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Magnetism and Matter Important Questions

A full-width premium study page for magnetic dipole moment, bar magnets, axial and equatorial fields, torque, potential energy, earth magnetism, magnetic materials, Curie law and hysteresis. Built for serious exam practice with corrected diagrams, solved examples and non-repeating question sets.

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Concept Foundation

Magnetism and Matter: What Students Must Understand First

This chapter is not only a formula chapter. The same result may change by a factor of 2, a sign, or a unit depending on whether the point is axial or equatorial, whether a quantity is pole strength or magnetic moment, and whether a material is diamagnetic, paramagnetic or ferromagnetic.

Magnetic Dipole Thinking

A bar magnet behaves like a dipole with moment M = m × 2l. A current loop behaves like a dipole with moment M = NIA.

Field Geometry

Axial and equatorial points are not interchangeable. Axial field has factor 2; equatorial field is opposite to the magnetic moment.

Material Behaviour

The sign and size of susceptibility explain attraction, repulsion, permeability, retentivity, coercivity and hysteresis loss.

Formula Bank

Important Formulas With Conceptual Use

Before substituting numbers, identify the physical model: pole model, short dipole, current loop, material response or hysteresis cycle. That choice decides the formula.

Bar magnet magnetic moment M = m × 2l Use when pole strength and magnetic length are given.
Current loop magnetic moment M = NIA The direction is given by the right-hand thumb rule.
Torque on a dipole τ = MB sinθ Use for a dipole in a uniform magnetic field.
Potential energy U = -MB cosθ Minimum at θ = 0°, maximum at θ = 180°.
Axial field of short dipole Baxial = (μ0/4π)(2M/r3) Remember the factor 2.
Equatorial field of short dipole Bequatorial = (μ0/4π)(M/r3) Direction is opposite to M.
Magnetisation M = magnetic moment / volume This M is magnetisation in material context.
Susceptibility χ = M/H Sign of χ classifies material response.
Relative permeability μr = 1 + χ Valid for linear magnetic materials.
Magnetic induction in matter B = μ0(H + M) Links applied field and magnetisation.
Curie law χ = C/T Use absolute temperature in kelvin.
Hysteresis loss loss per unit volume per cycle = area of B-H loop For power loss, multiply by volume and frequency.

Corrected Diagrams

Magnetism and Matter Diagrams Students Should Draw Correctly

These SVG diagrams are lightweight and exam-focused. They correct the most common visual mistakes: wrong moment direction, unclear current direction, incomplete domain alignment and poorly labelled hysteresis loops.

Current loop magnetic moment: M = NIA
applied reference direction area vector and M M = NIA area A current I Curl fingers with current; thumb gives M.
Bar magnet as a magnetic dipole
S N M inside: S → N magnetic length 2l M = m × 2l south pole north pole
Diamagnetic, paramagnetic and ferromagnetic response
Diamagnetic applied B induced M opposite B χ < 0 Paramagnetic applied B weak alignment with B small χ > 0 Ferromagnetic applied B strong domain alignment large χ > 0
Ferromagnetic domain theory during magnetisation
increasing external field H Before random domains, net M ≈ 0 Partial alignment domains rotate and grow Strong alignment large net magnetisation
Physically labelled B-H hysteresis loop
H B +Br at H = 0 -Br -Hc +Hc positive saturation negative saturation loop area = energy loss per unit volume per cycle

Magnetic Materials

Diamagnetic, Paramagnetic and Ferromagnetic Materials

Material TypeSusceptibilityResponseExamplesExam Trap
DiamagneticSmall negativeWeakly repelled; induced moment opposite to fieldBi, Cu, waterχ is not zero; it is negative.
ParamagneticSmall positiveWeakly attracted; follows Curie law approximatelyAl, Pt, O2Use temperature in kelvin for χ = C/T.
FerromagneticVery large positiveStrong domain alignment and hysteresisFe, Co, NiDomain theory and hysteresis matter.

Solved Derivations

Five Strong Solved Examples

1Magnetic field on the axial line of a short bar magnet

For a short dipole of moment M, a point P lies on the axis at distance r from the centre, where r is much larger than the magnet length.

  1. Treat the bar magnet as a magnetic dipole because r >> l.
  2. The axial field contributions from the two poles combine along the axis.
  3. The short-dipole result is Baxial = (μ0/4π)(2M/r3).
  4. The direction on the north-side axial line is along M; on the opposite side it is still along the local axial field direction of the dipole.

Use this formula only when the observation distance is large compared with magnet length.

2Magnetic field on the equatorial line of a short bar magnet

For a short dipole of moment M, a point P lies on the equatorial line at distance r from the centre.

  1. The distances from both poles are nearly equal when r >> l.
  2. The axial components cancel and the transverse components add.
  3. The magnitude is Bequatorial = (μ0/4π)(M/r3).
  4. The direction is opposite to the magnetic moment.

This is the formula where students often forget that the direction is opposite to M.

3Torque and potential energy of a magnetic dipole

A dipole of moment M makes angle θ with a uniform magnetic field B.

  1. The field exerts equal and opposite forces on the two poles, forming a couple.
  2. The torque magnitude is τ = MB sinθ.
  3. Potential energy is U = -M·B = -MB cosθ.
  4. Stable equilibrium is θ = 0° with U = -MB; unstable equilibrium is θ = 180° with U = +MB.

Torque is zero at both 0° and 180°, so stability must be decided using energy.

4Magnetic moment of a current loop

A coil has N turns, current I and area A.

  1. Each turn acts like a magnetic dipole.
  2. For one turn, magnetic moment is IA.
  3. For N identical turns, M = NIA.
  4. Direction is perpendicular to the plane by the right-hand thumb rule.

For non-circular coils, A is the actual area enclosed by each turn.

5Hysteresis loss per unit volume per cycle

A ferromagnetic material is taken once around a complete B-H loop.

  1. During a cycle, magnetisation lags behind the magnetising field.
  2. The work done per unit volume in a small change is H dB.
  3. Over one complete cycle, the energy dissipated per unit volume is the closed-loop integral ∮H dB.
  4. Graphically, this equals the area enclosed by the B-H loop.

For AC cores, power loss = loop area × volume × frequency.

Extracted Book Questions

Extracted Book Questions With Step-by-Step Solutions

The visible book-page questions are preserved and polished below. One visibly cropped question is included with an incomplete-data note instead of a fabricated answer.

B1Force between unequal magnetic poles

Question: The force between two magnetic poles in air is 9.604 mN. If one pole is 10 times stronger than the other, calculate the pole strength of each. Distance between poles is 10 cm.

Solution:
  1. Let weaker pole strength be m; stronger pole strength = 10m.
  2. Use F = (μ0/4π)(m1m2/r2) because the poles are separated in air.
  3. 9.604 × 10-3 = 10-7(10m2)/(0.10)2 = 10-4m2.
  4. m2 = 96.04, so m = 9.8 A m and 10m = 98 A m.
Final Answer: 9.8 A m and 98 A m
B2Distance changed for equal poles

Question: Two equal magnetic poles placed 5 cm apart attract each other with a force of 14.4 × 10-4 N. How far apart should they be held so that the force becomes 1.6 × 10-4 N?

Solution:
  1. For the same pair of poles, F ∝ 1/r2.
  2. F1/F2 = r22/r12 = 14.4/1.6 = 9.
  3. r2 = 3r1 = 15 cm = 0.15 m.
Final Answer: 0.15 m
B3Two magnets in a vertical glass tube

Question: Two identical magnets of length 10 cm and weight 50 gf each are arranged with like poles facing in a vertical glass tube. The upper magnet floats above the lower one, and the distance between the nearest poles is 3 mm. Determine the pole strength.

Solution:
  1. The magnetic repulsion supports the weight of the upper magnet: W = 50 gf = 0.49 N.
  2. For finite magnets, include all four pole interactions: W = (μ0m2/4π)[1/d2 + 1/(d+2l)2 - 2/(d+l)2].
  3. With d = 0.003 m and 2l = 0.10 m, substitution gives m ≈ 6.7 A m.
Final Answer: Approximately 6.7 A m
B4Pole strength from magnetic moment

Question: A bar magnet of magnetic moment 5 A m2 has poles 0.20 m apart. Calculate the pole strength.

Solution:
  1. Use M = m × 2l because magnetic moment equals pole strength times magnetic length.
  2. m = M/(2l) = 5/0.20 = 25 A m.
Final Answer: 25 A m
B5Semicircular magnet

Question: A steel wire of length l has magnetic moment M. It is bent into a semicircular arc. What is the new magnetic moment?

Solution:
  1. The pole strength remains the same because the magnetised wire is only bent.
  2. Original pole strength m = M/l.
  3. For a semicircle, l = πR, so pole separation is diameter 2R = 2l/π.
  4. M' = m(2l/π) = (M/l)(2l/π).
Final Answer: 2M/π
B6Equatorial field of a finite bar magnet

Question: Calculate the magnetic induction at a point 4 cm from the centre and along the equator of a bar magnet of length 6 cm and magnetic moment 0.26 A m2.

Solution:
  1. Here r = 0.04 m and l = 0.03 m.
  2. Use the finite equatorial formula B = (μ0/4π) M/(r2+l2)3/2.
  3. r2+l2 = 0.0025 and (0.0025)3/2 = 1.25 × 10-4.
  4. B = 10-7 × 0.26/(1.25 × 10-4) = 2.08 × 10-4 T.
Final Answer: 2.08 × 10-4 T
B7Moment from axial and equatorial fields

Question: The magnetic field at a point 10 cm away from a magnetic dipole is 2.0 × 10-4 T. Find the magnetic moment if the point is (a) end-on and (b) broadside-on.

Solution:
  1. For end-on position, B = (μ0/4π)(2M/r3).
  2. M = Br3/(2 × 10-7) = 1.0 A m2.
  3. For broadside-on position, B = (μ0/4π)(M/r3).
  4. M = Br3/10-7 = 2.0 A m2.
Final Answer: (a) 1.0 A m2, (b) 2.0 A m2
B8Order of Earth's magnetic field

Question: Earth's field is approximated by a dipole of moment 8 × 1022 J T-1 at its centre. Check the order of magnitude.

Solution:
  1. At the magnetic equator, B = (μ0/4π)M/R3.
  2. Using R = 6.4 × 106 m gives B ≈ 3.1 × 10-5 T.
  3. This is the observed order of Earth's magnetic field.
Final Answer: Correct order: about 10-5 T
B9Earth dipole moment from equatorial field

Question: The earth's magnetic field at the equator is approximately 0.4 G. Estimate the earth's dipole moment.

Solution:
  1. 0.4 G = 4 × 10-5 T.
  2. At the equator, M = BR3/10-7.
  3. M = (4 × 10-5)(6.4 × 106)3/10-7 ≈ 1.04 × 1023 A m2.
Final Answer: 1.04 × 1023 A m2
B10Field due to a short bar magnet

Question: Calculate the magnetic field due to a bar magnet 2 cm long and pole strength 100 A m at a point 10 cm from each pole.

Solution:
  1. The point is equidistant from the two poles, so it lies on the equatorial line.
  2. M = m(2l) = 100 × 0.02 = 2 A m2.
  3. B = 10-7M/d3 = 10-7 × 2/(0.10)3.
Final Answer: 2 × 10-4 T
B11Field at geomagnetic poles

Question: The magnetic moment of the assumed dipole at the earth's centre is 8.0 × 1022 A m2. Calculate B at the geomagnetic poles. Radius of earth is 6400 km.

Solution:
  1. At the pole, B = (μ0/4π)(2M/R3).
  2. B = 2 × 10-7 × 8.0 × 1022/(6.4 × 106)3.
  3. B ≈ 6.0 × 10-5 T.
Final Answer: 60 μT
B12Earth dipole moment from equatorial field

Question: The magnetic field at a point on the magnetic equator is 3.1 × 10-5 T. Taking earth's radius as 6400 km, calculate the magnetic moment of the assumed dipole at the centre.

Solution:
  1. Use M = BR3/10-7.
  2. M = (3.1 × 10-5)(6.4 × 106)3/10-7.
  3. M ≈ 8.1 × 1022 A m2.
Final Answer: 8.1 × 1022 A m2
B13Earth field at poles from equator value

Question: If the earth's magnetic field is 3.4 × 10-5 T at the magnetic equator, what is its value at the geomagnetic poles?

Solution:
  1. For a dipole at the same distance, axial field is twice the equatorial field.
  2. Bpole = 2Bequator = 6.8 × 10-5 T.
Final Answer: 6.8 × 10-5 T
B14Magnet in two inclined magnetic fields

Question: A bar magnet is acted upon by two magnetic fields inclined at 60°. One field is 1.2 × 10-2 T. The magnet is in stable equilibrium at 15° with this field. Calculate the other field.

Solution:
  1. The magnet aligns with the resultant field.
  2. Resolve perpendicular to the resultant: B2 sin(60° - 15°) = B1 sin15°.
  3. B2 = 1.2 × 10-2 × sin15°/sin45° = 4.4 × 10-3 T.
Final Answer: 4.4 × 10-3 T
B15Two crossed magnets

Question: Two magnets of magnetic moments M and M√3 are joined to form a cross. The combination is suspended in a uniform magnetic field. In equilibrium, the magnet of moment M makes angle θ with the field. Determine θ.

Solution:
  1. The resultant magnetic moment must lie along the field.
  2. For perpendicular components, tanθ = (M√3)/M = √3.
  3. Therefore θ = 60°.
Final Answer: 60°
B16Torque on a bar magnet

Question: A magnet of length 20 cm and pole strength 2 × 10-5 A m is placed in earth's magnetic field B = 2 × 10-5 T. Find torque when the magnet is parallel and perpendicular to the field.

Solution:
  1. M = m(2l) = 2 × 10-5 × 0.20 = 4 × 10-6 A m2.
  2. τ = MB sinθ.
  3. For θ = 0°, τ = 0. For θ = 90°, τ = 4 × 10-6 × 2 × 10-5.
Final Answer: 0 and 8 × 10-11 N m
B17Work to rotate a circular coil

Question: A circular coil of 100 turns and radius 0.05 m carries 0.1 A. Calculate the work required to turn it through 180° in B = 1.5 T. The plane is initially at right angles to the field.

Solution:
  1. A = πr2 = 7.854 × 10-3 m2.
  2. M = NIA = 100 × 0.1 × 7.854 × 10-3 = 0.07854 A m2.
  3. A 180° rotation from stable to unstable changes energy by 2MB.
  4. W = 2 × 0.07854 × 1.5 ≈ 0.236 J.
Final Answer: 0.24 J
B18Work done in rotating a coil

Question: A circular coil of 50 turns and diameter 8 cm carries 2 A. How much work is done in rotating it through 180° in a uniform field of 0.1 T?

Solution:
  1. r = 0.04 m and A = πr2 = 5.027 × 10-3 m2.
  2. M = NIA = 50 × 2 × 5.027 × 10-3 = 0.5027 A m2.
  3. W = 2MB = 2 × 0.5027 × 0.1 = 0.1005 J.
Final Answer: 0.10 J
B19Torque on solenoid

Question: A solenoid of length 50 cm has 1000 turns per metre and area 2 × 10-4 m2. It is placed at 30° to B = 0.32 T. Find torque when I = 2 A.

Solution:
  1. Total turns N = nL = 1000 × 0.50 = 500.
  2. M = NIA = 500 × 2 × 2 × 10-4 = 0.20 A m2.
  3. τ = MB sin30° = 0.20 × 0.32 × 0.5.
Final Answer: 0.032 N m
B20Moment, equilibrium and potential energy

Question: A short bar magnet at 30° to a field of 0.16 T experiences torque 0.032 N m. Estimate its magnetic moment and the potential energy in stable and unstable equilibrium.

Solution:
  1. τ = MB sinθ, so M = 0.032/(0.16 × 0.5) = 0.4 A m2.
  2. Stable equilibrium is θ = 0°, U = -MB = -0.064 J.
  3. Unstable equilibrium is θ = 180°, U = +MB = +0.064 J.
Final Answer: M = 0.4 A m2; U = -0.064 J and +0.064 J
B21Magnetic moment of a coil

Question: A circular coil of 300 turns and diameter 14 cm carries 15 A. Find its magnetic moment.

Solution:
  1. r = 0.07 m and A = πr2 = 0.015394 m2.
  2. M = NIA = 300 × 15 × 0.015394.
Final Answer: 69.3 A m2
B22Earth as a current loop

Question: The magnetic dipole moment of earth is 6.4 × 1021 A m2. If it were due to a current loop around the magnetic equator, find the current. R = 6400 km.

Solution:
  1. Use M = IA = IπR2.
  2. I = 6.4 × 1021/[π(6.4 × 106)2].
  3. I ≈ 4.97 × 107 A.
Final Answer: 5.0 × 107 A
B23Revolving electron magnetic moment

Question: An electron revolves in an orbit of radius 0.53 Å with frequency 6.8 × 109 MHz. Calculate the equivalent magnetic moment.

Solution:
  1. Convert f = 6.8 × 1015 Hz and r = 0.53 × 10-10 m.
  2. Equivalent current I = ef.
  3. M = IA = efπr2.
  4. M ≈ 9.6 × 10-24 A m2.
Final Answer: 9.6 × 10-24 A m2
B24Electron in orbit

Question: An electron revolves in an orbit of radius 0.5 Å with frequency 1010 MHz. Calculate the equivalent magnetic moment.

Solution:
  1. f = 1016 Hz and r = 0.5 × 10-10 m.
  2. M = efπr2.
  3. M = (1.6 × 10-19)(1016)π(0.5 × 10-10)2.
Final Answer: 1.26 × 10-23 A m2
M1Rowland ring magnetic field

Question: A Rowland ring of mean radius 18 cm has 3500 turns wound on a ferromagnetic core of relative permeability 800. Find B for current 1.2 A.

Solution:
  1. Use B = μ0μrNI/(2πr).
  2. B = (4π × 10-7)(800)(3500)(1.2)/(2π × 0.18).
Final Answer: 3.73 T
M2Intensity of magnetisation from cgs moment

Question: A magnet weighs 75 g and has magnetic moment 2000 cgs units. If density is 7.5 g/cc, calculate intensity of magnetisation.

Solution:
  1. Volume = mass/density = 75/7.5 = 10 cc.
  2. Intensity of magnetisation = magnetic moment/volume = 2000/10.
Final Answer: 200 cgs units
M3Flux density in an iron ring

Question: Calculate magnetic flux density in an iron ring of mean radius 0.25 m with 1500 turns, current 5 A and relative permeability 1600.

Solution:
  1. Use B = μ0μrNI/(2πr).
  2. B = (4π × 10-7)(1600)(1500)(5)/(2π × 0.25).
Final Answer: 9.6 T
M4Cropped iron rod question

Question: An iron rod of area 0.2 cm2 is subjected to a magnetising field of 1200 A m-1. The remaining part of the question is not visible.

Solution:
  1. The visible information gives H and area only.
  2. A unique answer needs the missing target quantity and any required material data such as χ, μr, length or volume.
  3. If χ were given, M = χH; if volume were given, magnetic moment = MV.
Final Answer: Incomplete data in the uploaded image
M5Magnetic intensity near a pole

Question: Calculate H at a distance 20 cm from a pole of strength 40 A m in air. Find B at the same point.

Solution:
  1. H = m/(4πr2) = 40/[4π(0.20)2] = 79.54 A m-1.
  2. B = μ0H = 4π × 10-7 × 79.54.
Final Answer: H = 79.54 A m-1; B = 1.0 × 10-4 T
M6Relative permeability of toroid core

Question: A toroid has 2000 turns, inner radius 11 cm and outer radius 12 cm. B = 2.5 T when I = 0.7 A. Find relative permeability.

Solution:
  1. Use mean radius r = 0.115 m.
  2. μr = B(2πr)/(μ0NI).
  3. μr = 2.5 × 2π × 0.115/[(4π × 10-7)(2000)(0.7)].
Final Answer: 1027 approximately
M7Permeability from susceptibility

Question: The susceptibility of annealed iron at saturation is 5500. Find the permeability.

Solution:
  1. μr = 1 + χ = 5501.
  2. μ = μ0μr = 4π × 10-7 × 5501.
Final Answer: 6.9 × 10-3 T m A-1
M8Relative permeability and susceptibility

Question: In a material, B = 1.6 T and H = 1000 A m-1. Calculate μr and χ.

Solution:
  1. μr = B/(μ0H).
  2. μr = 1.6/[(4π × 10-7)(1000)] ≈ 1.27 × 103.
  3. χ = μr - 1 ≈ 1.27 × 103.
Final Answer: μr ≈ 1.27 × 103, χ ≈ 1.27 × 103
M9Mu-metal permeability

Question: The maximum permeability of μ-metal is 0.126 T m A-1. Find maximum relative permeability and susceptibility.

Solution:
  1. μr = μ/μ0 = 0.126/(4π × 10-7) ≈ 1.00 × 105.
  2. χ = μr - 1 ≈ 1.00 × 105.
Final Answer: Both approximately 1.00 × 105
M10Curie law for aluminium

Question: For H = 2 × 103 A m-1, aluminium at 280 K has magnetisation 4.8 × 10-2 A m-1. Find χ at 280 K, and χ and M at 320 K.

Solution:
  1. χ1 = M/H = 4.8 × 10-2/(2 × 103) = 2.4 × 10-5.
  2. For a paramagnetic material, χT = constant.
  3. χ2 = χ1T1/T2 = 2.1 × 10-5.
  4. M2 = χ2H = 4.2 × 10-2 A m-1.
Final Answer: 2.4 × 10-5; 2.1 × 10-5; 4.2 × 10-2 A m-1
M11Hysteresis loss per unit volume per cycle

Question: An iron sample of mass 8.4 kg is cycled at 50 Hz. It dissipates 3.2 × 104 J in 30 min. Density is 7200 kg m-3. Find energy dissipated per unit volume per cycle.

Solution:
  1. Volume V = m/ρ = 8.4/7200 = 1.1667 × 10-3 m3.
  2. Total cycles = ft = 50 × 1800 = 90000.
  3. Loss per unit volume per cycle = E/(Vft).
  4. Loss = 3.2 × 104/(1.1667 × 10-3 × 90000).
Final Answer: 304.8 J m-3 cycle-1
M12Magnetic moment of iron rod in solenoid

Question: An iron rod of volume 10-4 m3 and relative permeability 1000 is inside a long solenoid of 5 turns/cm. Current is 0.5 A. Find magnetic moment of the rod.

Solution:
  1. n = 5 turns/cm = 500 m-1, so H = nI = 250 A m-1.
  2. χ = μr - 1 = 999.
  3. Magnetisation M = χH = 249750 A m-1.
  4. Magnetic moment = MV = 249750 × 10-4.
Final Answer: 25 A m2
M13Bar magnet material quantities

Question: A bar magnet has pole strength 4.5 A m, magnetic length 12 cm and cross-section 0.9 cm2. Find magnetisation, H at centre and B at centre.

Solution:
  1. Area A = 0.9 × 10-4 m2.
  2. Magnetisation I = m/A = 4.5/(0.9 × 10-4) = 5 × 104 A m-1.
  3. Using the pole model at the centre, H ≈ (m/4π)(2/l2) with l = 0.06 m gives about 199 A m-1.
  4. B = μ0(H + I) ≈ 6.26 × 10-2 T.
Final Answer: 5 × 104 A m-1; 199 A m-1; 6.26 × 10-2 T

NEET Practice

NEET Conceptual MCQs

These original NEET-level questions target formula selection, directions, material classification and common one-line traps.

NC1The magnetic moment of a bar magnet is directed

Question: The magnetic moment of a bar magnet is directed

  1. A) from north to south inside the magnet
  2. B) from south to north inside the magnet
  3. C) radially outward from both poles
  4. D) perpendicular to the magnetic axis
Answer: B
Explanation: The magnetic dipole moment of a bar magnet is conventionally from south pole to north pole inside the magnet.
NC2The SI unit of magnetic pole strength is

Question: The SI unit of magnetic pole strength is

  1. A) A m2
  2. B) A m
  3. C) T m
  4. D) N m
Answer: B
Explanation: Since M = m(2l), m has unit A m2/m = A m.
NC3For a short magnetic dipole at the same distance, the axial field magnitude is

Question: For a short magnetic dipole at the same distance, the axial field magnitude is

  1. A) half the equatorial field
  2. B) equal to the equatorial field
  3. C) twice the equatorial field
  4. D) four times the equatorial field
Answer: C
Explanation: Baxial = 2(μ0/4π)M/r3, while Bequatorial = (μ0/4π)M/r3.
NC4Torque on a magnetic dipole in a uniform magnetic field is maximum when

Question: Torque on a magnetic dipole in a uniform magnetic field is maximum when

  1. A) θ = 0°
  2. B) θ = 45°
  3. C) θ = 90°
  4. D) θ = 180°
Answer: C
Explanation: τ = MB sinθ is maximum when sinθ = 1.
NC5A diamagnetic material has

Question: A diamagnetic material has

  1. A) large positive susceptibility
  2. B) small negative susceptibility
  3. C) zero permeability
  4. D) permanent domains strongly aligned
Answer: B
Explanation: Diamagnetism is weak and induced opposite to the applied magnetic field, so χ is small and negative.
NC6Curie law for a paramagnetic substance states that

Question: Curie law for a paramagnetic substance states that

  1. A) χ ∝ T
  2. B) χ ∝ 1/T
  3. C) χ is independent of T
  4. D) χ ∝ T2
Answer: B
Explanation: For a paramagnetic material in the Curie-law range, χ = C/T.
NC7The area of a B-H hysteresis loop represents

Question: The area of a B-H hysteresis loop represents

  1. A) retentivity only
  2. B) coercivity only
  3. C) energy loss per unit volume per cycle
  4. D) magnetic moment per unit volume
Answer: C
Explanation: The loop area equals hysteresis energy loss per unit volume in one complete cycle.
NC8For a linear isotropic magnetic material, the correct relation is

Question: For a linear isotropic magnetic material, the correct relation is

  1. A) μr = 1 - χ
  2. B) μr = 1 + χ
  3. C) μrχ = 1
  4. D) B = μ0M only
Answer: B
Explanation: Using B = μ0(H+M) and M = χH, B = μ0(1+χ)H, so μr = 1 + χ.
NC9A good permanent magnet should have

Question: A good permanent magnet should have

  1. A) high retentivity and high coercivity
  2. B) low retentivity and low coercivity
  3. C) high permeability and zero coercivity
  4. D) large hysteresis loss and low retentivity
Answer: A
Explanation: High retentivity helps it retain magnetisation; high coercivity resists demagnetisation.
NC10A transformer core is usually made of a material with

Question: A transformer core is usually made of a material with

  1. A) wide hysteresis loop
  2. B) narrow hysteresis loop and high permeability
  3. C) high coercivity and low permeability
  4. D) negative susceptibility
Answer: B
Explanation: A narrow loop reduces heat loss, and high permeability helps produce strong flux.
NC11At stable equilibrium of a magnetic dipole in uniform B, its potential energy is

Question: At stable equilibrium of a magnetic dipole in uniform B, its potential energy is

  1. A) maximum
  2. B) minimum
  3. C) zero for all M
  4. D) undefined
Answer: B
Explanation: U = -MB cosθ is minimum at θ = 0°, where the dipole aligns with the field.
NC12Retentivity is the value of

Question: Retentivity is the value of

  1. A) B or M left when H becomes zero
  2. B) H required to make B maximum
  3. C) B required to make H zero
  4. D) current per unit length
Answer: A
Explanation: Retentivity is the residual magnetisation or flux density after the magnetising field is removed.
NC13Coercivity is the reverse field needed to

Question: Coercivity is the reverse field needed to

  1. A) make magnetisation maximum
  2. B) reduce residual magnetisation to zero
  3. C) increase magnetic length
  4. D) reverse current in a loop
Answer: B
Explanation: Coercive field Hc is the reverse field required to bring B or M back to zero after retentivity.
NC14If a bar magnet is cut into two equal halves perpendicular to its length, each part has

Question: If a bar magnet is cut into two equal halves perpendicular to its length, each part has

  1. A) same moment as original
  2. B) zero pole strength
  3. C) half the magnetic moment approximately
  4. D) double pole strength
Answer: C
Explanation: Pole strength remains nearly the same but magnetic length halves, so magnetic moment halves.
NC15Magnetisation is defined as

Question: Magnetisation is defined as

  1. A) magnetic moment per unit volume
  2. B) magnetic field per unit current
  3. C) pole strength per unit area only
  4. D) force per pole strength
Answer: A
Explanation: Magnetisation M is magnetic dipole moment per unit volume of the material.
NC16For a current loop, the magnetic moment is perpendicular to the plane of the loop according to

Question: For a current loop, the magnetic moment is perpendicular to the plane of the loop according to

  1. A) left-hand palm rule
  2. B) right-hand thumb rule
  3. C) Fleming's left-hand rule
  4. D) Lenz's law only
Answer: B
Explanation: Curl the fingers along current; the thumb gives the direction of area vector and magnetic moment.
NC17A ferromagnetic material below Curie temperature mainly differs from a paramagnet because it has

Question: A ferromagnetic material below Curie temperature mainly differs from a paramagnet because it has

  1. A) no atomic dipoles
  2. B) strong magnetic domains
  3. C) negative susceptibility
  4. D) zero permeability
Answer: B
Explanation: Ferromagnetism arises from exchange interaction and domain formation.
NC18The horizontal component of Earth's magnetic field is related to dip angle δ by

Question: The horizontal component of Earth's magnetic field is related to dip angle δ by

  1. A) BH = B cosδ
  2. B) BH = B sinδ
  3. C) BH = B tanδ
  4. D) BH = B/ tanδ
Answer: A
Explanation: If δ is the angle made by the total field with the horizontal, its horizontal component is B cosδ.
NC19Which material is weakly repelled by a magnet?

Question: Which material is weakly repelled by a magnet?

  1. A) iron
  2. B) cobalt
  3. C) bismuth
  4. D) aluminium
Answer: C
Explanation: Bismuth is diamagnetic; aluminium is paramagnetic and iron/cobalt are ferromagnetic.
NC20The magnetic field on the equatorial line of a short bar magnet is directed

Question: The magnetic field on the equatorial line of a short bar magnet is directed

  1. A) along the magnetic moment
  2. B) opposite to the magnetic moment
  3. C) perpendicular to the magnetic moment
  4. D) randomly
Answer: B
Explanation: At an equatorial point, the resultant field is opposite to the magnetic moment direction.

NEET Practice

NEET Numerical MCQs

These questions are calculation-focused but intentionally short, matching the speed and accuracy NEET demands.

NN1A coil has 200 turns, area 4 × 10-3 m2 and current 2 A. Its magnetic moment is

Question: A coil has 200 turns, area 4 × 10-3 m2 and current 2 A. Its magnetic moment is

  1. A) 0.4 A m2
  2. B) 1.6 A m2
  3. C) 4.0 A m2
  4. D) 16 A m2
Answer: B
Explanation: M = NIA = 200 × 2 × 4 × 10-3 = 1.6 A m2.
NN2A bar magnet has M = 6 A m2 and pole separation 0.30 m. Pole strength is

Question: A bar magnet has M = 6 A m2 and pole separation 0.30 m. Pole strength is

  1. A) 10 A m
  2. B) 20 A m
  3. C) 30 A m
  4. D) 60 A m
Answer: B
Explanation: m = M/(2l) = 6/0.30 = 20 A m.
NN3A short dipole of M = 0.5 A m2 is observed on its axial line at r = 0.10 m. B is

Question: A short dipole of M = 0.5 A m2 is observed on its axial line at r = 0.10 m. B is

  1. A) 1 × 10-4 T
  2. B) 5 × 10-5 T
  3. C) 1 × 10-3 T
  4. D) 2 × 10-5 T
Answer: A
Explanation: B = 2 × 10-7M/r3 = 2 × 10-7 × 0.5/10-3 = 1 × 10-4 T.
NN4For the same dipole as NN3, the equatorial field at 0.10 m is

Question: For the same dipole as NN3, the equatorial field at 0.10 m is

  1. A) 1 × 10-4 T
  2. B) 5 × 10-5 T
  3. C) 2 × 10-4 T
  4. D) 5 × 10-6 T
Answer: B
Explanation: B = 10-7M/r3 = 5 × 10-5 T.
NN5A dipole of M = 0.5 A m2 in B = 0.2 T has maximum torque

Question: A dipole of M = 0.5 A m2 in B = 0.2 T has maximum torque

  1. A) 0.01 N m
  2. B) 0.05 N m
  3. C) 0.10 N m
  4. D) 0.20 N m
Answer: C
Explanation: τmax = MB = 0.5 × 0.2 = 0.10 N m.
NN6For M = 0.4 A m2 and B = 0.16 T, energy difference between unstable and stable orientations is

Question: For M = 0.4 A m2 and B = 0.16 T, energy difference between unstable and stable orientations is

  1. A) 0.032 J
  2. B) 0.064 J
  3. C) 0.128 J
  4. D) 0.256 J
Answer: C
Explanation: ΔU = (+MB) - (-MB) = 2MB = 2 × 0.4 × 0.16 = 0.128 J.
NN7If magnetisation is 60 A m-1 for H = 2 × 103 A m-1, χ is

Question: If magnetisation is 60 A m-1 for H = 2 × 103 A m-1, χ is

  1. A) 0.003
  2. B) 0.03
  3. C) 3
  4. D) 30
Answer: B
Explanation: χ = M/H = 60/2000 = 0.03.
NN8If χ = 999, relative permeability is

Question: If χ = 999, relative permeability is

  1. A) 998
  2. B) 999
  3. C) 1000
  4. D) 1001
Answer: C
Explanation: μr = 1 + χ = 1000.
NN9A paramagnetic sample has χ = 3 × 10-5 at 300 K. At 600 K, χ becomes

Question: A paramagnetic sample has χ = 3 × 10-5 at 300 K. At 600 K, χ becomes

  1. A) 1.5 × 10-5
  2. B) 3 × 10-5
  3. C) 6 × 10-5
  4. D) 9 × 10-5
Answer: A
Explanation: By Curie law, χT = constant, so doubling T halves χ.
NN10A solenoid has n = 1000 m-1 and I = 2 A. The magnetising field H is

Question: A solenoid has n = 1000 m-1 and I = 2 A. The magnetising field H is

  1. A) 500 A m-1
  2. B) 1000 A m-1
  3. C) 2000 A m-1
  4. D) 4000 A m-1
Answer: C
Explanation: For a long solenoid, H = nI = 2000 A m-1.
NN11A sample loses 900 J in volume 0.01 m3 over 300 cycles. Loss per unit volume per cycle is

Question: A sample loses 900 J in volume 0.01 m3 over 300 cycles. Loss per unit volume per cycle is

  1. A) 0.3 J m-3
  2. B) 3 J m-3
  3. C) 30 J m-3
  4. D) 300 J m-3
Answer: D
Explanation: Loss = E/(Vn) = 900/(0.01 × 300) = 300 J m-3.
NN12For μr = 500 and H = 100 A m-1, B is approximately

Question: For μr = 500 and H = 100 A m-1, B is approximately

  1. A) 2π × 10-2 T
  2. B) 2π × 10-3 T
  3. C) 4π × 10-5 T
  4. D) 5 T
Answer: A
Explanation: B = μ0μrH = 4π × 10-7 × 500 × 100 = 2π × 10-2 T.
NN13Two equal poles of strength 20 A m are 0.10 m apart in air. Force between them is

Question: Two equal poles of strength 20 A m are 0.10 m apart in air. Force between them is

  1. A) 4 × 10-3 N
  2. B) 4 × 10-4 N
  3. C) 2 × 10-3 N
  4. D) 8 × 10-3 N
Answer: A
Explanation: F = 10-7m2/r2 = 10-7 × 400/0.01 = 4 × 10-3 N.
NN14If distance between two poles is doubled, the force becomes

Question: If distance between two poles is doubled, the force becomes

  1. A) twice
  2. B) half
  3. C) one-fourth
  4. D) four times
Answer: C
Explanation: Pole force follows inverse-square dependence, so F' = F/4.
NN15A single-turn circular loop of radius 0.20 m has magnetic moment 0.628 A m2. Current is approximately

Question: A single-turn circular loop of radius 0.20 m has magnetic moment 0.628 A m2. Current is approximately

  1. A) 1 A
  2. B) 2 A
  3. C) 5 A
  4. D) 10 A
Answer: C
Explanation: M = IA = Iπr2. A = π(0.20)2 ≈ 0.126 m2; I ≈ 0.628/0.126 = 5 A.
NN16For M = 2 A m2, B = 0.5 T and θ = 60°, potential energy is

Question: For M = 2 A m2, B = 0.5 T and θ = 60°, potential energy is

  1. A) -0.5 J
  2. B) -1.0 J
  3. C) +0.5 J
  4. D) +1.0 J
Answer: A
Explanation: U = -MB cosθ = -2 × 0.5 × 0.5 = -0.5 J.
NN17If Earth's horizontal field is 3.0 × 10-5 T and dip angle is 60°, total field is

Question: If Earth's horizontal field is 3.0 × 10-5 T and dip angle is 60°, total field is

  1. A) 1.5 × 10-5 T
  2. B) 3.0 × 10-5 T
  3. C) 6.0 × 10-5 T
  4. D) 9.0 × 10-5 T
Answer: C
Explanation: BH = B cosδ, so B = 3.0 × 10-5/0.5 = 6.0 × 10-5 T.
NN18If equatorial field due to Earth's dipole is 3.2 × 10-5 T, polar field is

Question: If equatorial field due to Earth's dipole is 3.2 × 10-5 T, polar field is

  1. A) 1.6 × 10-5 T
  2. B) 3.2 × 10-5 T
  3. C) 6.4 × 10-5 T
  4. D) 9.6 × 10-5 T
Answer: C
Explanation: At the same distance for a dipole, polar field is twice equatorial field.
NN19A wire magnet of moment M is bent into a semicircle. The new moment is

Question: A wire magnet of moment M is bent into a semicircle. The new moment is

  1. A) M/π
  2. B) 2M/π
  3. C) πM/2
  4. D) M
Answer: B
Explanation: Arc length l = πR and pole separation is 2R, so M' = (M/l)(2l/π) = 2M/π.
NN20If μ = 4π × 10-4 T m A-1, relative permeability is

Question: If μ = 4π × 10-4 T m A-1, relative permeability is

  1. A) 10
  2. B) 100
  3. C) 1000
  4. D) 10000
Answer: C
Explanation: μr = μ/μ0 = (4π × 10-4)/(4π × 10-7) = 1000.

JEE Main

JEE Main Questions With Step-by-Step Solutions

The JEE Main set mixes numerical substitution, proportional reasoning, units and conceptual checks.

JM1Axial field comparison

Question: A short bar magnet has magnetic moment 1.5 A m2. Find the ratio of fields at 10 cm and 20 cm on the axial line.

Solution:
  1. Axial field varies as 1/r3 for the same dipole.
  2. B10/B20 = (20/10)3 = 8.
Final Answer: 8:1
JM2Dipole moment from torque

Question: A magnet placed at 30° in a uniform field of 0.25 T experiences torque 0.05 N m. Find its magnetic moment.

Solution:
  1. Use τ = MB sinθ.
  2. M = 0.05/(0.25 × 0.5) = 0.4 A m2.
Final Answer: 0.4 A m2
JM3Energy required for rotation

Question: A magnetic dipole of moment 0.8 A m2 is rotated slowly from θ = 0° to θ = 90° in a field 0.5 T. Find external work.

Solution:
  1. External work equals increase in potential energy.
  2. U = -MB cosθ.
  3. ΔU = 0 - (-MB) = MB = 0.8 × 0.5.
Final Answer: 0.4 J
JM4Pole strength from force

Question: Two equal magnetic poles repel with force 2.5 × 10-3 N at separation 0.20 m. Find each pole strength.

Solution:
  1. F = 10-7m2/r2.
  2. m2 = Fr2/10-7 = 2.5 × 10-3 × 0.04/10-7 = 1000.
  3. m = 31.6 A m.
Final Answer: 31.6 A m
JM5Finite equatorial field

Question: A bar magnet has length 8 cm and moment 0.96 A m2. Find B at an equatorial point 6 cm from the centre.

Solution:
  1. Here l = 0.04 m and r = 0.06 m.
  2. B = 10-7M/(r2+l2)3/2.
  3. r2+l2 = 0.0052.
  4. (0.0052)3/2 ≈ 3.75 × 10-4.
Final Answer: 2.56 × 10-4 T
JM6Loop moment

Question: A square coil of side 10 cm has 50 turns and carries 4 A. Find its magnetic moment.

Solution:
  1. Area = (0.10)2 = 0.01 m2.
  2. M = NIA = 50 × 4 × 0.01.
Final Answer: 2 A m2
JM7Permeability and susceptibility

Question: In a medium, B = 0.12 T when H = 800 A m-1. Find μr and χ.

Solution:
  1. μr = B/(μ0H).
  2. μr = 0.12/[(4π × 10-7)800] ≈ 119.
  3. χ = μr - 1 ≈ 118.
Final Answer: μr ≈ 119, χ ≈ 118
JM8Curie-law temperature change

Question: A paramagnetic salt has χ = 4.0 × 10-4 at 300 K. At what temperature will χ become 3.0 × 10-4?

Solution:
  1. For Curie law, χT = constant.
  2. T2 = χ1T1/χ2 = (4.0 × 10-4 × 300)/(3.0 × 10-4).
Final Answer: 400 K
JM9Hysteresis power loss

Question: A core of volume 2 × 10-3 m3 has hysteresis loss 150 J m-3 per cycle. If frequency is 50 Hz, find power loss.

Solution:
  1. Energy per cycle = loss density × volume = 150 × 2 × 10-3 = 0.30 J.
  2. Power = energy per cycle × frequency = 0.30 × 50.
Final Answer: 15 W
JM10Earth field components

Question: At a place, total Earth's field is 5 × 10-5 T and dip angle is 37°. Find horizontal and vertical components. Use sin37° = 0.6, cos37° = 0.8.

Solution:
  1. BH = B cosδ = 5 × 10-5 × 0.8.
  2. BV = B sinδ = 5 × 10-5 × 0.6.
Final Answer: BH = 4 × 10-5 T, BV = 3 × 10-5 T
JM11Current for a required moment

Question: A circular coil of radius 5 cm has 200 turns. What current gives magnetic moment 1 A m2?

Solution:
  1. M = NIA = NIπr2.
  2. I = M/(Nπr2) = 1/[200π(0.05)2].
Final Answer: 0.637 A
JM12Magnet cut along its length

Question: A bar magnet of moment M is cut into two equal pieces along its length. What is the moment of each piece?

Solution:
  1. Length stays same but cross-section halves.
  2. Pole strength becomes half.
  3. M' = (m/2)(2l) = M/2.
Final Answer: M/2
JM13Magnet cut perpendicular to length

Question: A bar magnet of moment M is cut into two equal pieces perpendicular to its length. What is the moment of each piece?

Solution:
  1. Pole strength remains approximately the same.
  2. Magnetic length becomes half.
  3. M' = m(2l/2) = M/2.
Final Answer: M/2
JM14Two crossed magnetic moments

Question: Two magnets of moments 3 A m2 and 4 A m2 are fixed perpendicular to each other. Find resultant moment and its angle with the 3 A m2 magnet.

Solution:
  1. Resultant M = √(32 + 42) = 5 A m2.
  2. tanθ = 4/3.
Final Answer: 5 A m2, θ = tan-1(4/3)
JM15Magnetic field from orbital electron

Question: An electron revolves with frequency f in a circular orbit of radius r. Write the expression for orbital magnetic moment.

Solution:
  1. The orbit is equivalent to current I = ef.
  2. Area = πr2.
  3. M = IA.
Final Answer: M = efπr2
JM16Equilibrium orientation

Question: A dipole is free in a uniform magnetic field. Which orientations are equilibrium and which is stable?

Solution:
  1. Equilibrium requires τ = MB sinθ = 0.
  2. So θ = 0° and θ = 180°.
  3. U = -MB cosθ is minimum at 0° and maximum at 180°.
Final Answer: Stable: 0°; unstable: 180°
JM17Solenoid magnetising field

Question: A long solenoid has 800 turns/m and carries 1.5 A. A material with χ = 200 is inserted. Find magnetisation.

Solution:
  1. H = nI = 800 × 1.5 = 1200 A m-1.
  2. M = χH = 200 × 1200.
Final Answer: 2.4 × 105 A m-1
JM18Magnetic induction in material

Question: For H = 500 A m-1 and M = 1.5 × 105 A m-1, find B.

Solution:
  1. B = μ0(H + M).
  2. H is negligible compared with M but should still be included.
  3. B = 4π × 10-7(150500).
Final Answer: 0.189 T approximately
JM19Axial field moment estimate

Question: At 20 cm on the axial line of a short magnet, B = 5 × 10-5 T. Estimate M.

Solution:
  1. B = 2 × 10-7M/r3.
  2. M = Br3/(2 × 10-7) = 5 × 10-5 × (0.20)3/(2 × 10-7).
Final Answer: 2 A m2
JM20Loop in a field

Question: A coil with magnetic moment 0.25 A m2 is in a field 0.8 T. Find torque at 30° and potential energy at 60°.

Solution:
  1. τ = MB sin30° = 0.25 × 0.8 × 0.5.
  2. U = -MB cos60° = -0.25 × 0.8 × 0.5.
Final Answer: τ = 0.10 N m, U = -0.10 J

JEE Advanced

JEE Advanced Style Questions

These are higher-level but not artificially difficult. They require careful reasoning about directions, limiting cases, graph interpretation and material choice.

JA1Multiple correct: dipole fields

Question: For a short magnetic dipole of moment M, choose the correct statements about points at the same distance r.

Key statements / solution:
  • Axial field magnitude is twice equatorial field magnitude.
  • Equatorial field is opposite to the magnetic moment.
  • Axial field is along the magnetic moment on the north-side axial line.
  • Both magnitudes vary as 1/r3.
Reasoning:
  1. The standard dipole field formulas give the factor 2 and 1/r3 dependence.
  2. Direction must be read from the vector field, not just magnitudes.
Answer: All four statements are correct
JA2Combined dipole and torque

Question: Two perpendicular magnetic moments 2M and M are rigidly fixed. The combination is placed in a uniform B field. Find the maximum torque and stable angle made by the resultant with the 2M arm.

Key statements / solution:
  • Resultant moment R = √[(2M)2 + M2] = √5 M.
  • Maximum torque = RB = √5 MB.
  • Stable equilibrium occurs when resultant is along B.
  • tanθ = M/(2M) = 1/2.
Reasoning:
  1. Treat the two fixed magnets as one resultant dipole.
  2. Use vector addition for perpendicular magnetic moments.
  3. Maximum torque depends on the resultant moment, not on either arm alone.
Answer: τmax = √5 MB, θ = tan-1(1/2)
JA3Matrix match: materials

Question: Match the material type with susceptibility and behaviour.

Key statements / solution:
  • Diamagnetic → χ < 0 → weak repulsion.
  • Paramagnetic → small χ > 0 → weak attraction and Curie law.
  • Ferromagnetic → large χ > 0 → domain alignment and hysteresis.
  • Transformer core → high permeability with small hysteresis area.
Reasoning:
  1. The key discriminator is the sign and size of χ, plus whether domains and hysteresis are important.
Answer: Correct matching shown in solution
JA4Energy path independence

Question: A dipole of moment M is slowly rotated in a uniform magnetic field from angle α to β. Show the work done by an external agent.

Key statements / solution:
  • Potential energy is U = -MB cosθ.
  • For slow rotation, external work = ΔU.
  • W = -MB cosβ - (-MB cosα).
Reasoning:
  1. The magnetic force is conservative for a fixed uniform field.
  2. Therefore the work depends on initial and final angles, not on the path.
Answer: W = MB(cosα - cosβ)
JA5Finite magnet limit

Question: For a finite bar magnet, the equatorial field is B = (μ0/4π)M/(r2+l2)3/2. Show how the short magnet formula appears.

Key statements / solution:
  • If r >> l, then r2 + l2 ≈ r2.
  • (r2+l2)3/2 ≈ r3.
  • So B ≈ (μ0/4π)M/r3.
Reasoning:
  1. The short magnet approximation is a limiting case.
  2. It is valid only when observation distance is much larger than half-length.
Answer: Short-dipole equatorial formula recovered
JA6Multiple correct: hysteresis loop

Question: Which statements are correct for a B-H loop?

Key statements / solution:
  • Retentivity is read at H = 0.
  • Coercivity is read at B = 0.
  • Loop area gives energy loss per unit volume per cycle.
  • A transformer core should use a material with small loop area.
Reasoning:
  1. These are direct interpretations of the B-H loop used in material selection.
Answer: All four statements are correct
JA7Magnetising field and induction

Question: A material obeys M = χH with χ = 499. It is inside a long solenoid with n = 1200 m-1, I = 2 A. Find H, M and B.

Key statements / solution:
  • H = nI = 2400 A m-1.
  • M = χH = 499 × 2400 = 1.1976 × 106 A m-1.
  • B = μ0(H+M) = μ0(500H).
Reasoning:
  1. Do not confuse magnetisation M with magnetic moment.
  2. In material problems, B is found from H + M.
Answer: H = 2400 A m-1, M = 1.1976 × 106 A m-1, B ≈ 1.51 T
JA8Paragraph style: compass deflection

Question: A small magnet is placed on the axial line of a compass needle. The magnet's field at the compass is equal to the horizontal component of Earth's field. Find the deflection direction condition and angle.

Key statements / solution:
  • The compass aligns with the resultant of Earth's horizontal field and magnet field.
  • If the two fields are perpendicular and equal, tanθ = Bmagnet/BH = 1.
  • θ = 45° toward the magnet's field.
Reasoning:
  1. A compass does not align with either field separately.
  2. It aligns with the vector resultant of the two horizontal fields.
Answer: 45°, when the magnet field is perpendicular to Earth's horizontal field
JA9Multiple correct: cutting magnets

Question: A uniformly magnetised bar magnet is cut into two equal parts. Choose correct statements.

Key statements / solution:
  • If cut perpendicular to length, each part has nearly half the magnetic moment.
  • If cut along length, each part also has nearly half the magnetic moment.
  • Each piece is still a complete magnet with both poles.
  • Isolated north or south poles are not obtained.
Reasoning:
  1. Cutting changes dimensions or pole strength, but each fragment remains a dipole.
Answer: All four statements are correct
JA10Curie law with graph thinking

Question: A graph of 1/χ versus T for a paramagnetic material is plotted. What should be its form under Curie law, and how is C obtained?

Key statements / solution:
  • Curie law: χ = C/T.
  • Therefore 1/χ = T/C.
  • The graph of 1/χ versus T is a straight line through origin.
  • Slope = 1/C.
Reasoning:
  1. Rearrange the law into the plotted variables.
  2. The constant is obtained from the reciprocal of the slope.
Answer: Straight line through origin; C = 1/slope

Assertion-Reason

Assertion-Reason Questions

AR1A permanent magnet should have high coercivity.

Assertion: A permanent magnet should have high coercivity.
Reason: High coercivity means the material is difficult to demagnetise.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: A
Explanation: Both statements are true and the reason directly explains why high coercivity is useful.
AR2The axial field of a short dipole is twice the equatorial field at the same distance.

Assertion: The axial field of a short dipole is twice the equatorial field at the same distance.
Reason: The axial and equatorial fields have the same direction.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: C
Explanation: The magnitude relation is true, but the direction statement is false; equatorial field is opposite to magnetic moment.
AR3Diamagnetic materials are weakly repelled by a magnetic field.

Assertion: Diamagnetic materials are weakly repelled by a magnetic field.
Reason: Their induced magnetic moment is opposite to the applied field.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: A
Explanation: The opposite induced moment causes weak repulsion.
AR4A transformer core should have a wide hysteresis loop.

Assertion: A transformer core should have a wide hysteresis loop.
Reason: A wide loop reduces heat loss per cycle.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: D
Explanation: The reason is false as stated; a narrow loop reduces hysteresis loss.
AR5For a current loop, M = NIA.

Assertion: For a current loop, M = NIA.
Reason: The direction of M is given by the right-hand thumb rule.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: B
Explanation: Both are true, but the direction rule does not explain the magnitude formula.
AR6At θ = 180°, a dipole in a uniform field is in unstable equilibrium.

Assertion: At θ = 180°, a dipole in a uniform field is in unstable equilibrium.
Reason: At θ = 180°, potential energy U = +MB is maximum.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: A
Explanation: Maximum potential energy corresponds to unstable equilibrium.
AR7Curie law applies to ferromagnets at all temperatures.

Assertion: Curie law applies to ferromagnets at all temperatures.
Reason: Paramagnetic susceptibility varies inversely with absolute temperature.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: D
Explanation: The assertion is false; the reason is true for paramagnetic materials in the Curie-law range.
AR8Coercivity is read where the hysteresis loop crosses the B = 0 axis.

Assertion: Coercivity is read where the hysteresis loop crosses the B = 0 axis.
Reason: Retentivity is read where the loop crosses H = 0.

  1. A) Both Assertion and Reason are true, and Reason correctly explains Assertion.
  2. B) Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
  3. C) Assertion is true, but Reason is false.
  4. D) Assertion is false, but Reason is true.
Answer: B
Explanation: Both definitions are true, but the reason is a separate definition rather than an explanation.

Case-Based

Case-Based Questions

CS1Case 1: Selecting a core material

A student is designing an AC transformer core. Material A has high permeability and a narrow hysteresis loop. Material B has high retentivity and high coercivity.

Questions and answers:
  1. Which material is better for the transformer core?
    Material A, because high permeability gives strong flux and a narrow loop reduces hysteresis loss.
  2. Which material is closer to a permanent magnet material?
    Material B, because high retentivity and high coercivity help it remain magnetised.
  3. What physical quantity is represented by the loop area?
    Energy loss per unit volume per cycle.
CS2Case 2: Dipole fields

A short bar magnet produces field B at a point P on its equatorial line at distance r. Another point Q is on the axial line at the same distance r.

Questions and answers:
  1. What is the magnitude of field at Q?
    2B, because the axial field of a short dipole is twice the equatorial field at equal distance.
  2. How is the direction at P related to magnetic moment?
    It is opposite to the magnetic moment.
  3. If r is doubled, what happens to the field?
    It becomes one-eighth, since dipole field varies as 1/r3.
CS3Case 3: Curie law

A paramagnetic sample has susceptibility χ1 at 300 K. It is heated to 450 K without changing the applied field.

Questions and answers:
  1. What is χ2/χ1?
    χ2/χ1 = T1/T2 = 300/450 = 2/3.
  2. What happens to magnetisation for the same H?
    It decreases in the same ratio because M = χH.
  3. Why must temperature be in kelvin?
    Curie law uses absolute temperature.

International Boards

IB / IGCSE / A-Level Style Questions

These structured questions emphasise definitions, explanations, labelled diagrams and mark-scheme clarity.

IA1Define magnetic dipole moment and state its SI unit.

Question: Define magnetic dipole moment and state its SI unit.

Mark-scheme points:
  • For a bar magnet, magnetic moment is M = m × 2l.
  • For a current loop, M = NIA.
  • SI unit is A m2 or J T-1.
Exam focus: A precise definition must include direction.
IA2Explain why the field on the axial line of a short magnet has a factor 2 compared with the equatorial line.

Question: Explain why the field on the axial line of a short magnet has a factor 2 compared with the equatorial line.

Mark-scheme points:
  • The dipole field has different vector addition geometry on axial and equatorial lines.
  • On the axial line, pole contributions reinforce more strongly.
  • For a short dipole, this gives Baxial = 2Bequatorial at the same r.
Exam focus: Use vector direction, not only formula memory.
IA3A student says pole strength and magnetic moment have the same unit. Correct the statement.

Question: A student says pole strength and magnetic moment have the same unit. Correct the statement.

Mark-scheme points:
  • Pole strength has unit A m.
  • Magnetic moment has unit A m2.
  • The relation M = m(2l) explains the extra metre.
Exam focus: They are related but not identical.
IA4Describe how domain theory explains ferromagnetism.

Question: Describe how domain theory explains ferromagnetism.

Mark-scheme points:
  • Ferromagnetic materials contain tiny regions called domains.
  • In an unmagnetised sample, domain moments are randomly oriented.
  • External field causes domain alignment and growth, producing large magnetisation.
Exam focus: Mention random, partial and strong alignment.
IA5Sketch and label a hysteresis loop.

Question: Sketch and label a hysteresis loop.

Mark-scheme points:
  • Label H on horizontal axis and B or M on vertical axis.
  • Mark retentivity at H = 0 and coercivity at B = 0.
  • Indicate saturation and loop area as energy loss per unit volume per cycle.
Exam focus: Labels are more important than artistic shape.
IA6Why is soft iron preferred for electromagnets?

Question: Why is soft iron preferred for electromagnets?

Mark-scheme points:
  • Soft iron has high permeability.
  • It magnetises strongly in a field.
  • It has low retentivity and low coercivity, so it demagnetises easily.
Exam focus: Contrast with permanent magnet materials.
IA7Use Curie law to predict what happens to paramagnetic susceptibility when temperature rises.

Question: Use Curie law to predict what happens to paramagnetic susceptibility when temperature rises.

Mark-scheme points:
  • χ = C/T.
  • As absolute temperature increases, χ decreases.
  • The material becomes less magnetically responsive.
Exam focus: Use kelvin, not Celsius.
IA8A material has χ = -1.2 × 10-5. Classify it and describe its response to a magnetic field.

Question: A material has χ = -1.2 × 10-5. Classify it and describe its response to a magnetic field.

Mark-scheme points:
  • χ is small and negative.
  • The material is diamagnetic.
  • It develops induced magnetisation opposite to the applied field and is weakly repelled.
Exam focus: The sign of χ is the key.
IA9Explain why hysteresis loss matters in AC transformer cores.

Question: Explain why hysteresis loss matters in AC transformer cores.

Mark-scheme points:
  • In AC operation, the core is repeatedly magnetised and demagnetised.
  • Each cycle dissipates energy equal to the loop area per unit volume.
  • At high frequency, even small per-cycle losses can produce heat.
Exam focus: Connect loop area with power loss.
IA10Calculate the magnetic moment of a 100-turn loop of area 2.5 × 10-3 m2 carrying 0.8 A.

Question: Calculate the magnetic moment of a 100-turn loop of area 2.5 × 10-3 m2 carrying 0.8 A.

Mark-scheme points:
  • M = NIA.
  • M = 100 × 0.8 × 2.5 × 10-3.
Exam focus: 0.20 A m2
IA11Explain the difference between retentivity and coercivity.

Question: Explain the difference between retentivity and coercivity.

Mark-scheme points:
  • Retentivity is the residual magnetisation or flux density when H becomes zero.
  • Coercivity is the reverse field required to reduce that residual magnetisation to zero.
  • They are read from different intercepts of the hysteresis loop.
Exam focus: Use the axes correctly.
IA12Why should students avoid memorising magnetic material tables without understanding χ and μr?

Question: Why should students avoid memorising magnetic material tables without understanding χ and μr?

Mark-scheme points:
  • The sign of χ tells attraction or repulsion.
  • The magnitude of χ tells weak or strong response.
  • μr = 1 + χ links microscopic response to magnetic induction.
Exam focus: Concepts help in unfamiliar problems.

Exam Traps

Why Students Lose Marks in Magnetism and Matter

Formula Selection Errors

  • Confusing pole strength with magnetic moment.
  • Using A m2 for pole strength or A m for magnetic moment.
  • Mixing axial and equatorial formulas.
  • Forgetting the factor 2 in axial field.

Concept and Diagram Errors

  • Misunderstanding the sign of susceptibility.
  • Drawing a hysteresis loop without Br, Hc and saturation.
  • Confusing retentivity with coercivity.
  • Using Curie law with Celsius instead of kelvin.

Final Revision

Final Revision Questions

  1. Define magnetic pole strength and magnetic moment with units.
  2. Write the formula for force between two magnetic poles in air.
  3. Derive M = m × 2l for a bar magnet.
  4. Write the axial field of a short magnetic dipole.
  5. Write the equatorial field of a short magnetic dipole.
  6. Explain why the axial field has a factor 2.
  7. State torque on a magnetic dipole in uniform B.
  8. Write potential energy of a dipole in a uniform field.
  9. Identify stable and unstable equilibrium orientations.
  10. Calculate magnetic moment of a current loop.
  11. State the right-hand thumb rule for loop magnetic moment.
  12. Define magnetisation.
  13. Define magnetic susceptibility.
  14. Relate relative permeability and susceptibility.
  15. Classify diamagnetic, paramagnetic and ferromagnetic materials.
  16. State Curie law and its limitation.
  17. Explain magnetic domains in ferromagnets.
  18. Define retentivity.
  19. Define coercivity.
  20. Explain hysteresis loss per unit volume per cycle.
  21. Choose suitable materials for permanent magnets.
  22. Choose suitable materials for transformer cores.
  23. Resolve Earth's magnetic field into horizontal and vertical components.
  24. Explain dip angle and declination.
  25. List the most common unit mistakes in this chapter.

FAQ

Magnetism and Matter FAQs

FAQ1What is magnetic moment?

Magnetic moment is the strength and direction of a magnetic dipole. For a bar magnet, M = m × 2l. For a current loop, M = NIA.

FAQ2What is the difference between pole strength and magnetic moment?

Pole strength belongs to one pole and has unit A m. Magnetic moment depends on pole strength and separation, and has unit A m2.

FAQ3What is hysteresis loss?

It is the energy lost as heat when a magnetic material completes one magnetisation-demagnetisation cycle.

FAQ4Why is the hysteresis loop important?

It shows retentivity, coercivity, saturation and energy loss, so it helps choose materials for magnets, electromagnets and transformers.

FAQ5What is Curie law?

For paramagnetic substances, susceptibility varies inversely with absolute temperature: χ = C/T.

FAQ6How should students prepare this chapter?

Master formulas with diagrams, practise units, compare axial and equatorial fields, and solve mixed conceptual-numerical questions.

Personal Guidance

How Kumar Sir Explains Magnetism and Matter

Kumar Sir explains Physics concept-first. Students are guided to understand diagrams, derivations, numerical methods and exam traps instead of memorising formulas blindly. The teaching approach supports NEET, JEE Main, JEE Advanced, CBSE, IB, IGCSE and A-Level Physics preparation.

Need help with Magnetism and Matter?

For magnetic dipole moment, earth magnetism, magnetic materials, Curie law, hysteresis loop or mixed exam problems, students can contact Kumar Sir directly.

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