Moving Charges and Magnetism

Oersted's experiment showed that an electric current produces a magnetic effect, linking electricity and magnetism for the first time. A region where a moving charge or a magnet experiences a force is said to have a magnetic field, denoted B. When a charge q moves with velocity v through a region with both electric field E and magnetic field B, the total electromagnetic force on it is the Lorentz force.

• Electric force: qE — acts along E, independent of motion.
• Magnetic force: q(v × B) — depends on velocity, and is always perpendicular to both v and B.

The magnitude of the magnetic force alone is F = qvB sinθ, where θ is the angle between v and B. Its direction is found using the right-hand rule for the cross product v × B (then reversed if q is negative). Because this force is always perpendicular to velocity, it can never do work on the charge — it changes only the direction of motion, never the speed or kinetic energy.

If v is perpendicular to a uniform B (θ = 90°), the magnetic force supplies exactly the centripetal force needed for circular motion: qvB = mv²/r, giving a radius r = mv/(qB). The time for one revolution, T = 2πm/(qB), and hence the frequency ν_c = qB/(2πm), depend only on the charge-to-mass ratio and B — not on the speed v or radius r. This speed-independence is exactly what makes the cyclotron work.

Worked Example 1: A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves at 4 × 10⁶ m/s perpendicular to a field of 0.5 T. Find the radius and period of its circular path.

1. r = mv/(qB) = (1.67 × 10⁻²⁷ × 4 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.5) = 6.68 × 10⁻²¹/8 × 10⁻²⁰ ≈ 0.0835 m = 8.35 cm.
2. T = 2πm/(qB) = 2π × 1.67 × 10⁻²⁷/8 × 10⁻²⁰ ≈ 1.31 × 10⁻⁷ s.

When crossed electric and magnetic fields act together, a charge travels undeviated only if qE = qvB, i.e. v = E/B. Such a velocity selector passes only particles of one particular speed and rejects the rest, regardless of the particle's mass or charge magnitude.

Worked Example 2: In a velocity selector, E = 3 × 10⁵ V/m and B = 0.2 T. Only particles with v = E/B = 3 × 10⁵/0.2 = 1.5 × 10⁶ m/s pass through undeflected.

The cyclotron uses this speed-independence of ν_c: charged particles spiral outward between two D-shaped electrodes (dees) in a magnetic field, gaining energy each time they cross the gap where an oscillating electric field is applied at frequency equal to ν_c. The radius grows with speed, but the time per revolution stays fixed, so the same oscillator frequency keeps accelerating the particle turn after turn — until relativistic mass increase at very high speeds throws the timing out of sync, which limits the maximum energy attainable.

Just as a point charge sets up an electric field, a small current element I dl sets up a magnetic field dB at a point P. The Biot-Savart law gives its magnitude as directly proportional to I, dl and sinθ (θ being the angle between dl and the position vector r̂ from the element to P), and inversely proportional to r². The direction of dB is along dl × r̂, found by the right-hand rule, and is always perpendicular to the plane containing dl and r.

The proportionality constant μ₀/4π uses the permeability of free space, μ₀ = 4π × 10⁻⁷ T·m/A. The law is structurally similar to Coulomb's law but for current elements instead of point charges, and it obeys the same inverse-square dependence on distance.

Applying the Biot-Savart law to every current element of a circular loop of radius R carrying current I, and using the symmetry of the loop, gives the field at the centre of the loop (for N turns): B = μ₀NI/(2R), directed along the axis.

More generally, at a point P on the axis at distance x from the centre, only the axial components of dB from each element survive (the perpendicular components cancel in pairs by symmetry), giving B = μ₀NIR²/[2(R² + x²)^(3/2)]. Setting x = 0 recovers the centre-of-loop formula. For x ≫ R, this field falls off as 1/x³, exactly like the axial field of an electric dipole — showing that a current loop behaves as a magnetic dipole at large distances, an idea developed further later in the chapter.

Worked Example: A circular coil of radius 10 cm has 50 turns and carries a current of 2 A. Find the magnetic field at its centre.

1. B = μ₀NI/(2R) = (4π × 10⁻⁷ × 50 × 2)/(2 × 0.1)
2. Numerator = 4π × 10⁻⁷ × 100 = 1.2566 × 10⁻⁴
3. B = 1.2566 × 10⁻⁴/0.2 = 6.28 × 10⁻⁴ T.

Ampere's circuital law states that the line integral of B around any closed loop equals μ₀ times the total current enclosed by that loop: ∮B·dl = μ₀I_enc. It plays the same role for magnetism that Gauss's law plays for electrostatics, and is especially powerful when the current distribution has enough symmetry to pull B out of the integral.

Infinite straight wire: By symmetry, B has the same magnitude at every point on a circle of radius a centred on the wire, and is tangential to it. So ∮B·dl = B(2πa) = μ₀I, giving B = μ₀I/(2πa). Field lines are concentric circles around the wire, direction given by the right-hand thumb rule.

Solenoid: A closely-wound helical coil, long compared to its diameter, has a nearly uniform field inside (along the axis, away from the ends) and a negligible field outside. Applying Ampere's law to a rectangular loop straddling the solenoid's wall gives B = μ₀nI inside, where n is the number of turns per unit length. This resembles the field of a bar magnet, with one end acting as a north pole and the other as a south pole.

Toroid: A solenoid bent into a closed ring. Applying Ampere's law to a circle of radius r running through the core gives B = μ₀NI/(2πr) inside the core, while the field is essentially zero both outside the toroid and in the hollow space enclosed by it, since those Amperian loops enclose zero net current.

Worked Example 1: Find B at 5 cm from a long straight wire carrying 10 A.

1. B = μ₀I/(2πa) = (2 × 10⁻⁷ × 10)/0.05 = 2 × 10⁻⁶/0.05 = 4 × 10⁻⁵ T.

Worked Example 2: A solenoid has 5000 turns per metre and carries 2 A. Find the field well inside it.

1. B = μ₀nI = 4π × 10⁻⁷ × 5000 × 2 = 4π × 10⁻³ ≈ 1.26 × 10⁻² T.

A current is a stream of moving charge carriers, each feeling a magnetic force qv_d × B. Summing this force over all the free charges in a straight conductor of length L carrying current I gives a single resultant force F = IL × B, of magnitude F = BIL sinθ, where θ is the angle between the wire and B. This is the basis of the electric motor effect and is used for the direction with the right-hand slap rule (or equivalently Fleming's left-hand rule for the motor effect).

Consider two long, straight, parallel wires separated by distance d, carrying currents I₁ and I₂. The field due to wire 1 at the location of wire 2 is B₁ = μ₀I₁/(2πd), directed perpendicular to wire 2. The force per unit length that wire 2 experiences is therefore f = I₂B₁ = μ₀I₁I₂/(2πd). By Newton's third law, wire 1 feels an equal and opposite force per unit length from wire 2.

If the currents flow in the same direction, the wires attract; if in opposite directions, they repel. This mutual force is used to define the SI unit of current: one ampere is the constant current which, if maintained in two straight parallel infinite conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of exactly 2 × 10⁻⁷ N per metre of length between them.

Worked Example 1: A 0.5 m wire carries 3 A in a field of 0.4 T, perpendicular to the wire. Find the force.

1. F = BIL = 0.4 × 3 × 0.5 = 0.6 N.

Worked Example 2: Two parallel wires 10 cm apart carry currents of 5 A and 8 A in the same direction. Find the force per unit length.

1. f = μ₀I₁I₂/(2πd) = (2 × 10⁻⁷ × 5 × 8)/0.1 = 8 × 10⁻⁶/0.1 = 8 × 10⁻⁵ N/m, attractive (same direction).

Consider a rectangular loop of area A carrying current I, placed in a uniform field B, with its plane inclined so that the normal to the loop makes angle θ with B. The forces on the two arms parallel to the field cancel (equal, opposite, along the same line), while the forces on the other two arms form a couple. Adding up these contributions gives a net torque of magnitude τ = NIAB sinθ for N turns, tending to rotate the loop so as to reduce θ — that is, to align the loop's normal with B. Crucially, the net force on the loop in a uniform field is always zero; only a torque acts.

It is useful to define the magnetic dipole moment of the loop as m = NIA, a vector of magnitude NIA directed along the normal to the loop (given by the right-hand curl rule: curl the fingers along the direction of current flow, and the thumb points along m). The torque can then be written compactly as τ = m × B, i.e. τ = mB sinθ.

The potential energy of this magnetic dipole in the field is U = −m·B = −mB cosθ. U is minimum (most stable) when m is parallel to B (θ = 0°), and maximum when antiparallel (θ = 180°). This torque-on-a-current-loop effect is exactly what makes an electric motor turn, and (with a suitable radial field) is what makes a moving coil galvanometer deflect proportionally to current — the subject of the next section.

Worked Example: A rectangular coil of 100 turns, area 0.02 m², carries a current of 5 A. Its plane is kept parallel to a uniform field of 0.3 T. Find the torque on it.

1. Plane parallel to B means the normal is perpendicular to B, so θ = 90° and sinθ = 1.
2. τ = NIAB sinθ = 100 × 5 × 0.02 × 0.3 × 1 = 3 N·m.

A moving coil galvanometer consists of a rectangular coil of many turns wound on a light frame, suspended in the radial magnetic field of a strong permanent magnet with curved pole pieces and a soft-iron cylindrical core between them. The core makes the field radial — always parallel to the plane of the coil — so that the deflecting torque NIAB is the same at every orientation of the coil, instead of varying as sinθ. This design choice is essential: it is what makes the instrument's scale linear.

When current I flows, the deflecting torque NIAB is opposed by the restoring torque kφ of a spring (k is the torsional constant, φ the deflection angle). At equilibrium, NIAB = kφ, so I = (k/NAB)φ — deflection is directly proportional to current, giving an evenly-spaced linear scale.

Two figures of merit describe the instrument's response. Current sensitivity is the deflection per unit current, I_s = φ/I = NAB/k. Voltage sensitivity is deflection per unit applied voltage, V_s = φ/V = NAB/(kR), where R is the coil's resistance. Increasing N or B (or decreasing k) raises current sensitivity, but simply increasing N also increases the coil's own resistance R, so voltage sensitivity does not necessarily improve in the same proportion.

A galvanometer alone can only measure small currents near its full-scale value and has non-zero resistance, so it must be modified to serve as a practical ammeter or voltmeter. For an ammeter, a small resistance (shunt) S is connected in parallel with the galvanometer so that most of the current bypasses the coil: S = I_gG/(I − I_g), where G is coil resistance, I_g the full-scale galvanometer current, and I the desired ammeter range. For a voltmeter, a large resistance R is connected in series: R = V/I_g − G, where V is the desired voltmeter range.

Worked Example: A galvanometer has N = 60 turns, coil area A = 3 × 10⁻³ m², is in a radial field B = 0.25 T, and has torsional constant k = 1.5 × 10⁻⁴ N·m/rad. Find the deflection produced by a current of 2 mA.

1. φ = NIAB/k = (60 × 2 × 10⁻³ × 3 × 10⁻³ × 0.25)/(1.5 × 10⁻⁴)
2. Numerator = 60 × 2 × 10⁻³ = 0.12; × 3 × 10⁻³ = 3.6 × 10⁻⁴; × 0.25 = 9 × 10⁻⁵.
3. φ = 9 × 10⁻⁵/1.5 × 10⁻⁴ = 0.6 rad.