Moving Charges and Magnetism — NEET Notes

Magnetic force on a moving charge — the Lorentz force

Every charged particle in motion through a magnetic field experiences a sideways shove. The magnitude and direction of that shove are captured in a single vector equation due to H. A. Lorentz. If a charge q moves with velocity v through a region where the magnetic field is B, the magnetic force on it is the cross-product F = qv × B. When an electric field E is also present, the two contributions add: F = q(E + v × B). This combined expression is what physicists call the Lorentz force.

Three features of the magnetic part deserve to be memorised before any calculation. First, the force vanishes if v and B are parallel — the cross-product is zero. It is maximum, qvB, when v is perpendicular to B. Second, because F is always perpendicular to v, it does no work on the charge; the kinetic energy and therefore the speed remain constant. The magnetic field changes direction, never magnitude. Third, the sign of the charge matters — the force on a positive charge is opposite to that on a negative one moving with the same velocity in the same field. NEET 2021 tested this exact algebra by giving F, v and partial B and asking the student to solve for the missing component.

F = q (E + v × B)

The Lorentz force law — every electromagnetic interaction on a charge

The SI unit of B follows directly: from F = qvB sinθ, one tesla is the field that exerts one newton on a one-coulomb charge moving at one metre per second perpendicular to the field. One tesla equals 10⁴ gauss; the earth's surface field is about 0.5 G or 5 × 10⁻⁵ T, while a strong neodymium magnet reaches roughly 1 T.

For a current-carrying conductor, the same law summed over the moving charges gives the force on a straight wire of length L carrying current I in a uniform field: F = IL × B, where L points along the conventional current. NEET 2023 (Q.42) asked exactly this — a wire of length L along the x-axis in B = (−2î + 3ĵ + 4k̂) T — and the magnitude came out to 5IL after a clean cross-product.

Motion in a magnetic field — circles and helices

Since the magnetic force does no work, a charged particle in a uniform B keeps a constant speed. What changes is direction, and the geometry of that change is entirely fixed by the angle between v and B. Two limiting cases organise every problem in this section.

If v is perpendicular to B, the force qvB acts as a centripetal force pointing toward a centre, and the particle moves in a circle. Equating qvB = mv²/r gives the radius r = mv/(qB). The angular speed is ω = qB/m, the period is T = 2πm/(qB), and the frequency ν = qB/(2πm). All three depend only on the particle's specific charge q/m and the field — never on the particle's speed. This isochronism is what made the cyclotron possible.

If v has a component v∥ along B and a component v⊥ perpendicular to it, the parallel component is unaffected (no force along B) while the perpendicular component drives circular motion. The trajectory is a helix wound around the field lines: the radius of the helix is r = mv⊥/(qB) and the pitch — the axial distance per revolution — is p = v∥ × T = 2πmv∥/(qB). NEET problem 4.11 in NCERT itself works through this geometry for an electron in a 6.5 G field.

The cyclotron — using B to keep time

E. O. Lawrence's cyclotron, built in 1932, exploits the isochronism of circular motion in a magnetic field to accelerate charged particles to high energies. Two hollow D-shaped electrodes ("dees") sit between the poles of a strong magnet, separated by a small gap. An alternating voltage between them oscillates at a fixed frequency. A charged particle released near the centre is accelerated across the gap, drifts in a semicircle inside one dee at speed determined by qvB = mv²/r, and arrives back at the gap just as the polarity has flipped — accelerating it again. With each crossing the radius grows; the period does not. After many cycles the particle spirals outward at high speed and is extracted.

Three formulas pay back the time spent memorising them. The radius of the orbit at any stage is r = mv/(qB); the period is T = 2πm/(qB); the cyclotron frequency is ν = qB/(2πm). The maximum kinetic energy attainable, set by the radius R of the dees, is KE_max = q²B²R²/(2m). The cyclotron fails for highly relativistic particles, because mass effectively increases with speed and the resonance breaks — but for protons up to roughly 25 MeV the classical picture is excellent.

Biot-Savart law — field from a current element

To find the field produced by a current rather than the force on a charge, two laws are available. The first, Biot-Savart, treats every infinitesimal segment of current as a source. If a length element dl carries current I, the field it produces at a point P located by the displacement r from the element is

dB = (μ₀/4π) · I (dl × r̂) / r²

Biot-Savart law — the magnetic analogue of Coulomb's law

Here r̂ is the unit vector from the element to the field point, r is the distance, and the constant μ₀ = 4π × 10⁻⁷ T·m·A⁻¹ is the permeability of free space. The total field at P is found by integrating over the entire current-carrying circuit. Three features parallel Coulomb's law: the inverse-square dependence, linearity in the source, and the principle of superposition. Two features differ: the source dl is a vector (current has a direction); and the cross-product means dB is perpendicular to both the element and the line of sight. NEET 2022 (Q.4) tested precisely these distinctions in a statement-based question — the analogy with Coulomb's law is real but not exact.

The classic application is the long straight wire. Integrating Biot-Savart along an infinite wire carrying current I gives the field at perpendicular distance d as B = μ₀I/(2πd), with direction given by the right-hand grip rule — curl the fingers of the right hand around the wire so the thumb points along I; the fingers point along B. NEET 2021 (Q.6) used this formula to compute the force on an electron moving parallel to a current-carrying wire, getting 8 × 10⁻²⁰ N.

Magnetic field on the axis of a circular loop

A circular loop of radius R carrying current I produces, on its axis at distance x from the centre, a field directed along the axis with magnitude

Two limits are worth remembering. At the centre of the loop (x = 0), B = μ₀I/(2R). Far from the loop (x ≫ R), B ≈ μ₀IR²/(2x³) — the field falls off as 1/x³, the signature of a magnetic dipole. The direction is given by the right-hand rule: curl the fingers along the current, the thumb gives B. For a coil of N turns, multiply by N. NEET 2023 (Q.43) tested a variant in which a long wire is bent into a semicircle plus two straight sections — the straight pieces and the arc each contributed a Biot-Savart term and the answer needed careful sign work.

The magnetic moment of the loop is m = IA, where A = πR². The same loop, viewed from far away on its axis, behaves like a magnetic dipole of moment m — the bridge to the next chapter on Magnetism and Matter.

Ampere's circuital law — symmetry as a shortcut

The Biot-Savart law works in principle for any geometry, but the integrals get painful. Ampere's circuital law is the integral form of the magnetostatic equivalent of Gauss's law, and it lets us evaluate B in a single line whenever symmetry permits. The statement: the line integral of B around any closed loop equals μ₀ times the net current enclosed by the loop.

∮ B · dl = μ₀ I_enc

Ampere's circuital law — the shortcut for symmetric currents

The trick is to choose an "Amperian loop" along which B is either constant in magnitude and parallel to dl, or perpendicular to dl (so the dot product vanishes). For the infinite straight wire, the natural loop is a circle of radius d centred on the wire. By symmetry B is constant in magnitude on this circle and tangent to it, so the integral becomes B · 2πd = μ₀I, giving B = μ₀I/(2πd) — the same result as Biot-Savart, in one line.

For a thick cable of radius R carrying current uniformly distributed across its cross-section, Ampere's law gives two regimes. Outside (r ≥ R), B = μ₀I/(2πr), falling as 1/r. Inside (r < R), only the current enclosed by a loop of radius r — which is I · (r²/R²) — contributes, giving B = μ₀Ir/(2πR²), rising linearly with r. The field thus rises linearly to a peak at the surface, then falls as 1/r outside. NEET 2021 (Q.25) and NEET 2022 (Q.49) both tested this exact profile.

Solenoid and toroid — the engineered fields

A solenoid is a long helical coil of wire. When the length is much greater than the radius and the turns are packed closely, the field inside is uniform, parallel to the axis, and the field outside is negligible. Apply Ampere's law to a rectangular loop with one side of length L inside the solenoid (parallel to the axis) and the other outside (where B ≈ 0); the sides perpendicular to the axis contribute zero because B ⊥ dl. The enclosed current is nLI, where n is the number of turns per unit length and I the current. The result is the workhorse formula:

NEET 2020 (Q.92) gave 100 turns over 50 cm and 2.5 A — substituting n = 200 m⁻¹ yields B = 4π × 10⁻⁷ × 200 × 2.5 ≈ 6.28 × 10⁻⁴ T. NEET 2022 (Q.29) repeated the pattern: 100 turns per mm (= 10⁵ m⁻¹) and 1 A gave B = 12.56 × 10⁻² T. Both questions reward the student who can convert turns-per-millimetre into turns-per-metre without mistakes.

A toroid is a solenoid bent into a closed ring. Applying Ampere's law to a circular loop of radius r inside the toroidal core gives B · 2πr = μ₀NI, so B = μ₀NI/(2πr), where N is the total number of turns. The field is confined to the interior; outside the toroid B is zero, both for loops outside the outer surface and for loops inside the central hole.

Force between two parallel currents — defining the ampere

Two parallel wires carrying steady currents experience a force on each other. Wire 1 produces, at the location of wire 2, a field B₁ = μ₀I₁/(2πd) pointing perpendicular to wire 1. Wire 2, of length L, in this field experiences a force F = I₂LB₁ = μ₀I₁I₂L/(2πd). The force per unit length is therefore

The geometry of the cross-products gives the famous rule: parallel currents attract, antiparallel currents repel — the opposite of what charges do. The definition of the ampere is built directly on this formula: one ampere is the steady current which, when flowing in two infinitely long, parallel, straight conductors placed one metre apart in vacuum, produces a force of exactly 2 × 10⁻⁷ N on each metre of length. Every other electrical unit follows: the coulomb is an ampere-second, the volt is a joule per coulomb, the ohm follows from V = IR.

NEET 2017 (Q.156) tested a three-wire configuration in which a middle wire feels the resultant of two equal-and-perpendicular pulls — the answer was √2 μ₀I²/(2πd). NEET 2016 (Q.155) asked for the net force on a square loop placed coplanar with a long wire, exploiting attraction on the near side and repulsion on the far side; the difference of two 1/r terms gave the answer.

Torque on a current loop — birth of the magnetic dipole

A rectangular loop carrying current I in a uniform field B experiences no net force — the forces on opposite sides cancel — but it does experience a net torque. Two sides of length b carry currents perpendicular to B; they feel forces of magnitude IbB pointing opposite ways, separated by the loop's width a. The torque about the loop's centre is τ = (IbB) × a × sinθ = IAB sinθ, where A = ab is the area and θ is the angle between the area-vector n̂ and the field B.

Generalised to N turns, the torque is τ = NIAB sinθ. In vector form, define the magnetic dipole moment as m = NI A n̂, with n̂ given by the right-hand rule (curl fingers along current, thumb gives n̂). Then τ = m × B — identical in form to the electric-dipole torque τ = p × E. The loop tends to rotate until m aligns with B; this is the equilibrium orientation.

The potential energy of the dipole in the field is U(θ) = −m·B = −mB cosθ. The work required to rotate from θ₁ to θ₂ is W = mB(cosθ₁ − cosθ₂). NEET 2017 (Q.138) asked for the work to flip a 250-turn coil by 180°: W = 2mB = 2NIAB ≈ 9.1 μJ — pure plug-and-chug once the formula is memorised.

NEET 2021 (Q.50) tested a wire of fixed length 12a wound into either an equilateral triangle or a square. Both shapes have the same wire length, but different number of turns N (since N = total length / perimeter) and different enclosed area A — and the product NA happens to give 3Ia² for both shapes. The lesson: write m = NIA and don't skip steps.

The moving coil galvanometer

The moving coil galvanometer is the laboratory instrument that turns the τ = NIAB equation into a usable measurement. A multi-turn rectangular coil pivots in the radial field between cylindrical pole pieces of a permanent magnet; a soft iron cylinder sits inside the coil to keep the field radial. Because the field is always radial, the plane of the coil is always parallel to B (θ = 90°) — so the magnetic torque is simply τ_mag = NIAB, independent of the angle of deflection.

A torsion spring provides a restoring torque proportional to the angular deflection φ: τ_spring = kφ. In equilibrium NIAB = kφ, giving

φ = (NAB / k) · I

Galvanometer deflection — linear in the current

The linearity is what makes the instrument quantitative. Two figures of merit summarise its performance. The current sensitivity is deflection per unit current, φ/I = NAB/k. The voltage sensitivity is deflection per unit voltage, φ/V = NAB/(kR), where R is the coil's resistance. The two are not interchangeable: increasing N raises both, but raising R raises only current sensitivity (the numerator) while reducing voltage sensitivity (the denominator grows faster than the numerator if R grows faster than N).

Two important conversions follow. A galvanometer becomes an ammeter by connecting a small shunt resistance r_s in parallel; most of the current bypasses the coil so the meter can read large currents. It becomes a voltmeter by connecting a large series resistance R; the meter then draws negligible current and the voltage drop across the series combination equals the measured voltage. The shunt is small; the series resistance is large — a confusion students often invert under exam pressure.

Quick reference — the four B-field formulas

Four configurations cover almost every NEET problem on the field produced by a steady current. They are summarised below, each with its derivation route and a numerical anchor.