1. Chapter at a Glance
Moving charges produce magnetic fields, and magnetic fields push on moving charges. This chapter uses the Biot–Savart law and Ampere's circuital law to find the magnetic field of wires, loops and solenoids. It then covers the Lorentz force, circular and helical motion of charges, forces between current-carrying wires, torque on a current loop, and the moving-coil galvanometer. A short section on magnetism and matter (magnetic dipoles and materials) is added at the end, as NEET includes it. Expect two questions from this area.
- Field of a straight wire, a circular loop and a solenoid.
- Radius and period of a charge moving in a magnetic field.
- Force between parallel wires; torque on a coil.
- Converting a galvanometer into an ammeter or a voltmeter.
- Dia-, para- and ferromagnetic materials.
2. Biot–Savart Law and Its Uses
- The field dB is perpendicular to both dl and r (direction by the right-hand rule).
- There is no field along the line of the current element itself (θ = 0).
- The SI unit of B is the tesla (T); 1 gauss = 10⁻⁴ T.
| Source | Magnetic field | NEET Note |
|---|---|---|
| Long straight wire (distance r) | B = μ₀I/(2πr) | Circles around the wire; B ∝ 1/r |
| Centre of a circular loop (N turns) | B = μ₀NI/(2R) | Along the axis |
| Axis of a loop (distance x) | B = μ₀NIR²/[2(R² + x²)3/2] | Falls as 1/x³ far away |
| Long solenoid (n turns per metre) | B = μ₀nI | Uniform inside; about half at the ends |
Hold a wire in your right hand with the thumb along the current; the curled fingers show the direction of the magnetic field lines. For a loop, curl the fingers along the current and the thumb points along the field at the centre.
3. Ampere's Circuital Law
- It plays the same role for magnetism that Gauss's law plays for electrostatics: it gives B quickly when there is symmetry.
- Applying it to a circle around a long wire gives B × 2πr = μ₀I, so B = μ₀I/(2πr).
- Applying it to a rectangular loop through a solenoid gives B = μ₀nI inside and nearly zero outside.
4. Force on Moving Charges
- The magnetic force is perpendicular to the velocity, so it does no work and cannot change the speed or kinetic energy.
- A charge moving along the field (θ = 0) feels no force.
- Perpendicular entry: circular path with r = mv/(qB), period T = 2πm/(qB), frequency f = qB/(2πm). T does not depend on speed.
- Entry at angle θ: helical path, radius r = mv sinθ/(qB), pitch p = v cosθ × T.
- Velocity selector: crossed E and B fields let through, undeviated, only charges with v = E/B.
5. Force on Currents and Torque on a Loop
- Force on a straight conductor: F = IlB sinθ (Fleming's left-hand rule gives the direction).
- Two parallel wires: force per unit length F/l = μ₀I₁I₂/(2πd). Currents in the same direction attract; opposite currents repel.
- The ampere is defined from this force: 1 A in each of two wires 1 m apart gives 2 × 10⁻⁷ N m⁻¹.
- Current loop: magnetic moment m = NIA (direction by the right-hand rule). In a uniform field, net force = 0 and torque τ = mB sinθ = NIAB sinθ.
6. Moving-Coil Galvanometer, Ammeter and Voltmeter
- A coil in a radial magnetic field feels torque NIAB, balanced by the spring torque kφ, so I = (k/NAB)φ: deflection is proportional to current.
- Current sensitivity = φ/I = NAB/k; voltage sensitivity = φ/V = NAB/(kG).
- Increasing the number of turns raises the current sensitivity but not always the voltage sensitivity, because the resistance also rises.
| Conversion | What to add | Formula | NEET Note |
|---|---|---|---|
| Ammeter (range I) | Small shunt S in parallel | S = IgG/(I − Ig) | Ideal ammeter: R = 0 |
| Voltmeter (range V) | Large R in series | R = V/Ig − G | Ideal voltmeter: R = ∞ |
7. Magnetism and Matter (NEET Essentials)
- A bar magnet behaves like a magnetic dipole of moment m; a current loop and a solenoid are equivalent to bar magnets.
- Field of a bar magnet: on the axis B = (μ₀/4π)(2m/r³); on the equatorial line B = (μ₀/4π)(m/r³) — the same form as an electric dipole.
- Torque on a magnet τ = mB sinθ; potential energy U = −mB cosθ.
- Magnetic field lines form closed loops: isolated magnetic poles (monopoles) do not exist. This is Gauss's law for magnetism: net magnetic flux through any closed surface is zero.
- Magnetisation M = magnetic moment per unit volume; susceptibility χ = M/H; relative permeability μr = 1 + χ.
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Susceptibility χ | Small, negative | Small, positive | Large, positive |
| μr | Slightly less than 1 | Slightly more than 1 | Very large (≫ 1) |
| In a non-uniform field | Moves to weaker field | Moves to stronger field | Strongly attracted |
| Effect of temperature | Almost none | χ ∝ 1/T (Curie's law) | Becomes paramagnetic above the Curie temperature |
| Examples | Bi, Cu, water, N₂ | Al, Na, O₂ | Fe, Co, Ni |
Superconductors are perfect diamagnets (χ = −1): they expel magnetic fields completely (Meissner effect). Soft iron (low retentivity, low coercivity) is used for electromagnets and transformer cores; steel (high coercivity) is used for permanent magnets.
8. Formula Master Table
| Quantity | Formula |
|---|---|
| Straight wire | B = μ₀I/2πr |
| Centre of a coil | B = μ₀NI/2R |
| Solenoid | B = μ₀nI |
| Force on a charge | F = qvB sinθ |
| Radius / period | r = mv/qB ; T = 2πm/qB |
| Pitch of a helix | p = (v cosθ)(2πm/qB) |
| Velocity selector | v = E/B |
| Force on a wire | F = IlB sinθ |
| Parallel wires | F/l = μ₀I₁I₂/2πd |
| Torque on a coil | τ = NIAB sinθ |
| Shunt / series resistance | S = IgG/(I − Ig) ; R = V/Ig − G |
| Permeability | μr = 1 + χ |
9. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Magnetic force | ⊥ to v | Does no work; speed constant |
| Circular motion | r ∝ mv/q | Period independent of speed |
| Parallel wires | Like currents attract | Opposite to electric charges |
| Current loop | Behaves like a magnet | m = NIA |
| Ammeter | Low-resistance shunt | Connected in series |
| Voltmeter | High series resistance | Connected in parallel |
| Diamagnet | χ < 0 | Repelled by magnets |
| Ferromagnet | Domains align | Curie temperature |
10. Mind Map
- Biot–Savart
- Ampere's law
- Wire, loop, solenoid
- qvB sinθ
- Circle, helix
- Velocity selector
- IlB sinθ
- Parallel wires
- Torque NIAB
- Galvanometer
- Ammeter / voltmeter
- Dia / para / ferro
11. Tips and Tricks
- For particles with the same kinetic energy, r = √(2mK)/(qB) ∝ √m/q.
- For particles accelerated through the same voltage V, r = √(2mV/q)/B.
- A semicircular arc of radius R gives B = μ₀I/(4R) at its centre (half of a full loop).
- Torque on a coil is maximum when its plane is parallel to B (θ = 90° between m and B).
- Thinking a magnetic field can speed up a charge.
- Putting the shunt in series or the multiplier in parallel.
- Forgetting that parallel currents attract.
- Mixing up the signs of χ for diamagnetic and paramagnetic materials.
12. Important One-liners
- A stationary charge produces only an electric field; a moving charge produces both electric and magnetic fields.
- The magnetic field inside a long solenoid is uniform.
- The time period of a charge in a uniform magnetic field does not depend on its speed.
- An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.
- Magnetic field lines always form closed loops.
- Diamagnetism is present in all materials but is masked in para- and ferromagnets.
- Above its Curie temperature, a ferromagnet becomes paramagnetic.
- Soft iron is used for electromagnets because it is easy to magnetise and demagnetise.
13. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take μ₀ = 4π × 10⁻⁷ T m A⁻¹, mp = 1.67 × 10⁻²⁷ kg and e = 1.6 × 10⁻¹⁹ C. Click to reveal each answer and working.
Average level
Find the magnetic field 8 cm from a long straight wire carrying 12 A.
Show answer & solution
Concept: B = μ₀I/(2πr) = 2 × 10⁻⁷ I/r.
- B = 2 × 10⁻⁷ × 12/0.08.
- B = 3 × 10⁻⁵ T.
Answer: 3 × 10⁻⁵ T
A circular coil of 40 turns and radius 8 cm carries a current of 3 A. Find the field at its centre.
Show answer & solution
Concept: B = μ₀NI/(2R).
- B = 4π × 10⁻⁷ × 40 × 3/(2 × 0.08).
- = 4π × 10⁻⁷ × 120/0.16 = 4π × 10⁻⁷ × 750.
- B ≈ 9.42 × 10⁻⁴ T.
Answer: about 9.4 × 10⁻⁴ T
A long solenoid has 2000 turns per metre and carries 3 A. Find the magnetic field inside it.
Show answer & solution
Concept: B = μ₀nI.
- B = 4π × 10⁻⁷ × 2000 × 3.
- B ≈ 7.54 × 10⁻³ T.
Answer: about 7.5 × 10⁻³ T
A 20 cm wire carrying 5 A is placed at 30° to a uniform field of 0.4 T. Find the force on it.
Show answer & solution
Concept: F = IlB sinθ.
- F = 5 × 0.2 × 0.4 × sin 30°.
- F = 0.4 × 0.5 = 0.2 N.
Answer: 0.2 N
A proton moving at 3 × 10⁶ m s⁻¹ enters a 0.25 T field at right angles. Find the radius of its path.
Show answer & solution
Concept: r = mv/(qB).
- r = 1.67 × 10⁻²⁷ × 3 × 10⁶ / (1.6 × 10⁻¹⁹ × 0.25).
- = 5.01 × 10⁻²¹ / 4 × 10⁻²⁰.
- r ≈ 0.125 m = 12.5 cm.
Answer: about 12.5 cm
Advanced level
Two long parallel wires 12 cm apart carry 4 A and 6 A in the same direction. Find the force on 1 m of each wire and state its nature.
Show answer & solution
Concept: F/l = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷ I₁I₂/d.
- F/l = 2 × 10⁻⁷ × 4 × 6/0.12.
- = 2 × 10⁻⁷ × 200 = 4 × 10⁻⁵ N m⁻¹.
- Currents are in the same direction, so the force is attractive (equal on both wires, by Newton's third law).
Answer: 4 × 10⁻⁵ N per metre, attractive
A galvanometer of resistance 40 Ω gives full-scale deflection for 2 mA. Find the resistance needed, and how it is connected, to convert it into (a) an ammeter reading up to 1 A, (b) a voltmeter reading up to 20 V.
Show answer & solution
Concept: shunt S = IgG/(I − Ig); series R = V/Ig − G.
- (a) S = 0.002 × 40/(1 − 0.002) = 0.08/0.998 ≈ 0.080 Ω, connected in parallel.
- (b) R = 20/0.002 − 40 = 10000 − 40 = 9960 Ω, connected in series.
Answer: (a) about 0.08 Ω in parallel; (b) 9960 Ω in series
A rectangular coil of 100 turns, 5 cm × 4 cm, carries 2 A in a uniform field of 0.25 T. Find its magnetic moment and the maximum torque on it. For which orientation is the torque maximum?
Show answer & solution
Concept: m = NIA; τmax = mB.
- A = 0.05 × 0.04 = 2 × 10⁻³ m².
- m = 100 × 2 × 2 × 10⁻³ = 0.4 A m².
- τmax = 0.4 × 0.25 = 0.1 N m, when the plane of the coil is parallel to B (m ⊥ B).
Answer: 0.4 A m²; 0.1 N m (plane of coil parallel to B)
A proton moving at 4 × 10⁵ m s⁻¹ enters a 0.2 T field with its velocity at 60° to the field. Find the radius and pitch of its helical path.
Show answer & solution
Concept: split v into v sinθ (circular part) and v cosθ (along B).
- T = 2πm/(qB) = 2π × 1.67 × 10⁻²⁷/(1.6 × 10⁻¹⁹ × 0.2) ≈ 3.28 × 10⁻⁷ s.
- r = m v sin60°/(qB) = 1.67 × 10⁻²⁷ × 3.46 × 10⁵/(3.2 × 10⁻²⁰) ≈ 1.81 × 10⁻² m.
- Pitch = v cos60° × T = 2 × 10⁵ × 3.28 × 10⁻⁷ ≈ 6.56 × 10⁻² m.
Answer: radius about 1.8 cm; pitch about 6.6 cm
In a velocity selector, the electric field is 3 × 10⁴ V m⁻¹ and the magnetic field is 0.15 T, at right angles. Which speed passes through undeflected? What happens to a faster particle?
Show answer & solution
Concept: no deflection when qE = qvB.
- v = E/B = 3 × 10⁴/0.15 = 2 × 10⁵ m s⁻¹ (independent of charge and mass).
- For a faster particle, qvB > qE, so the magnetic force wins and it is deflected in the direction of the magnetic force.
Answer: 2 × 10⁵ m s⁻¹; faster particles bend in the direction of the magnetic force
14. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
The work done by a magnetic field on a moving charge is:
- qvB
- Zero
- qvB sinθ
- Positive only for positive charges
Show solution
The magnetic force is always perpendicular to the velocity, so F·v = 0 and the work done is zero. qvB is the size of the force, not the work it does.
Answer: (B)
Two parallel wires carry currents in opposite directions. They:
- Attract
- Repel
- Exert no force
- Rotate
Show solution
Wires with opposite currents repel each other. Wires with currents in the same direction attract. Note that this is the reverse of the rule for charges.
Answer: (B)
A proton and an α-particle enter a uniform magnetic field perpendicularly with the same speed. The ratio of their radii rp : rα is:
- 1 : 1
- 1 : 2
- 2 : 1
- 1 : 4
Show solution
r = mv/(qB) ∝ m/q. For the proton m/q = 1/1; for the α-particle m/q = 4/2 = 2. So rp : rα = 1 : 2.
Answer: (B)
To convert a galvanometer into a voltmeter, we connect:
- A low resistance in parallel
- A high resistance in series
- A low resistance in series
- A high resistance in parallel
Show solution
A voltmeter must draw very little current, so a large resistance is added in series. (A) describes an ammeter.
Answer: (B)
The susceptibility of a diamagnetic substance is:
- Small and positive
- Small and negative
- Large and positive
- Zero
Show solution
A diamagnet is weakly magnetised opposite to the applied field, so χ is small and negative. A small positive χ is paramagnetic, and a large positive χ is ferromagnetic.
Answer: (B)
If the current in a long solenoid is doubled and its number of turns per metre is halved, the field inside:
- Doubles
- Halves
- Stays the same
- Becomes four times
Show solution
B = μ₀nI. The product n × I = (n/2)(2I) is unchanged, so B stays the same.
Answer: (C)
An electron and a proton move with the same kinetic energy perpendicular to a uniform magnetic field. Which statement is correct?
- Both have equal radii
- The electron's path has the larger radius
- The proton's path has the larger radius
- Both move in straight lines
Show solution
r = √(2mK)/(qB). The charges are equal in size, so r ∝ √m. The proton is much heavier, so it has the larger radius. Both paths are circles, not straight lines.
Answer: (C)
A wire carrying current I is bent into a semicircle of radius R, with the straight leads running radially away from the centre. The magnetic field at the centre is:
- μ₀I/(2R)
- μ₀I/(4R)
- μ₀I/(2πR)
- Zero
Show solution
Radial leads point straight at the centre, so they give no field there (θ = 0 in Biot–Savart). The semicircle gives half the field of a full loop: μ₀I/(4R). (A) is the full-loop value, and (C) is the formula for a long straight wire.
Answer: (B)
15. Quick Revision Summary
- Wire μ₀I/2πr; coil centre μ₀NI/2R; solenoid μ₀nI.
- F = qvB sinθ does no work; r = mv/qB; T = 2πm/qB.
- Parallel wires: F/l = μ₀I₁I₂/2πd; like currents attract.
- Torque τ = NIAB sinθ; ammeter = shunt, voltmeter = series R.
- Dia χ < 0, para small +, ferro large +; Curie temperature.