Why direction dominates this chapter
Consider the force on a moving charge: F = qv × B. The magnitude qvB sinθ is trivial. The direction requires a cross product, and then a sign flip if the charge is negative. A student who has the magnitude right and the direction wrong scores zero, and in a multiple-choice paper the opposite direction is always among the options.
The practical implication is that time spent drilling the right-hand rule until it is automatic returns more marks than time spent on additional numerical practice. That is unusual among physics chapters and worth acting on.
How JEE actually asks Magnetism
NTA does not publish chapter-wise weightage, so any figure online is a coaching estimate from past papers; confirm the syllabus and pattern for your session in the official NTA information bulletin.
Past papers show three families. Force on moving charges and current-carrying conductors supplies the direction-heavy questions. Field calculations for the standard geometries - long wire, circular loop, solenoid, toroid - supply the substitutions. Magnetic materials and Earth's magnetism supply short recall questions that are easy marks if prepared and pure loss if skipped.
Key concepts, compressed
- ✓A magnetic field exerts a force only on moving charges, and only on the component of velocity perpendicular to the field.
- ✓Because F is perpendicular to v, the force changes direction but never speed - so magnetic forces do no work.
- ✓A charge moving perpendicular to a uniform field travels in a circle of radius r = mv/qB with period T = 2πm/qB.
- ✓The Biot-Savart law gives the field of a current element; Ampere's circuital law is its symmetric shortcut, analogous to Gauss's law in electrostatics.
- ✓A current loop is a magnetic dipole: in a uniform field it feels a torque τ = mB sinθ but no net force.
- ✓Materials are diamagnetic (weakly repelled), paramagnetic (weakly attracted) or ferromagnetic (strongly attracted, with hysteresis).
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Force on a moving charge | F = qvB sinθ | direction from v × B |
| Radius of circular path | r = mv/qB | |
| Period of circular motion | T = 2πm/qB | independent of speed |
| Velocity selector | v = E/B | electric and magnetic forces balance |
| Force on a wire | F = BIL sinθ | |
| Force between parallel wires | F/L = μ₀I₁I₂/2πd | same direction attracts |
| Long straight wire | B = μ₀I/2πr | |
| Centre of a circular loop | B = μ₀I/2R | |
| Centre of a semicircular arc | B = μ₀I/4R | half the full loop |
| Inside a solenoid | B = μ₀nI | n is turns per unit length |
| Magnetic moment of a loop | m = NIA | |
| Torque on a loop | τ = NIAB sinθ = mB sinθ | |
| Potential energy of a dipole | U = −mB cosθ | minimum when aligned |
The five mistakes that cost the most marks
- ✓Forgetting that the magnetic force does no work. Any answer claiming a magnetic field changed a particle's speed or kinetic energy is wrong.
- ✓Swapping B = μ₀I/2πr for a wire with B = μ₀I/2R at a loop centre. They differ by a factor of π and by what r means.
- ✓Applying the right-hand rule to a negative charge without reversing the result. The rule gives the direction for conventional positive current.
- ✓Treating n in B = μ₀nI as the total number of turns. It is turns per unit length, so a total-turns question needs dividing by the length first.
- ✓Assuming a current loop in a uniform field experiences a net force. It experiences a torque only; a net force requires a non-uniform field.
Practice set 1: force on moving charges
1. A charge of 2 μC moves at 10⁵ m/s perpendicular to a 0.5 T magnetic field. What force acts on it?
0.1 N. F = qvB sinθ = 2 × 10⁻⁶ × 10⁵ × 0.5 × sin90° = 2 × 10⁻⁶ × 10⁵ × 0.5 = 0.1 N. The direction is given by v × B, perpendicular to both. Since sin90° = 1, this is the maximum possible force for that speed and field.
2. The same charge now moves parallel to the field. What force acts on it?
Zero. With θ = 0, sinθ = 0 and F = qvB sinθ vanishes. A charge moving along the field direction is entirely unaffected by it - which is why a general velocity is resolved into components parallel and perpendicular to B, giving a helical path rather than a circle.
3. Can a magnetic field change the speed of a charged particle?
No. The force qv × B is always perpendicular to the velocity, so it does no work and the kinetic energy is constant. It changes direction continuously, which is why the particle moves in a circle at unchanging speed. An electric field, by contrast, can and does change speed.
4. A particle of mass 2 × 10⁻²⁶ kg and charge 1.6 × 10⁻¹⁹ C moves at 10⁵ m/s perpendicular to a 0.5 T field. What is the radius of its path?
0.025 m, or 2.5 cm. r = mv/qB = (2 × 10⁻²⁶ × 10⁵)/(1.6 × 10⁻¹⁹ × 0.5) = 2 × 10⁻²¹/(8 × 10⁻²⁰) = 0.025 m. Note the structure: heavier or faster particles curve less, while larger charge or stronger field curves them more. That reasoning alone answers many multiple-choice variants.
5. Does the period of that circular motion depend on the particle's speed?
No. T = 2πm/qB contains no v - a faster particle travels a proportionally larger circle in the same time. This speed-independence is exactly what makes the cyclotron work: the accelerating voltage can be reversed at a fixed frequency regardless of how fast the particle has become.
6. In a velocity selector, the electric field is 10⁴ V/m and the magnetic field is 0.1 T. Which speed passes through undeflected?
10⁵ m/s. The electric force qE and magnetic force qvB must balance, so v = E/B = 10⁴/0.1 = 10⁵ m/s. Note that the selected speed is independent of both the charge and the mass, which is why a velocity selector is placed before a mass spectrometer - it makes the subsequent deflection depend on mass alone.
Practice this now
Practice set 2: force on current-carrying conductors
7. A 1 m wire carrying 2 A lies perpendicular to a 0.5 T field. What force acts on it?
1 N. F = BIL sinθ = 0.5 × 2 × 1 × 1 = 1 N, directed perpendicular to both the wire and the field. This is the same physics as question 1 - a current is simply many moving charges - and the two formulas are related by I = nAqv.
8. The same wire is now aligned parallel to the field. What force acts on it?
Zero, since sinθ = 0. A current-carrying wire parallel to a magnetic field experiences no force at all. Options offering a non-zero value rely on the assumption that any current in any field must feel something.
9. Two parallel wires carry current in the same direction. Do they attract or repel?
They attract. Each wire sits in the field of the other, and applying the right-hand rule twice gives forces pointing toward each other. Currents in opposite directions repel. This is the reverse of the intuition from electrostatics, where like charges repel - so it is worth deriving once rather than assuming.
10. Two parallel wires 0.1 m apart each carry 10 A. What is the force per unit length between them?
2 × 10⁻⁴ N/m. F/L = μ₀I₁I₂/2πd = (4π × 10⁻⁷ × 10 × 10)/(2π × 0.1). The π cancels: (2 × 10⁻⁷ × 100)/0.1 = 2 × 10⁻⁵/0.1 = 2 × 10⁻⁴ N/m. Cancelling π against 2π before substituting saves time and avoids a rounding error.
11. How is the ampere defined in terms of this force?
Historically, one ampere was the current which, flowing in two infinitely long parallel wires one metre apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre between them. That definition is why μ₀ has the exact value 4π × 10⁻⁷. The SI ampere is now defined via the elementary charge, but the older definition is still what JEE examines.
Practice set 3: fields produced by currents
12. What is the magnetic field 0.1 m from a long straight wire carrying 10 A?
2 × 10⁻⁵ T. B = μ₀I/2πr = (4π × 10⁻⁷ × 10)/(2π × 0.1). Cancelling π: (2 × 10⁻⁷ × 10)/0.1 = 2 × 10⁻⁶/0.1 = 2 × 10⁻⁵ T. The field forms concentric circles around the wire, with direction given by the right-hand grip rule.
13. What is the field at the centre of a circular loop of radius 0.1 m carrying 10 A?
About 6.28 × 10⁻⁵ T. B = μ₀I/2R = (4π × 10⁻⁷ × 10)/(2 × 0.1) = (4π × 10⁻⁶)/0.2 = 2π × 10⁻⁵ ≈ 6.28 × 10⁻⁵ T. Compare with question 12: the same current at the same distance gives a field π times larger at a loop centre than beside a straight wire, because every part of the loop contributes in the same direction.
14. What is the field inside a solenoid with 1000 turns per metre carrying 2 A?
About 2.51 × 10⁻³ T. B = μ₀nI = 4π × 10⁻⁷ × 1000 × 2 = 8π × 10⁻⁴ ≈ 2.51 × 10⁻³ T. The field inside a long solenoid is uniform and independent of position, and it depends on turns per unit length rather than total turns - which is the detail most often misread.
15. How do you determine the direction of the field around a straight current-carrying wire?
By the right-hand grip rule: point your right thumb along the conventional current and your fingers curl in the direction of B. The field circles the wire, so it has no start or end. For a loop, curl your fingers along the current and the thumb gives the field direction through the loop.
16. What is the magnetic field outside an ideal toroid?
Zero. Applying Ampere's law to a circular path outside the toroid encloses zero net current - the currents going in and out cancel - so the field vanishes. Inside, B = μ₀NI/2πr. This confinement is why toroidal coils are used where stray fields must be avoided.
17. State Ampere's circuital law and the condition for using it.
The line integral of B around any closed loop equals μ₀ times the current enclosed by that loop. Like Gauss's law it is always true, but it is only useful for calculation when symmetry allows B to be taken outside the integral - which in practice means a long wire, a solenoid or a toroid.
18. What is the field at the centre of a semicircular arc of radius R carrying current I?
μ₀I/4R - exactly half the value for a full loop, because only half the current path contributes and every element contributes in the same direction. Arc questions generalise this: an arc subtending angle θ at the centre gives B = μ₀Iθ/4πR, and the semicircle is the case θ = π.
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Practice set 4: torque and magnetic moment
19. A 100-turn coil of area 0.01 m² carries 1 A in a 0.5 T field, with its plane parallel to the field. What torque acts on it?
0.5 N·m. τ = NIAB sinθ = 100 × 1 × 0.01 × 0.5 × sin90° = 0.5 N·m. When the plane of the coil is parallel to the field, the normal to the coil is perpendicular to it, so θ = 90° and the torque is maximum. Reading the geometry the wrong way round is the usual error here.
20. What is the magnetic moment of that coil?
1 A·m². m = NIA = 100 × 1 × 0.01 = 1 A·m². The moment is a vector along the normal to the coil, with direction given by the right-hand rule applied to the current. Once you have m, the torque is simply mB sinθ, which is often the quicker route.
21. When is the torque on a current loop in a uniform field zero?
When the magnetic moment is parallel to the field - that is, when the plane of the coil is perpendicular to B. Then θ = 0 and sinθ = 0. This is the stable equilibrium position, and it is also where the potential energy U = −mB cosθ is at its minimum.
22. Why does a moving-coil galvanometer use a radial magnetic field?
So that the plane of the coil is always parallel to the field regardless of how far it has rotated, keeping sinθ = 1 and making the torque proportional to current alone. Without a radial field the deflection would not be linear in current, and the scale could not be uniform.
23. What is the potential energy of a magnetic dipole in a field, and where is it minimum?
U = −mB cosθ, minimum at θ = 0 where the moment is aligned with the field and U = −mB. It is maximum at θ = 180°, when anti-aligned, which is an unstable equilibrium. This mirrors the electric dipole exactly, and questions often ask for the work done rotating between two orientations - which is just the difference in U.
Practice set 5: magnetic materials and Earth's magnetism
24. What characterises a diamagnetic material?
It is weakly repelled by a magnetic field and has a small negative susceptibility. Diamagnetism arises from induced moments opposing the applied field, and it is present in all materials - it is simply masked when stronger para- or ferromagnetic effects are present. Bismuth, copper and water are standard examples.
25. What characterises a paramagnetic material?
Weak attraction and a small positive susceptibility. Its atoms have permanent magnetic moments that partially align with an applied field, though thermal motion opposes the alignment - so the effect weakens as temperature rises. Aluminium and oxygen are the usual examples.
26. What distinguishes a ferromagnetic material from a paramagnetic one?
Ferromagnets have very large positive susceptibility, retain magnetisation after the field is removed, and organise into domains. The domain structure is the key difference: within a domain, moments align spontaneously without any applied field, which is impossible in a paramagnet. Iron, cobalt and nickel are the standard examples.
27. What happens to a ferromagnetic material above its Curie temperature?
It becomes paramagnetic. Thermal agitation destroys the domain alignment, so the spontaneous magnetisation disappears and only the weak paramagnetic response remains. The transition is reversible - cooling below the Curie point restores ferromagnetism.
28. Why do magnetic field lines always form closed loops?
Because magnetic monopoles do not exist, so there is no source or sink for the field to begin or end at. Mathematically this is the statement that the net magnetic flux through any closed surface is zero - the magnetic counterpart of Gauss's law, and the reason breaking a bar magnet gives two complete magnets rather than isolated poles.
29. Should an electromagnet core be made of soft iron or steel, and why?
Soft iron. It has a narrow hysteresis loop, so it magnetises and demagnetises easily with little energy lost per cycle - exactly what an electromagnet needs. Steel has a wide loop and retains magnetisation, which makes it suitable for permanent magnets and unsuitable for anything that must switch off.
30. What are the three elements of Earth's magnetic field?
Declination, the angle between geographic and magnetic north; dip or inclination, the angle the field makes with the horizontal; and the horizontal component of the field. Together they specify the field completely at any location. Dip is zero at the magnetic equator and 90° at the magnetic poles.
How to study this chapter efficiently
- ✓Drill the right-hand rule physically until it is automatic. It is the single highest-return activity in this chapter and cannot be replaced by more numerical practice.
- ✓For every direction question, check the sign of the charge before quoting an answer. The rule gives conventional-current direction; a negative charge reverses it.
- ✓Write the two field formulas side by side and note what r means in each - distance from a wire versus radius of a loop.
- ✓Learn the material classifications as a three-row table: susceptibility sign and magnitude, behaviour in a field, and one example each.
- ✓Practise the arc-fraction generalisation. Semicircles, quarter circles and combinations of arcs are common and all follow from B = μ₀Iθ/4πR.
- ✓Do not skip Earth's magnetism and hysteresis. They are short recall topics that appear regularly and cost nothing to prepare.
Turn this into active practice
Direction errors are invisible when you read a worked solution - the arrow is already drawn - and obvious when you have to produce one yourself. That asymmetry is exactly why reading this chapter feels productive and does not transfer to the exam.
The JEE Magnetism quiz on QUFF generates fresh questions across force on charges and wires, field calculations, torque on loops and magnetic materials, marks them instantly and explains each answer. Do a timed set and sort your errors into direction errors, formula swaps and recall gaps - each has a different fix.