Moving Charges and Magnetism

Oersted's compass experiment (1820) showed that a current deflects a magnetic needle, and Ampere's later work established that all magnetism has moving charges behind it. So the chapter is built on one pair of statements: moving charges produce a magnetic field B, and a magnetic field exerts a force on moving charges. Like the electric field, B is a vector field defined at every point, it obeys superposition, and it is produced by currents. Unlike E, it acts only on charges that move.

A charge q moving with velocity v in a region where both an electric field E and a magnetic field B exist experiences the Lorentz force F = q(E + v × B). The magnetic part F_m = q(v × B) has three properties the examiner expects you to state. It depends on q, v and B and is proportional to each; it vanishes when the charge is at rest or moves parallel or antiparallel to B (sin θ = 0); and it is always perpendicular to both v and B, so it can bend the path but never change the speed. Because the force is perpendicular to the displacement at every instant, the work done by a magnetic force on a charge is always zero and its kinetic energy stays constant.

The direction comes from the right-hand rule for the cross product: point the fingers along v, curl them towards B, and the thumb gives v × B. For a positive charge the force is along the thumb; for an electron it is opposite. The SI unit of B is the tesla (T): a field of 1 T exerts a force of 1 N on a charge of 1 C moving at 1 m s⁻¹ perpendicular to the field, so 1 T = 1 N A⁻¹ m⁻¹. A smaller unit is the gauss, 1 G = 10⁻⁴ T; the earth's field is about 3.6 × 10⁻⁵ T.

A current is a stream of moving charges, so a wire carrying a current in a magnetic field feels a force. Take a straight conductor of length l and cross-section A with n free electrons per unit volume, each drifting with velocity v_d. The total mobile charge is q = (nAl)e, and the force on all of it is F = (nAle)(v_d × B). Since the current is I = neAv_d, this becomes F = I (l × B), where l is a vector of magnitude l along the direction of the current. The magnitude is F = BIl sin θ, θ being the angle between the wire and the field. The force is maximum (BIl) when the wire is perpendicular to the field and zero when it lies along the field.

For a wire of arbitrary shape, add the forces on its elements: F = ∑ I dl × B. In a uniform field this sum depends only on the vector joining the two ends, which gives two useful results: a closed loop in a uniform field experiences no net force (though it may experience a torque), and a curved wire between two points feels the same force as a straight wire joining them.

The direction can be found with Fleming's left-hand rule: hold the forefinger, middle finger and thumb of the left hand mutually perpendicular; forefinger along B, middle finger along the current, and the thumb gives the force. NCERT Example 4.1 is the standard application: a wire of mass 200 g and length 1.5 m carrying 2 A floats in a horizontal field when BIl = mg, giving B = 0.2 × 9.8/(2 × 1.5) = 0.65 T, with the field horizontal and perpendicular to the wire.

Because the magnetic force is always perpendicular to the velocity, it acts as a pure deflecting force: it changes the direction of v but not its magnitude. Consider a charge q moving with speed v perpendicular to a uniform field B. The force qvB is perpendicular to v and constant in magnitude, which is exactly the condition for uniform circular motion. Equating it to the centripetal force, qvB = mv²/r, so the radius is r = mv/(qB). The larger the momentum, the larger the circle; the stronger the field, the tighter the circle.

The angular frequency is ω = v/r = qB/m and the time period is T = 2πr/v = 2πm/(qB). Both are independent of the speed and of the radius: a fast particle moves on a bigger circle but completes it in the same time as a slow one. The frequency ν = qB/(2πm) is called the cyclotron frequency. This independence is what allowed the cyclotron to accelerate particles with an alternating voltage of fixed frequency.

If the velocity makes an angle θ with B, resolve it. The component v sin θ perpendicular to the field produces circular motion of radius r = mv sin θ/(qB); the component v cos θ along the field is unaffected, so the particle drifts along the field while it circles. The path is a helix with its axis along B. The distance moved along the field in one revolution is the pitch, p = (v cos θ) T = 2πmv cos θ/(qB). A charge moving parallel to B feels no force and continues in a straight line.

Two comparisons recur in the paper. For the same speed, r ∝ m/q, so an alpha particle (m = 4m_p, q = 2e) has twice the radius of a proton. For the same kinetic energy K, r = √(2mK)/(qB), so a proton and an alpha particle of equal energy have equal radii, while a deuteron has √2 times the proton's radius.

The Biot–Savart law is the magnetic counterpart of Coulomb's law: it gives the field produced by a small element of current. Consider a conductor carrying a steady current I and a small element dl of it, pointing along the current. At a point P whose position vector from the element is r, the field is dB = (μ₀/4π) · I (dl × r)/r³. In magnitude, dB = (μ₀/4π) · I dl sin θ/r², where θ is the angle between dl and r. The direction of dB is perpendicular to the plane containing dl and r, given by the right-hand rule for the cross product; for a straight wire this means the field circles the wire.

Points to state, because they carry marks: the field is directly proportional to the current, to the length of the element and to sin θ, and inversely proportional to the square of the distance. It is zero at points along the line of the element (θ = 0 or 180°) and maximum in the plane perpendicular to it (θ = 90°). The constant μ₀/4π = 10⁻⁷ T m A⁻¹, so μ₀ = 4π × 10⁻⁷ T m A⁻¹, the permeability of free space. The total field of a conductor is the vector sum (integral) of the contributions of all its elements; that is the principle of superposition applied to B.

Comparison with Coulomb's law is a standard 2-mark question. Both are long range and fall as 1/r²; both obey superposition. The differences: the electrostatic field is produced by a scalar source (charge) while the magnetic field is produced by a vector source (I dl); the electric field is along the line joining source and point, while the magnetic field is perpendicular to the plane containing the element and that line; and the magnetic field has the angle dependence sin θ, being zero along the element.

This is the chapter's first flagship derivation. Take a circular loop of radius R carrying current I, centred at the origin with its plane in the y–z plane, and a point P on the axis at distance x from the centre. Pick an element dl on the loop. The distance from it to P is r = √(x² + R²), and dl is perpendicular to r, so sin θ = 1 and dB = (μ₀/4π) · I dl/(x² + R²).

The direction of dB is perpendicular to both dl and r. Resolve it into a component dB cos α along the axis and dB sin α perpendicular to it, where α is the angle between r and the axis. For the diametrically opposite element the perpendicular component is reversed, so all perpendicular components cancel in pairs and only the axial components add. With cos α = R/√(x² + R²), the axial component is dB_x = (μ₀/4π) · I dl R/(x² + R²)^3/2. Summing dl over the whole loop gives 2πR, so B = μ₀IR²/[2(x² + R²)^3/2], directed along the axis.

Two limits matter. At the centre, x = 0 and B = μ₀I/(2R); for a coil of N turns, B = μ₀NI/(2R). Far away, x ≫ R and B ≈ μ₀IR²/(2x³) = (μ₀/4π) · 2m/x³ with m = IπR², which is the same form as the axial field of an electric dipole; that is why a current loop is called a magnetic dipole. The direction of the field on the axis follows the right-hand thumb rule: curl the fingers of the right hand along the current and the thumb points along B at the centre. A current circulating anticlockwise, seen by the observer, gives a field towards the observer (a north face).

Fields of arcs follow at once from the centre formula. A semicircle gives μ₀I/(4R); an arc subtending angle φ (in radians) gives μ₀Iφ/(4πR). A long straight wire touching the arc contributes nothing at the centre if the centre lies on its line.

Ampere's circuital law is to magnetism what Gauss's law is to electrostatics: a statement about the field summed round a closed path. It says that the line integral of B around any closed loop equals μ₀ times the total current passing through the surface bounded by the loop: ∮ B·dl = μ₀I_enc. The loop is called an Amperian loop; it is a mathematical curve, not a physical wire. Currents outside the loop contribute to B at points on the loop but not to the integral. The sign of each current is fixed by the right-hand rule: with the fingers curling along the direction of traversal, currents along the thumb count positive.

The law is most useful when symmetry makes B constant in magnitude and tangential all along the loop, so that ∮ B·dl = B × (length of loop). For a long straight wire carrying current I, choose a circle of radius r centred on the wire: B(2πr) = μ₀I, so B = μ₀I/(2πr). The field circles the wire, falls as 1/r, and its direction is given by the right-hand thumb rule (thumb along the current, fingers curl along B).

The same method handles a thick wire of radius a with current I spread uniformly over its cross-section. Outside (r ≥ a) the loop encloses all of I, so B = μ₀I/(2πr) as before. Inside (r < a) the enclosed current is I(r²/a²), so B(2πr) = μ₀Ir²/a², giving B = μ₀Ir/(2πa²). The field grows linearly from zero at the axis to a maximum μ₀I/(2πa) at the surface and then falls as 1/r. The B–r graph, a straight line up to r = a followed by a hyperbola-like fall, is asked frequently and should be drawn with the surface value marked.

Ampere's law is exact for steady currents but, like Gauss's law, it is easy to use only in symmetric cases: a long straight wire, a solenoid, a toroid. For a finite wire or a single loop one must go back to the Biot–Savart law.

A solenoid is a long insulated wire wound closely in the form of a helix, so that each turn behaves like a circular loop. Inside a long solenoid the fields of neighbouring turns add to give a strong, uniform field parallel to the axis; outside, the fields of the upper and lower parts of the turns oppose each other and the field is very weak, essentially zero for an ideal (infinitely long) solenoid. Near the ends the field lines spread out; the solenoid behaves like a bar magnet with a north pole at the end where the current appears anticlockwise to an observer facing it.

The derivation with Ampere's law is a standard 3-mark answer. Let the solenoid have n turns per unit length and carry current I. Choose a rectangular Amperian loop abcd with side ab of length h inside and parallel to the axis, side cd outside, and sides bc and da perpendicular to the axis. Evaluate ∮ B·dl along the four sides. Along ab, B is parallel to dl and uniform, contributing Bh. Along bc and da, B is perpendicular to dl inside and zero outside, so they contribute nothing. Along cd the field is zero. Hence ∮ B·dl = Bh. The loop encloses nh turns, each carrying I, so I_enc = nhI. Ampere's law gives Bh = μ₀nhI, that is B = μ₀nI.

The result is independent of the radius of the solenoid and of the distance of the point from the axis, which is the meaning of a uniform field. It is increased by winding more turns per metre, by raising the current, or by filling the core with a ferromagnetic material of relative permeability μ_r, in which case B = μ₀μ_rnI. At either end of a long solenoid the field drops to half, μ₀nI/2, because only half of the turns lie on one side. For a solenoid of length L with N turns in total, n = N/L, and in NCERT exercise 4.8 the five layers of 400 turns on an 80 cm former give n = 2000/0.8 = 2500 m⁻¹.

Two long straight parallel wires a and b, separated by distance d, carry currents I_a and I_b in the same direction. Wire a produces at the position of wire b a field B_a = μ₀I_a/(2πd), directed perpendicular to b and into the plane (for a to the left of b with both currents upward). A length L of wire b, carrying I_b in this field, experiences the force F_ba = I_bLB_a = μ₀I_aI_bL/(2πd). By the rule F = I (L × B) this force is directed towards a. Repeating the argument for the force of b on a gives an equal force towards b: the wires attract. If one current is reversed the force reverses and the wires repel. So parallel currents attract, antiparallel currents repel, which is the opposite of the rule for charges.

The force per unit length on either wire is f = μ₀I_aI_b/(2πd) = 2 × 10⁻⁷ I_aI_b/d in SI units. Putting I_a = I_b = 1 A and d = 1 m gives f = 2 × 10⁻⁷ N m⁻¹. This is the NCERT definition of the ampere: one ampere is that steady current which, when maintained in each of two very long straight parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces on each a force of 2 × 10⁻⁷ newton per metre of length. The coulomb follows as the charge carried past a point in one second by a current of one ampere.

Two things to note in the answer. The force on b due to a and the force on a due to b are equal and opposite here, as Newton's third law requires; NCERT points out that this need not hold in general for magnetic forces between isolated current elements. And because the force depends on the product of the currents and on 1/d, doubling both currents quadruples the force, while doubling the separation halves it.

A rectangular loop of sides a and b carrying current I is placed in a uniform field B. First take the field in the plane of the loop, parallel to the sides of length a. Those two sides carry current along or against B, so they feel no force. The sides of length b are perpendicular to B and carry current in opposite directions, so they feel equal and opposite forces F₁ = F₂ = IbB, directed into and out of the plane. Their lines of action are separated by a, so they form a couple of torque τ = F₁(a/2) + F₂(a/2) = IabB = IAB, where A = ab is the area. The net force is zero; the loop tends to rotate.

Now let the normal to the plane of the loop make an angle θ with B. The forces on the sides of length b are still IbB each, but the perpendicular distance between their lines of action is now a sin θ, so τ = IAB sin θ. The forces on the other two sides are equal, opposite and collinear, contributing nothing. For a coil of N turns, τ = NIAB sin θ. Defining the magnetic moment of the loop as m = NIA, with A along the normal by the right-hand rule (fingers along the current, thumb along m), the torque is τ = m × B. The torque is zero when the normal is along B (stable equilibrium at θ = 0, unstable at θ = 180°) and maximum, NIAB, when the plane of the loop is parallel to the field.

The magnetic moment has unit A m² and behaves exactly like the electric dipole moment: τ = p × E becomes τ = m × B. The result holds for a planar loop of any shape, which is why NCERT exercise 4.13(b) says the torque does not change when a circular coil is reshaped into an irregular loop of the same area. A circular loop is likewise a magnetic dipole: its axial field far away, B = (μ₀/4π) · 2m/x³, has the same form as the axial field of an electric dipole with 1/4πε₀ replaced by μ₀/4π and p by m. Unlike an electric dipole, however, a magnetic dipole cannot be split into two isolated poles.

The moving coil galvanometer (MCG) is the chapter's main application and its most reliable 5-mark question. Principle: a current-carrying coil placed in a magnetic field experiences a torque. Construction: a coil of many turns of fine insulated wire wound on a light rectangular frame is suspended by a phosphor-bronze strip (or pivoted between springs) between the cylindrical pole pieces of a permanent magnet. A soft-iron cylinder is fixed inside the coil without touching it. The pole pieces are concave and, together with the soft-iron core, make the field radial: at every position of the coil the plane of the coil is parallel to the field lines, so sin θ = 1 always. The core also concentrates the field, making it stronger.

Working: when a current I flows, the magnetic torque NIAB turns the coil, twisting the suspension. The suspension provides a restoring torque kφ proportional to the twist φ, k being the torsion constant. At equilibrium NIAB = kφ, so φ = (NAB/k) I. The deflection is directly proportional to the current, which is why the scale is linear; a radial field is essential for this, because otherwise the magnetic torque would carry the factor sin θ and the scale would not be uniform.

Current sensitivity is the deflection per unit current, I_s = φ/I = NAB/k. Voltage sensitivity is the deflection per unit voltage, V_s = φ/V = NAB/(kR), where R is the resistance of the coil. Both are raised by a large N, a large area, a strong field and a small k, which is why phosphor bronze, with its small torsion constant, is used. Increasing N to raise the current sensitivity does not necessarily raise the voltage sensitivity, because a longer wire also raises R; doubling N with the same wire doubles R and leaves V_s unchanged.

A galvanometer as such cannot measure current in a circuit: it is very sensitive, giving full-scale deflection for microamperes, and its resistance is not small enough to leave the circuit current unchanged. To convert it into an ammeter of range I, connect a small shunt resistance S in parallel so that only I_g passes through the coil: S = I_gG/(I − I_g). The ammeter then has a very small resistance GS/(G + S) and is connected in series. To convert it into a voltmeter of range V, connect a large resistance R in series: R = V/I_g − G. The voltmeter has a large resistance G + R and is connected in parallel across the element.