Physics for UPSC – Chapter 17: Moving Charges and Magnetism, Magnetic Force and Biot–Savart Law

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Physics for UPSC · Chapter 17

Moving Charges & Magnetism

Magnetic fields, forces, solenoids, current loops, cyclotrons, galvanometers, detailed numerical practice and PYQ concepts.

13 modulesDetailed Physics theory
7 numericalsWorked answers
20 MCQsExplanation included

Magnetic fields and the connection with moving charge

Magnetism appears when charges move and when elementary particles carry magnetic moments. A steady current in a conductor produces a magnetic field around the conductor, observed through its effect on a compass needle. The field B is a vector measured in tesla. Unlike the field from an isolated electric point charge, magnetic-field lines form closed loops. In a straight current-carrying wire, the lines circle the wire, whereas in a long ideal solenoid the interior field is approximately along its axis.

An ordinary magnetic compass aligns its needle with the local resultant magnetic field, which may include Earth’s field and nearby magnets. A field’s strength and its direction are separate quantities. A neutral current-carrying conductor can experience magnetic force even though its net electric charge is nearly zero. This is because its moving carriers interact with the magnetic field and transmit mechanical force to the conductor. Understanding how geometry changes field direction is essential before attempting calculations.

Lorentz force on a moving electric charge

The total electromagnetic force on charge q moving with velocity v is F=q(E+v×B). The magnetic contribution has magnitude |q|vB sinθ where θ is the angle between v and B. A positive particle’s force direction follows the right-hand cross-product rule; a negative charge experiences the opposite direction for the same velocity and field. For velocity parallel to B, magnetic force vanishes, and for velocity perpendicular to B its magnitude is greatest.

Pure magnetic force is perpendicular to instantaneous velocity, so its instantaneous power F·v is zero. In a field where no electric field or other work-producing force acts, magnetic force turns the trajectory without directly changing speed or kinetic energy. Electric fields differ because they may accelerate a charge along its direction of motion and do work. This distinction explains the different roles of the electric and magnetic fields in particle accelerators. Magnetic force should not be mistaken for gravitational attraction or mechanical friction.

Circular paths, helical trajectories and cyclotron frequency

If a charged particle enters a uniform magnetic field with velocity perpendicular to B, the magnetic force provides centripetal acceleration. Equating |q|vB=mv²/r yields orbit radius r=mv/(|q|B). A particle with twice the momentum has twice the radius at unchanged charge and field; a particle in twice the magnetic field has half the radius. For nonrelativistic motion the angular cyclotron frequency is ω=|q|B/m and the period T=2πm/(|q|B).

If velocity has a component parallel to B, that component remains constant while the perpendicular component rotates. The full path is a helix whose pitch depends on parallel speed and period. Magnetic field alone does not change the total speed in the ideal model. When particle speeds become relativistic, mass–velocity dynamics need treatment using relativistic momentum and the simple fixed-frequency cyclotron approximation ceases to remain exact. For objective questions, first identify v⊥, rather than incorrectly substituting total speed into a circular-radius relation.

Combined electric and magnetic fields: velocity selection

An ideal velocity selector has mutually perpendicular electric and magnetic fields arranged so their forces on an incoming charged particle oppose. With velocity perpendicular to both fields, straight-line transmission requires qE and qvB to cancel, so the selected speed is v=E/B. The speed ratio is independent of the charge magnitude and particle mass in the basic ideal condition. Charges with different speeds are deflected and may be stopped by mechanical apertures.

A full diagram is important because E and B perpendicular to one another are not sufficient unless the velocity direction also makes the forces oppose. Reversing charge sign reverses both electromagnetic forces under unchanged v, E and B, potentially preserving the cancellation. Instruments combining electric selection and magnetic deflection help determine charge-to-mass ratios. The expression E/B has dimensions (N/C)/(N/(A m)), which simplifies to speed; dimensional analysis checks the answer without recalling the entire apparatus.

Right-hand thumb rule and field of a long wire

For an approximately infinitely long straight wire carrying conventional current I, the magnetic-field magnitude at distance r in vacuum is B=μ₀I/(2πr), with μ₀ the permeability of free space. The field direction follows the right-hand thumb rule: align thumb with conventional current and curl the fingers in the direction of magnetic field lines. The lines are concentric circles in planes perpendicular to the wire; they are not radial lines pointing away from the wire.

Doubling the current doubles the field at fixed radial distance; doubling distance halves it. This inverse-distance relationship differs from the inverse-square electric field of a point charge. If the wire is not long compared with the observation distance to its ends, finite-segment geometry modifies the expression. A conductor bending into a coil creates a different spatial field distribution. UPSC questions about compass deflection often test the current direction and the position of the compass rather than require a numerical answer.

Biot–Savart law and the circular current loop

The Biot–Savart law gives the contribution of a current element I dℓ to magnetic field: dB=(μ₀/4π)I(dℓ×r̂)/r² in vacuum. Its vector direction is given by a cross product. Integrating contributions along a complete circular loop of radius R gives the magnetic field magnitude at its centre B=μ₀I/(2R) for one turn. For N closely placed turns with the same current, B=μ₀NI/(2R).

The field on the axis at distance x from a circular loop is B=μ₀NIR²/[2(R²+x²)^(3/2)], directed along the axis. The formula reduces to the centre result at x=0. Field directions from opposite portions of a symmetric loop combine in a way that cancels sideways components and reinforces the axial component. Candidates sometimes apply the straight-wire law to a circular loop and obtain the wrong radius dependence. The geometry of the source always determines which law or integrated result is appropriate.

Ampère’s circuital law and magnetostatic symmetry

For steady currents under magnetostatic conditions, Ampère’s circuital law in vacuum gives ∮B·dℓ=μ₀I_enclosed. The line integral is taken around a closed imagined path. In highly symmetric systems, such as an infinitely long wire or an ideal long solenoid, the field has a simple relation to the path, allowing B to be taken outside part of the integral. This converts an otherwise difficult field calculation into algebra.

The law remains a statement about the circulation of B, not a claim that field magnitude is uniform along every arbitrary curve. For time-varying electric fields, the more general Ampère–Maxwell law includes a displacement-current term. In the basic current-magnetism chapter, assuming steady currents permits the simpler form. An Amperian loop is a mathematical construction and need not coincide with a real wire. Only current passing through a spanning surface of the chosen closed path contributes to the enclosed-current term in that model.

Solenoids, toroids and ideal field regions

A solenoid is a coil with turns arranged along a common axis. In an ideal long, tightly wound solenoid carrying current I and n turns per unit length, the field inside is approximately B=μ₀nI in air or vacuum and is nearly parallel to the axis. Doubling current or the turns per unit length doubles this ideal field. The field outside an infinitely long ideal solenoid vanishes in the simple approximation, though a real finite coil has a nonzero fringe field.

A toroid bends the solenoid into a ring. The ideal toroidal field within the winding region at radius r is approximately μ₀NI/(2πr) in vacuum when symmetry and coil density support the approximation. Magnetic cores can change field strength and introduce nonlinear magnetisation, so simply inserting a material’s permeability into every formula can fail when saturation or hysteresis matters. For NDA and other UPSC tests, the reliable comparison is that the long solenoid field is approximately uniform far from ends, not perfectly uniform in all parts of any real solenoid.

Force on current-carrying conductors and motor direction

A straight wire segment of vector length L carrying current I in uniform B experiences F=I(L×B), with magnitude ILB sinθ. Maximum force occurs at 90° and zero force at 0° or 180°. The direction is determined by the vector product or the familiar Fleming left-hand rule used for motor force. This force acts on an ordinary current-carrying conductor even when the conductor is not itself a permanent magnet.

Opposite sides of a rectangular current loop in a magnetic field can experience equal opposite forces forming a couple that rotates the loop. A uniform field exerts no net translational force on a small closed loop but can exert torque. The torque under an ideal N-turn loop model is τ=NIAB sinθ. The force on a wire and torque on a loop are related effects but require different geometry factors. A question involving the force on a straight section should not be solved by substituting the magnetic dipole formula without explaining the coil.

Parallel wire interaction and magnetic dipole moment

Two long, straight, parallel wires attract when they carry currents in the same direction and repel when currents are opposite. The magnetic force per unit length in vacuum under the ideal parallel-wire model is F/L=μ₀I₁I₂/(2πd), where d is separation. This result follows by computing the B field of one wire at the location of the second and using F=ILB. The forces form an action–reaction pair on different wires.

A current loop behaves approximately like a magnetic dipole at distances large compared with its size. Its dipole moment vector is m=NIA n̂, where n̂ follows a right-hand orientation rule and points along the loop axis. In a uniform magnetic field a dipole can undergo torque m×B, tending toward alignment; in a nonuniform field it may also experience net force. The word magnetic dipole describes a field pattern and moment, not a literal pair of separable north and south magnetic charges.

Cyclotron operation, work and relativistic limits

A basic cyclotron uses a magnetic field to steer charged particles into circular paths while an alternating electric potential between electrodes accelerates them whenever they cross the gap. As energy increases, orbit radius grows in the nonrelativistic model while angular cyclotron frequency remains |q|B/m. Matching the driving frequency allows repeated acceleration until the particle reaches the outer orbit.

The magnetic field does not directly supply the energy gained by the particle because magnetic force is perpendicular to instantaneous velocity. The electric field in the accelerating gap does work. At sufficiently high speed the relativistic momentum relation alters angular frequency, reducing the accuracy of a simple fixed-frequency cyclotron. Neutral particles cannot be steered by electric and magnetic Lorentz forces in the same way as charged particles, though neutral particle magnetic moments can interact with gradients. This is why school-level cyclotron questions state nonrelativistic assumptions.

Moving-coil galvanometer, ammeter and voltmeter

A moving-coil galvanometer detects small current by torque on a current-carrying coil in a magnetic field. In a radial field design magnetic torque approximately equals NIAB, balanced by a mechanical restoring torque Cθ. At equilibrium θ=(NAB/C)I, making deflection proportional to current under suitable conditions. The galvanometer is a current detector, not intrinsically a high-range ammeter or voltmeter without additional circuit components.

To convert it into an ammeter, connect a low-resistance shunt parallel to the galvanometer so most current bypasses the delicate coil. To make a voltmeter, connect a large resistance in series so the measured potential difference drives only a small coil current. The shunt and multiplier resistances perform different functions, and interchanging them is a common MCQ trap. Real instrument resistance, sensitivity and maximum safe coil current must be considered when numerical conversion details are given.

Magnetic applications and a comprehensive solving strategy

Current-produced magnetism is used in electric motors, lifting electromagnets, loudspeakers, magnetic actuators and particle-control systems. The field in a device depends on winding geometry, number of turns, supplied current and magnetic core behaviour. An ideal formula may predict simple scaling but engineering applications include resistance heating, finite-end effects, saturation and energy loss.

For an exam problem, first identify whether the source is a point charge, straight wire, loop, solenoid or toroid. Next determine the relevant field direction. Then choose the correct relationship for charge force, conductor force, torque, field or orbital radius. Check whether the problem describes static charges, steady currents or changing fields; the last category belongs to electromagnetic induction. Always keep magnetic field B separate from electric field E and from magnetic dipole moment m. A dimensional check on the final expression often eliminates plausible-looking but incorrect options.

Essential formulas

Lorentz forceF=q(E+v×B)
Magnetic force magnitudeF=|q|vB sinθ
Orbit radiusr=mv⊥/(|q|B)
Cyclotron angular frequencyω=|q|B/m
Selectorv=E/B
Long straight wireB=μ₀I/(2πr)
Loop centreB=μ₀NI/(2R)
Long solenoidB≈μ₀nI
Wire forceF=ILB sinθ
Parallel wiresF/L=μ₀I₁I₂/(2πd)
Dipole momentm=NIA
Torqueτ=mB sinθ

Step-by-step solved examples

Example 1: Electron force

With v=2×10⁶ m/s, B=0.1 T, perpendicular entry: F=e v B≈3.20×10⁻¹⁴ N.

Example 2: Speed selector

E=4000 V/m and B=0.02 T: v=E/B=2×10⁵ m/s.

Example 3: Long wire

At I=10 A, r=0.05 m, B=(4π×10⁻⁷×10)/(2π×0.05)=4×10⁻⁵ T.

Example 4: Solenoid

n=1000 m⁻¹, I=2 A: B≈4π×10⁻⁷×2000=2.51 mT.

Example 5: Conductor force

I=5 A, L=0.4 m, B=0.3 T perpendicular: F=5×0.4×0.3=0.6 N.

Example 6: Loop centre field

One loop R=0.1 m, I=2 A gives B≈1.26×10⁻⁵ T.

Example 7: Current pair

Long 10 A wires separated 0.2 m have force per metre μ₀100/(2π0.2)=1×10⁻⁴ N/m.

UPSC past-paper question concepts

The topics below are paraphrased from indexed UPSC NDA General Ability Test Physics questions; they are not verbatim paper reproductions or claims about EPFO/APFC questions. Consult the original UPSC paper for its exact wording, year and options.

PYQ topic 1

Long solenoid field

An indexed NDA General Ability question asks about field uniformity inside a sufficiently long solenoid. Ideal interior B≈μ₀nI, whereas edge effects occur in real coils.

Official UPSC papers
PYQ topic 2

Fleming’s left-hand rule

A past NDA question identifies the direction rule for the force on a perpendicular current-carrying conductor. It is the motor-force left-hand rule, not the generator right-hand rule.

Official UPSC papers
PYQ topic 3

Turns and field in a solenoid

A past-paper concept compares solenoid fields when turn density changes. At fixed current in the ideal model B∝n, so doubling n doubles B.

Official UPSC papers

20 original MCQs with explanations

Q1. Magnetic force on a stationary charge
  1. zero
  2. qB
  3. qvB
  4. mg
Reveal correct answer and explanation

Answer A. Magnetic Lorentz force includes velocity.

Q2. Magnetic force for v parallel B
  1. maximum
  2. zero
  3. qE
  4. 2qvB
Reveal correct answer and explanation

Answer B. sin0°=0.

Q3. Magnetic force always points
  1. along v
  2. opposite v
  3. perpendicular to v
  4. towards Earth
Reveal correct answer and explanation

Answer C. F=q(v×B) is orthogonal to velocity.

Q4. Magnetic work on isolated charge
  1. positive
  2. negative
  3. zero
  4. always kinetic energy
Reveal correct answer and explanation

Answer C. F·v=0.

Q5. Orbit radius for v perpendicular B
  1. mv/(|q|B)
  2. |q|B/(mv)
  3. mvB
  4. mB/(qv)
Reveal correct answer and explanation

Answer A. Equate magnetic and centripetal forces.

Q6. Doubling B changes radius to
  1. 2r
  2. r/2
  3. r/4
  4. 4r
Reveal correct answer and explanation

Answer B. r inversely proportional to B.

Q7. Velocity selector transmits when
  1. v=EB
  2. v=B/E
  3. v=E/B
  4. v=E+B
Reveal correct answer and explanation

Answer C. Balance qE=qvB.

Q8. Magnetic field around long straight wire
  1. radial
  2. circular
  3. always zero
  4. longitudinal only
Reveal correct answer and explanation

Answer B. Right-hand thumb rule gives concentric circles.

Q9. Field magnitude of infinite wire varies as
  1. r²
  2. r
  3. 1/r
  4. 1/r²
Reveal correct answer and explanation

Answer C. B=μ0I/(2πr).

Q10. Circular loop central field
  1. μ0I/(2R)
  2. μ0IR
  3. μ0I/R²
  4. 2R/(μ0I)
Reveal correct answer and explanation

Answer A. Standard centre formula for one turn.

Q11. Ampère law ∮B·dl equals
  1. μ0I enclosed
  2. μ0/I
  3. ε0E
  4. zero always
Reveal correct answer and explanation

Answer A. Steady-current magnetostatic law.

Q12. Ideal long solenoid interior field
  1. μ0nI
  2. μ0n/I
  3. μ0I/n
  4. zero
Reveal correct answer and explanation

Answer A. Ampère result.

Q13. Doubling solenoid turn density
  1. halves B
  2. doubles B
  3. quadruples B
  4. does nothing
Reveal correct answer and explanation

Answer B. B∝n at fixed I.

Q14. Force on perpendicular wire segment
  1. IL/B
  2. IB/L
  3. ILB
  4. I/(LB)
Reveal correct answer and explanation

Answer C. F=ILB.

Q15. Same-direction parallel current wires
  1. attract
  2. repel
  3. do not interact
  4. always cross
Reveal correct answer and explanation

Answer A. Magnetic interaction pulls wires together.

Q16. Magnetic moment of N-turn coil
  1. NIA
  2. N/I
  3. I/(NA)
  4. NAB
Reveal correct answer and explanation

Answer A. m=NIA.

Q17. Torque on dipole parallel to B
  1. maximum
  2. zero
  3. mB
  4. infinite
Reveal correct answer and explanation

Answer B. sin0°=0.

Q18. Cyclotron’s accelerating force comes from
  1. magnetic field only
  2. electric field
  3. gravity only
  4. friction
Reveal correct answer and explanation

Answer B. Electric field does work in the gap.

Q19. Convert galvanometer to ammeter with
  1. series high resistor
  2. parallel shunt
  3. open circuit
  4. capacitor only
Reveal correct answer and explanation

Answer B. Low shunt bypasses current.

Q20. Convert galvanometer to voltmeter with
  1. parallel shunt
  2. high series resistance
  3. short circuit
  4. magnet removal
Reveal correct answer and explanation

Answer B. Series resistance limits coil current.