m2| and (m1 + m2). Any option outside that range is wrong on sight.
Answer: (B) m√2
The same two magnets are now placed parallel with their like poles pointing the same way. The resultant moment is
Given
m1 = m2 = m; θ = 0°.
Asked
The resultant moment.
Concept applied
At θ = 0° the vectors point the same way, so the parallelogram law reduces to ordinary addition. This is the maximum possible resultant.
Formula to use
mR = √(m2 + m2 + 2m2cos 0°)
Baby steps
- cos 0° = 1, so the cross term is at its largest.
- mR = √(m2 + m2 + 2m2).
- = √(4m2).
- mR = 2m.
- Direction: along the common axis of both magnets.
- This is the upper bound of the range, as expected for the aligned case.
Assumption
Like poles pointing the same way, so both moment vectors are parallel — not the N-to-S contact arrangement.
Shortcut trick
Read the physical description carefully. 'Like poles the same way' means parallel moments (2m). 'Unlike poles together in a line' also gives parallel moments. 'Like poles in contact' means antiparallel moments (zero). The wording, not the picture in your head, decides.
Answer: (D) 2m
Two identical magnets of moment m each are placed with their axes along the same line but with their moments in opposite directions. The resultant moment is
Given
m1 = m2 = m; θ = 180°.
Asked
The resultant moment.
Concept applied
Antiparallel equal vectors cancel exactly. This is the minimum of the allowed range, and it explains why a compact closed loop of magnets shows no external field.
Formula to use
mR = √(m2 + m2 + 2m2cos 180°)
Baby steps
- cos 180° = −1.
- mR = √(m2 + m2 − 2m2).
- = √(0).
- mR = zero.
- Physically the two dipoles cancel, so the pair produces no dipole field at large distances.
- Note this does not mean there is no field anywhere — close up the fields do not cancel; only the far dipole term vanishes.
Assumption
Identical magnets of exactly equal moment, coincident centres.
Shortcut trick
The three landmark cases are worth having by heart: 0° → 2m, 90° → m√2, 180° → 0. Most combination questions are one of these three in disguise.
Answer: (C) zero
Two identical magnets of moment m each have their axes inclined at 60°. The resultant moment is
Given
m1 = m2 = m; θ = 60°.
Asked
The resultant moment.
Concept applied
With equal magnitudes the formula simplifies neatly, and 60° is one of the standard angles where the answer is clean.
Formula to use
mR = √(2m2 + 2m2cos θ) = 2m cos(θ/2)
Baby steps
- cos 60° = 0.5.
- mR = √(m2 + m2 + 2m2 × 0.5).
- = √(m2 + m2 + m2) = √(3m2).
- mR = m√3 ≈ 1.73 m.
- Faster route: for equal moments, mR = 2m cos(θ/2) = 2m cos 30° = 2m(√3/2) = m√3. ✔
- Direction: along the bisector of the two axes, by symmetry.
Assumption
Equal moments; rigid joint at a common centre.
Shortcut trick
For equal moments always use mR = 2m cos(θ/2). It is one step instead of four, and it makes the direction obvious — always the bisector.
Answer: (C) m√3
Two identical magnets of moment m each are inclined at 120°. The resultant moment is
Given
m1 = m2 = m; θ = 120°.
Asked
The resultant moment.
Concept applied
Same simplified formula, now with an obtuse angle where the cross term becomes negative and partly cancels the sum.
Formula to use
mR = 2m cos(θ/2)
Baby steps
- Apply the equal-moment shortcut: mR = 2m cos(120°/2).
- = 2m cos 60°.
- cos 60° = 0.5.
- mR = 2m × 0.5 = m.
- Long check: √(m2 + m2 + 2m2cos 120°) = √(2m2 − m2) = √(m2) = m. ✔
- Neat result: two equal moments at 120° give a resultant of exactly the same magnitude as either one.
Assumption
Equal moments; common centre.
Shortcut trick
cos 120° = −0.5, not −0.866. Obtuse-angle cosines are where sign errors live — write the value down explicitly before substituting rather than doing it in your head.
Answer: (B) m
Two magnets of moments m and 2m are placed perpendicular to each other. The resultant moment is
Given
m1 = m; m2 = 2m; θ = 90°.
Asked
The resultant moment.
Concept applied
Perpendicular vectors of unequal magnitude combine by Pythagoras. The equal-moment shortcut does not apply here, so use the full formula.
Formula to use
mR = √(m12 + m22) at 90°
Baby steps
- cos 90° = 0, so the cross term vanishes.
- mR = √(m2 + (2m)2).
- (2m)2 = 4m2, so the bracket is m2 + 4m2 = 5m2.
- mR = m√5 ≈ 2.24 m.
- Range check: it must lie between |2m − m| = m and 2m + m = 3m. 2.24m does. ✔
- Direction: at angle α from the larger moment, with tan α = m/2m = 0.5, so α ≈ 26.6°.
Assumption
Rigid arrangement with coincident centres.
Shortcut trick
Square the whole quantity, not just the number. Writing (2m)2 = 2m2 instead of 4m2 gives m√3 — option (C), planted for exactly that slip.
Answer: (B) m√5
Two equal magnetic moments are inclined at 90°. The resultant makes an angle with the first moment of
Given
m1 = m2 = m; θ = 90°.
Asked
The angle α between the resultant and the first moment.
Concept applied
The direction of a parallelogram resultant is given by the standard tangent formula. With equal magnitudes, symmetry alone forces the answer to be the bisector.
Formula to use
tan α = m2 sin θ / (m1 + m2 cos θ)
Baby steps
- Substitute θ = 90°: sin 90° = 1, cos 90° = 0.
- tan α = m × 1 / (m + m × 0).
- tan α = m/m = 1.
- α = arctan(1) = 45°.
- Symmetry check: the two moments are equal, so the resultant must bisect the angle between them — half of 90° is 45°. ✔
- This bisector argument works for equal moments at any angle, giving α = θ/2.
Assumption
Equal moments, so the symmetry argument is valid.
Shortcut trick
For equal moments, skip the tangent formula entirely: the resultant always bisects, so α = θ/2. Only reach for tan α when the two moments are unequal.
Diagram
Answer: (B) 45°
A magnet of moment m is cut transversely into two equal halves, which are then placed with their axes at 60°. The resultant moment is
Given
Original moment m; transverse cut into two equal halves; halves inclined at 60°.
Asked
The resultant moment of the pair.
Concept applied
Two stages that must be done in order. First the cut halves each moment; only then do the vectors combine.
Formula to use
after cut: m1 = m2 = m/2; then mR = 2(m/2)cos(θ/2)
Baby steps
- Stage 1 — the cut: a transverse cut halves the length, so each half has moment m/2.
- Stage 2 — the combination: two equal moments of m/2 at 60°.
- Apply the equal-moment shortcut: mR = 2 × (m/2) × cos 30°.
- = m × (√3/2).
- mR = (m√3)/2 ≈ 0.87 m.
- Option (A) is the answer you get by forgetting the cut and using m for each half — the planted trap.
Assumption
Clean transverse cut; each half retains uniform magnetisation; rigid joint at 60°.
Shortcut trick
Two-stage problems always run modify first, combine second. Write down the intermediate value (m/2 here) explicitly before touching the vector formula — skipping that line is what produces the trap answer.
Answer: (B) (m√3)/2
Two identical magnets of moment m each are fixed at right angles and placed in a uniform field B with the resultant moment at 30° to the field. The torque on the system is
Given
Two moments m at 90°; resultant at 30° to B.
Asked
The torque on the combined system.
Concept applied
A rigid combination behaves as a single dipole of moment mR. Find mR first, then apply the ordinary torque formula using the angle between mR and B.
Formula to use
mR = m√2; τ = mR B sin θ
Baby steps
- Stage 1 — combine: two equal moments at 90° give mR = m√2.
- Stage 2 — torque: the angle between mR and B is given as 30°.
- τ = mR B sin 30° = m√2 × B × 0.5.
- τ = mB√2 / 2 ≈ 0.707 mB.
- Note that the 90° between the magnets and the 30° to the field are different angles serving different purposes — do not mix them.
- Option (C) omits the sin 30°; option (D) treats the combination as 2m.
Assumption
The magnets are rigidly fixed together, so the system rotates as one body.
Shortcut trick
Two angles in one question means two separate stages. The inter-magnet angle builds mR; the field angle gives the torque. Label them before you start or they will get swapped.
Answer: (B) mB√2/2
Two magnets of moments 3 A m2 and 4 A m2 are combined at various angles. The resultant moment can take any value between
Given
m1 = 3 A m2; m2 = 4 A m2; angle variable from 0° to 180°.
Asked
The range of possible resultants.
Concept applied
Vector addition of two fixed magnitudes produces a resultant bounded by the aligned case (sum) and the opposed case (difference). Every intermediate value is achievable by choosing a suitable angle.
Formula to use
|m1 − m2| ≤ mR ≤ (m1 + m2)
Baby steps
- Maximum occurs at θ = 0° (parallel): mR = 3 + 4 = 7 A m2.
- Minimum occurs at θ = 180° (antiparallel): mR = |3 − 4| = 1 A m2.
- Since mR varies continuously with θ, all values between are attainable.
- Range: 1 to 7 A m2.
- Option (A) assumes complete cancellation, which requires equal moments — these are unequal, so zero is unreachable.
- Check at 90°: √(9 + 16) = 5 A m2, comfortably inside the range. ✔
Assumption
The magnitudes are fixed and only the angle between them varies.
Shortcut trick
Use this range as a validity filter on every combination question. Compute |m1 − m2| and (m1 + m2) first, then reject any option outside — it often leaves only one survivor.
Answer: (B) 1 and 7 A m2
D · Bent and reshaped magnet geometry
8 questionsA thin bar magnet of moment m is bent into a semicircle without changing its length. The new magnetic moment is
Given
Straight magnet of length L and moment m = qmL, bent into a semicircle of the same arc length.
Asked
The new magnetic moment.
Concept applied
Magnetic moment is pole strength times the straight-line distance between the poles, not the length of material. Bending shortens that straight-line distance while leaving the pole strength alone.
Formula to use
m′ = qm × (straight distance between poles)
Baby steps
- Original: m = qmL, with the poles a distance L apart.
- After bending, the material length L becomes the arc of a semicircle: L = πR, so R = L/π.
- The two poles now sit at opposite ends of a diameter, a straight-line distance 2R apart.
- 2R = 2L/π.
- Pole strength qm is unchanged — the cross-section has not altered.
- m′ = qm × 2L/π = (2/π)(qmL) = 2m/π ≈ 0.64 m.
Assumption
Uniform thin magnet; bending does not disturb the magnetisation or the pole strength.
Shortcut trick
The governing idea in every bending question: pole strength is fixed, separation changes. Compute the new straight-line separation, divide by the old one, and multiply m by that ratio. Here 2R/L = 2/π.
Answer: (B) 2m/π
The same magnet is instead bent into a complete circle so that its two ends meet. Its magnetic moment becomes
Given
Bar magnet of moment m bent into a closed circle, poles brought into contact.
Asked
The new magnetic moment.
Concept applied
When the ends meet, the N and S poles coincide. The straight-line separation between them becomes zero, so the moment vanishes — the same reason a toroid has no poles.
Formula to use
m′ = qm × (separation) = qm × 0 = 0
Baby steps
- Bending into a full circle brings the N end round to touch the S end.
- The straight-line distance between the poles is therefore zero.
- m′ = qm × 0 = zero.
- Physically the N and S poles neutralise each other at the joint.
- The result is a closed magnetic circuit with no external field — exactly the toroid situation of Example 5.1(c).
- It is a magnetic configuration with no north pole and no south pole, which NCERT confirms is possible.
Assumption
The ends meet cleanly and the magnetisation follows the curve all the way round.
Shortcut trick
Zero-moment configurations are a favourite conceptual target: the closed ring, the toroid, and the infinite straight conductor. All three have no poles, and all three answer 'must every magnetic configuration have poles?' with no.
Answer: (D) zero
A bar magnet of moment m is bent at its midpoint so that the two halves make an angle of 90° with each other. The resultant moment is
Given
Bar magnet of moment m bent at its midpoint; arms at 90°.
Asked
The resultant moment.
Concept applied
Bending at the midpoint creates two half-magnets of moment m/2 each. Their moment vectors run head to tail along the arms, so the angle between the vectors is 180° minus the angle between the arms.
Formula to use
m′ = 2(m/2)cos[(180°−θ)/2] = m sin(θ/2)
Baby steps
- Each half has moment m/2 after the bend (length halved, pole strength unchanged).
- The moments run S→N along each arm, i.e. head to tail, so the angle between the two moment vectors is 180° − 90° = 90°.
- Combine two equal moments of m/2 at 90°: m′ = 2(m/2)cos(90°/2) = m cos 45°.
- m′ = m/√2 ≈ 0.707 m.
- General result for arms at angle θ: m′ = m sin(θ/2).
- Check: θ = 180° (straight) gives m sin 90° = m ✔; θ = 0° (folded flat) gives 0 ✔.
Assumption
Sharp bend at the exact midpoint; each half retains uniform magnetisation.
Shortcut trick
Store the general result m′ = m sin(θ/2), where θ is the angle between the arms. It handles every midpoint-bend question in one line, and the two limiting checks confirm you have not confused arms with vectors.
Answer: (B) m/√2
The same magnet is bent at its midpoint so that the halves make an angle of 60°. The resultant moment is
Given
Bar magnet of moment m bent at midpoint; arms at θ = 60°.
Asked
The resultant moment.
Concept applied
Same geometry as the previous question, applied at a different angle. Using the stored formula makes it a one-line substitution.
Formula to use
m′ = m sin(θ/2)
Baby steps
- Each half has moment m/2.
- Apply the general result with θ = 60°: m′ = m sin(60°/2) = m sin 30°.
- sin 30° = 0.5.
- m′ = m/2.
- Long check: the angle between the moment vectors is 180° − 60° = 120°; two moments of m/2 at 120° give 2(m/2)cos 60° = m × 0.5 = m/2. ✔
- Note the answer is smaller than for the 90° bend — a tighter fold brings the poles closer together, reducing the moment.
Assumption
Midpoint bend; uniform magnetisation retained in each arm.
Shortcut trick
Sanity-check the trend rather than the number: tighter fold ⇒ smaller moment. Going 180° → 90° → 60° → 0° should give m → 0.71m → 0.5m → 0. If your answers do not fall monotonically, you have used the wrong angle.
Answer: (A) m/2
A bar magnet of moment m is bent into an arc subtending an angle θ (in radians) at the centre of curvature. Its new moment is
Given
Magnet of length L bent into an arc subtending θ radians.
Asked
The general expression for the new moment.
Concept applied
Compute the chord length of the arc, since that is the straight-line pole separation. The ratio of chord to arc gives the factor by which the moment shrinks.
Formula to use
arc L = Rθ; chord = 2R sin(θ/2); m′/m = chord/L
Baby steps
- Arc length is fixed at L, so L = Rθ, giving R = L/θ.
- The chord joining the two ends of an arc of half-angle θ/2 is 2R sin(θ/2).
- Substitute R: chord = 2(L/θ)sin(θ/2).
- The poles are separated by this chord, and qm is unchanged.
- m′ = qm × chord = m × [2 sin(θ/2)/θ].
- Check with the semicircle, θ = π: m′ = m × 2 sin(π/2)/π = 2m/π. ✔ Matches the earlier result.
Assumption
θ is in radians; uniform circular arc; pole strength unchanged.
Shortcut trick
Every bending problem reduces to chord ÷ original length. Semicircle gives 2/π ≈ 0.64; quarter circle gives 2√2/π ≈ 0.90; full circle gives 0. Compute the chord and the rest is a ratio.
Answer: (B) m × 2 sin(θ/2)/θ
A bar magnet of moment m is bent into a quarter circle. Its new moment is approximately
Given
Magnet of length L bent into an arc subtending θ = π/2 radians.
Asked
The new moment.
Concept applied
Direct application of the general arc result. A gentler bend removes less separation, so the moment falls only slightly.
Formula to use
m′ = m × 2 sin(θ/2)/θ, θ = π/2
Baby steps
- Substitute θ = π/2 radians.
- Half angle: θ/2 = π/4, and sin(π/4) = 0.7071.
- Numerator: 2 × 0.7071 = 1.4142.
- Denominator: θ = π/2 = 1.5708.
- m′ = m × 1.4142/1.5708 = 0.90 m.
- Consistency: a quarter circle is a gentler bend than a semicircle, so 0.90m must exceed the semicircle's 0.64m. ✔
Assumption
Uniform arc; pole strength unchanged by bending.
Shortcut trick
Option (A) is the semicircle answer, placed here to catch anyone who recalls '2/π' without checking which arc it belongs to. Always verify the angle before reaching for a memorised number.
Answer: (B) 0.90 m
A magnet is bent into a semicircle. The axial field it produces at a fixed far point changes by a factor of
Given
Magnet bent into a semicircle; field measured at the same distant point on the axis.
Asked
The factor by which the axial field changes.
Concept applied
The far axial field is directly proportional to the magnetic moment, so any change in m carries straight through to B. Distance and constants are unchanged.
Formula to use
Baxial = (μ0/4π)(2m/r3) ⇒ B ∝ m
Baby steps
- The bending changes the moment from m to 2m/π.
- The observation distance r is unchanged, as are μ0 and the geometry factor.
- Since Baxial ∝ m, the field scales by exactly the same factor as the moment.
- Factor = (2m/π) ÷ m = 2/π ≈ 0.64.
- So the field falls to about 64% of its original value.
- Option (B) is the result of squaring the factor — a natural but wrong instinct, since B depends on m linearly, not quadratically.
Assumption
The far-field dipole approximation still holds for the bent magnet, which requires r to be much larger than the magnet's size.
Shortcut trick
Decide the power first. B ∝ m1 and B ∝ r−3. A change in moment passes through linearly; only distance changes get cubed.
Answer: (A) 2/π
When a bar magnet is bent into an arc, which quantity remains unchanged?
Given
A bar magnet reshaped by bending, with no material removed.
Asked
Which quantity is unaffected.
Concept applied
Pole strength depends on the cross-sectional area and the degree of magnetisation, neither of which bending alters. Everything else in the list depends on the pole separation, which bending does change.
Formula to use
m = qm × separation — only the separation changes
Baby steps
- Pole strength qm is set by the cross-section and the magnetisation of the material.
- Bending changes neither of these, so qm is unchanged.
- The straight-line separation between the poles decreases — that is what bending does. Option (D) changes.
- Since m = qm × separation, the moment decreases. Option (A) changes.
- Since Baxial ∝ m, the field decreases too. Option (C) changes.
- Answer: pole strength.
Assumption
No material is removed or added; the magnetisation is not disturbed by the bending process.
Shortcut trick
Sort the whole chapter's operations by what they preserve. Bending keeps qm, changes separation. Transverse cutting keeps qm, halves length. Longitudinal cutting keeps length, halves qm. Identify which factor moves and the answer follows.
Answer: (B) its pole strength
E · Toroid and pole-free configurations
7 questionsA current-carrying toroid has
Given
An ideal current-carrying toroid.
Asked
Its pole structure.
Concept applied
Poles exist only where a net magnetic moment produces an external dipole field. A toroid's field is entirely internal and its net moment is zero, so it has nothing that could be called a pole.
Formula to use
net magnetic moment = 0 ⇒ no dipole field ⇒ no poles
Baby steps
- The field lines of a toroid are closed circles confined inside the windings.
- No field lines emerge into the surrounding space.
- A pole is identified by field lines emerging from or converging into a region — here there are none.
- The circulating current elements are arranged symmetrically around the ring, so their moments cancel and the net moment is zero.
- Therefore the toroid has no poles at all.
- NCERT's Example 5.1(c): must every magnetic configuration have a north and a south pole? Not necessarily — true only if the source has a net non-zero magnetic moment.
Assumption
Ideal closely wound toroid with negligible leakage field.
Shortcut trick
Poles are a consequence of a net moment, not a fundamental requirement. Zero net moment ⇒ no poles. This single idea covers the toroid, the closed magnetic ring, and the infinite straight wire.
Answer: (C) no poles at all
The magnetic field in the region outside an ideal toroid is
Given
An ideal closely wound toroid carrying a steady current.
Asked
The external field.
Concept applied
By symmetry and Ampere's law, any circular path drawn outside the toroid encloses zero net current — the current goes in one way and comes back the other. So the field there is zero.
Formula to use
∮B·dl = μ0Ienclosed = 0 outside
Baby steps
- Take an Amperian loop encircling the toroid from outside, or lying in the hole at the centre.
- For a loop outside the toroid, the winding current crosses the enclosed surface once in each direction, so the net enclosed current is zero.
- By Ampere's law the circulation of B around that loop is therefore zero.
- Combined with the symmetry of the arrangement, this forces B = zero outside.
- NCERT confirms it in Example 5.3(c): magnetic lines are completely confined within a toroid.
- This confinement is why a toroid is used where stray fields must be avoided.
Assumption
Ideal toroid, closely and uniformly wound, so that leakage is negligible.
Shortcut trick
The toroid is the magnetic equivalent of a shielded box: all field inside, none outside. Contrast it with a solenoid, which does have an external return field — that is the difference Example 5.3 panels (c) and (d) are testing.
Diagram
Answer: (B) zero
Besides the toroid, NCERT gives another example of a magnetic configuration with no poles, namely
Given
NCERT's answer to Example 5.1(c).
Asked
The second pole-free example.
Concept applied
An infinite straight wire produces circular field lines that close on themselves around the wire. There is no dipole moment and no direction that could be called north or south.
Formula to use
B = μ0I/2πr, circular lines, no net dipole moment
Baby steps
- The field of a long straight conductor consists of concentric circles around the wire.
- These lines are closed and never emerge from a region in a way that would identify a pole.
- There is no net magnetic dipole moment associated with an infinite straight current.
- So the configuration has no poles.
- NCERT: 'This is not so for a toroid or even for a straight infinite conductor.'
- Reject the others: a bar magnet, a solenoid and a current loop all have non-zero net moments and therefore do have poles.
Assumption
An idealised infinite straight conductor carrying a steady current.
Shortcut trick
Memorise NCERT's pair together — toroid and infinite straight conductor. Questions ask for either one, and knowing them as a pair means you recognise the answer whichever is named.
Answer: (B) a straight infinite current-carrying conductor
Closed loops of static magnetic field lines are permitted only when the loop
Given
The rule governing closed magnetostatic field-line loops.
Asked
The condition under which such a loop is legal.
Concept applied
NCERT states the rule in Example 5.3(b) and applies it in 5.3(c). A closed magnetic field line must link a current — that is what Ampere's law requires for a non-zero circulation.
Formula to use
∮B·dl = μ0I ≠ 0 requires enclosed current
Baby steps
- If a closed field line existed with no current through it, then ∮B·dl would be non-zero while Ienclosed = 0.
- That directly contradicts Ampere's circuital law.
- So a closed magnetostatic loop must enclose a region across which a current is passing.
- This is why the toroid diagram is legal — each internal field-line loop encircles the current-carrying windings.
- And it is why the loops-in-empty-space diagram of Example 5.3(b) is wrong.
- By contrast, electrostatic field lines can never form closed loops at all — neither in empty space nor around charges.
Assumption
Static fields; time-varying fields introduce induced effects studied in Chapter 6.
Shortcut trick
Three-way distinction worth holding: magnetic loops are legal if they enclose current; magnetic loops in empty space are illegal; electric loops are always illegal. Diagram questions test all three.
Answer: (B) encloses a region through which current passes
The net magnetic moment of an ideal toroid is
Given
An ideal toroid of N turns carrying current I.
Asked
Its net magnetic moment.
Concept applied
Each turn does carry a moment, but the turns face outward in all directions around the ring. Summing them vectorially around the full circle gives zero, exactly as equal vectors spread uniformly around a circle sum to nothing.
Formula to use
∑mturns = 0 by circular symmetry
Baby steps
- Each individual turn of the toroid is a small current loop with moment IA.
- The moment of each turn points along the local axis of the ring, i.e. tangentially around the toroid.
- As you go round the ring, these directions rotate through a full 360°.
- Vectors of equal magnitude distributed uniformly through 360° sum to zero.
- So the net magnetic moment of the toroid is zero, which is exactly why it has no poles and no external field.
- Contrast with a solenoid, where every turn's moment points the same way and they add to NIA.
Assumption
Ideal toroid, uniformly wound, perfectly circular.
Shortcut trick
Solenoid versus toroid in one line: solenoid = turns aligned, moments add, poles appear. Toroid = turns rotated round a circle, moments cancel, no poles. Same turns, different geometry, opposite conclusions.
Answer: (B) zero
Must every magnetic configuration have a north pole and a south pole?
Given
NCERT's Example 5.1(c).
Asked
Whether poles are universally required.
Concept applied
The absence of monopoles guarantees that poles come in pairs, but it does not guarantee that poles exist at all. A configuration with zero net moment has neither.
Formula to use
poles exist ⇔ net magnetic moment ≠ 0
Baby steps
- Poles are identified by the dipole field a configuration produces at a distance.
- That dipole field exists only if the net magnetic moment is non-zero.
- A toroid and an infinite straight conductor both have zero net moment, and neither has poles.
- So the answer is no — poles appear only when there is a net non-zero magnetic moment.
- NCERT's exact wording: 'Not necessarily. True only if the source of the field has a net non-zero magnetic moment.'
- Option (C) confuses two separate facts: no monopoles means poles cannot be isolated, not that poles must be present.
Assumption
'Pole' is understood in the usual dipole sense.
Shortcut trick
Distinguish the two statements carefully. 'Poles always come in pairs' is true. 'Every configuration has poles' is false. Option (C) welds them together and is the most-chosen wrong answer.
Answer: (B) No — only if the source has a net non-zero magnetic moment
Which statement correctly contrasts a solenoid with a toroid?
Given
An ordinary finite solenoid and an ideal toroid, both carrying current.
Asked
The correct contrast between them.
Concept applied
A solenoid's turns all point the same way, so their moments add and the field must return through the outside space. A toroid's turns are arranged in a circle, so the moments cancel and the field closes internally.
Formula to use
solenoid: m = NIA ≠ 0 | toroid: mnet = 0
Baby steps
- Solenoid: all turn-moments are parallel, so the net moment is NIA, non-zero.
- A non-zero moment produces an external dipole field, with the field lines curving out of one face and into the other — hence a north and a south pole.
- NCERT Example 5.3(d) confirms the external field must exist: lines cannot be cut off at the ends.
- Toroid: turn-moments are distributed round a circle and cancel, so the net moment is zero.
- Zero moment means no external field and no poles, and the lines close entirely inside.
- Answer: option (B).
Assumption
A finite solenoid (with real ends) and an ideal, uniformly wound toroid.
Shortcut trick
The distinction traces back to one question: do the turn-moments add or cancel? Everything else — external field, poles, confinement — follows from that single answer.
Answer: (B) A solenoid has an external field and poles; an ideal toroid has neither
F · Electrostatic analogy, constants and conventions
7 questionsTo convert an electric dipole formula into its magnetic counterpart, the replacement for 1/4πε0 is
Given
NCERT's electrostatic analog rules in section 5.2.4.
Asked
The replacement for the electrostatic constant.
Concept applied
NCERT gives three simultaneous substitutions that turn any electric dipole result into a magnetic one: E → B, p → m, and 1/4πε0 → μ0/4π.
Formula to use
E → B, p → m, 1/4πε0 → μ0/4π
Baby steps
- Take the electric axial field: E = 2p/4πε0r3.
- Replace E by B and p by m.
- Replace the constant 1/4πε0 by μ0/4π.
- Result: B = μ02m/4πr3 — which is NCERT Eq. 5.5. ✔
- The same substitution turns τ = p × E into τ = m × B, and U = −p·E into U = −m·B.
- Numerically μ0/4π = 10−7 T m A−1, whereas 1/4πε0 = 9 × 109 N m2 C−2.
Assumption
Short dipoles in the far-field region, where the analogy is exact.
Shortcut trick
Learn the rule, not the six resulting formulas. Chapter 5 has no new dipole physics — it is Chapter 1 with three symbols swapped. That halves what you have to memorise.
Answer: (A) μ0/4π
In the dipole analogy table, the magnetic counterpart of p × E is
Given
NCERT's Table 5.1, the dipole analogy.
Asked
The magnetic counterpart of the electric torque expression.
Concept applied
Apply the substitution rule term by term. p × E is a torque; replacing p with m and E with B gives the magnetic torque, with the cross product structure preserved.
Formula to use
τelectric = p × E → τmagnetic = m × B
Baby steps
- Identify the quantity: p × E is the torque on an electric dipole.
- Substitute p → m.
- Substitute E → B.
- The cross product structure is unchanged, giving m × B.
- Option (A) is the energy counterpart, matching −p·E → −m·B.
- Note the distinction: torque uses a cross product (giving a vector), energy uses a dot product (giving a scalar).
Assumption
Uniform external field in both cases.
Shortcut trick
Cross versus dot tells you which quantity you are looking at. Cross product ⇒ torque, a vector. Dot product ⇒ energy, a scalar. Options (A) and (B) are placed together to test exactly this.
Answer: (B) m × B
Ampere's hypothesis states that
Given
The hypothesis NCERT attributes to Ampere in section 5.2.2.
Asked
Its content.
Concept applied
Ampere proposed that there is no separate 'magnetic substance' — magnetism in matter arises entirely from currents, whether in wires or in the orbital and spin motion of atomic electrons.
Formula to use
bar magnet ≡ a large number of circulating currents
Baby steps
- NCERT: 'We mentioned Ampere's hypothesis that all magnetic phenomena can be explained in terms of circulating currents.'
- The evidence is that a bar magnet and a solenoid produce indistinguishable far fields.
- In a solenoid the circulating currents are conduction currents in wire.
- In iron they are the orbital and spin motions of electrons inside atoms.
- So a bar magnet may be thought of as a large number of circulating currents.
- Answer: option (B). Reject (A): the hypothesis explicitly removes the need for monopoles rather than requiring them.
Assumption
Classical framework as presented in NCERT; the quantum details of spin lie beyond the chapter.
Shortcut trick
This hypothesis is the through-line of the whole chapter. It explains the bar-magnet/solenoid equivalence, why cutting never yields a monopole, and why diamagnetism arises from disturbed electron orbits. If a question asks 'why', this is usually the answer.
Diagram
Answer: (B) all magnetic phenomena can be explained in terms of circulating currents
The value of μ0/4π is
Given
μ0 = 4π × 10−7 T m A−1.
Asked
The value of μ0/4π.
Concept applied
The 4π in μ0 is deliberately built in so that dividing by 4π leaves a clean power of ten — which is why dipole formulas are written with μ0/4π grouped together.
Formula to use
μ0/4π = (4π × 10−7)/4π = 10−7
Baby steps
- Write μ0 = 4π × 10−7 T m A−1.
- Divide by 4π: the 4π cancels exactly.
- μ0/4π = 10−7 T m A−1.
- This is why the axial field formula is written as (μ0/4π)(2m/r3) rather than μ02m/4πr3 in problem work — the grouped constant is simply 10−7.
- Option (B) is μ0 itself; option (C) is the electrostatic constant 1/4πε0.
- NCERT prints this in the remarks column of its physical-quantities table.
Assumption
SI units.
Shortcut trick
Always substitute the grouped constant 10−7, never μ0 = 4π × 10−7 followed by a division. Carrying π through the arithmetic and cancelling it later is a reliable way to lose a factor.
Answer: (A) 10−7 T m A−1
In NCERT's treatment, the zero of magnetic potential energy is fixed at the orientation where
Given
U = −mB cos θ, with the constant of integration set to zero.
Asked
Where the zero of potential energy sits.
Concept applied
The zero of potential energy is a matter of convention. NCERT takes the integration constant as zero, which places U = 0 at θ = 90° — the needle perpendicular to the field.
Formula to use
U = −mB cos θ; U = 0 when cos θ = 0, i.e. θ = 90°
Baby steps
- Integrate the torque to obtain Um = ∫τ(θ)dθ = ∫mB sin θ dθ = −mB cos θ + C.
- NCERT takes the constant of integration C to be zero.
- Then U = 0 requires cos θ = 0, i.e. θ = 90°.
- NCERT: 'Taking the constant of integration to be zero means fixing the zero of potential energy at θ = 90°, i.e., when the needle is perpendicular to the field.'
- This convention is what makes U = −mB at θ = 0° and U = +mB at θ = 180°.
- Note that at θ = 90° the torque is simultaneously at its maximum — zero energy does not mean zero torque.
Assumption
NCERT's convention; a different choice of C would shift all energies by a constant without changing any work calculation.
Shortcut trick
Energies are only meaningful as differences, so the convention never affects a work answer. But it does affect any question asking for U at a stated angle — and NEET asks those, so the θ = 90° anchor is worth knowing.
Answer: (B) m is perpendicular to B
Does a bar magnet exert a torque on itself due to its own field?
Given
NCERT's Example 5.4(c).
Asked
Whether a magnet twists itself.
Concept applied
A source cannot act on itself. NCERT states there is no force or torque on an element due to the field produced by that element itself — though a different element of the same object can exert one.
Formula to use
self-field exerts no net force or torque on its own source
Baby steps
- Consider a small element of the magnet and the field that element itself produces.
- NCERT: 'There is no force or torque on an element due to the field produced by that element itself.'
- Summed over the whole magnet, the self-torque is therefore zero — the magnet does not spin itself up.
- Answer: No.
- The important refinement NCERT adds: there is a force on an element due to other elements of the same wire or magnet.
- For the special case of a straight wire, even that force is zero.
Assumption
Rigid magnet; no external field present.
Shortcut trick
Do not overstate the result. The correct statement is not 'a magnet exerts no force on itself' but the sharper 'no element is acted on by its own field, though other elements of the same body can act on it.' Subjective answers are marked on that distinction.
Answer: (B) No — no element experiences a force or torque from the field it produces itself
NCERT avoids calling magnetic field lines 'lines of force' because
Given
The footnote in NCERT section 5.2.1 and Example 5.4(a).
Asked
The reason for avoiding the term.
Concept applied
The name suggests a charge would be pushed along the line. In fact the magnetic force is F = qv × B, which is always perpendicular to B, so a charge is never pushed along a field line.
Formula to use
F = qv × B ⊥ B always
Baby steps
- A field line's tangent gives the direction of B at that point.
- The magnetic force on a moving charge is F = qv × B.
- A cross product is perpendicular to both its factors, so F is always perpendicular to B.
- Therefore the force is never along the field line — it is at right angles to it.
- Calling them 'lines of force' would wrongly suggest charges are pushed along them.
- NCERT's footnote: unlike electrostatics, the field lines in magnetism do not indicate the direction of the force on a moving charge. Example 5.4(a) answers the same point with a plain No.
Assumption
A charge moving with non-zero velocity component perpendicular to B; a charge at rest feels no magnetic force at all.
Shortcut trick
Contrast the two cases explicitly. Electric: force is along E, so 'lines of force' is fair. Magnetic: force is perpendicular to B, so the name misleads. That contrast is the expected justification in a written answer.
Answer: (B) the magnetic force on a moving charge is not along them
What Tier 3 is really for
These types appear less often than Tier 1 and 2, but they are cheap marks when prepared and total losses when not. There is no partial credit for half-remembering that a semicircular bend gives 2m/π. You either have the geometry or you do not.
The four results worth memorising outright, because deriving them under time pressure is not realistic:
• Equal moments at angle θ: mR = 2m cos(θ/2), direction along the bisector
• Magnet bent at its midpoint, arms at θ: m′ = m sin(θ/2)
• Magnet bent into an arc of θ radians: m′ = m × 2 sin(θ/2)/θ — semicircle 0.64m, quarter circle 0.90m, full circle 0
• Revolving charge: m = qvr/2, and m/L = e/2me for an electron
Two syllabus notes. The Bohr magneton derivation (Q16) is not in the rationalised NCERT Chapter 5, though the NEET syllabus lists the revolving-electron moment — it appears again in the Gap Content pack. And NCERT's own dimensions table misprints the magnetic moment as [L−2A]; the correct answer is [L2A], as the printed unit A m2 in the same row confirms. Answer keys occasionally reproduce the typo, so know both.
Group C and D are where two-stage errors live. Cut first, then combine. Bend first, then compute the field. Writing the intermediate value on its own line costs three seconds and prevents the single most common mistake in this tier.