CLASS 12 PHYSICS · CHAPTER 4 · STUDY NOTES
Moving Charges and Magnetism
An exploration of the relationship between electricity and magnetism, covering the Lorentz force, Biot-Savart law, Ampere's circuital law, and their applications in devices like solenoids and galvanometers.
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Introduction to Electromagnetism
The connection between electricity and magnetism was first observed by Hans Christian Oersted in 1820, marking the birth of electromagnetism.Predict: how will each compass turn when the current reverses?
Current out of the page gives anticlockwise B. The orange tip is each compass’s north end: it follows the local tangent, not a line pointing straight at the wire. Earth’s field is neglected. A dot is towards you; a cross is away.
- Conventional current
- The direction in which positive charge would flow; in a metal it is opposite to electron drift.
Oersted’s compass experiment shows the link: a current-carrying wire changes the direction of a nearby compass needle. Around a long straight wire, the magnetic field is tangent to circles centred on the wire.
Use a dot (⊙) for a vector towards you, out of the page, and a cross (⊗) for a vector away from you, into the page. These symbols represent the tip and tail of an arrow.
Curl the fingers of your right hand in the field direction around the wire; your thumb follows conventional current. Reverse the current and the field direction reverses.
Let’s look a little closer.
Go deeper: Reason it through
A compass responds to the total local field. The circular pattern is clearest when the wire’s field dominates Earth’s field; the needle is not simply attracted straight towards the wire.
Take a moment. What do you think?
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Question
A wire’s conventional current points into the page. Which way does its magnetic field circle?
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Magnetic Force and Lorentz Force
The Lorentz force describes the total force experienced by a charged particle moving through both electric and magnetic fields.Predict: what changes when the moving charge becomes negative?
q = 2 μC, v = 200 m/s, B = 0.30 T along +y; θ = 60°. |F_mag| = |q|vB sinθ = 103.9 μN, out of the page (+z). Magnetic force is perpendicular to v, so F_mag·v = 0: it changes direction, not kinetic energy.
- Lorentz force
- The combined electric and magnetic force on a charge: F⃗ = q(E⃗ + v⃗ × B⃗).
The electric force is qE⃗. The magnetic force is q(v⃗ × B⃗), perpendicular to both velocity and magnetic field. The sign of q reverses the magnetic force direction.
The magnetic force magnitude is |q|vB sinθ, where θ is the angle between velocity and field. It vanishes for a stationary particle and for parallel or antiparallel motion.
Because F⃗mag · v⃗ = 0, the magnetic part does no work on a point charge: it can turn the velocity without changing kinetic energy. An electric field can change the speed.
Go deeper: Reason it through
Add electric and magnetic forces as vectors. Their magnitudes can be added only when the vectors point in the same direction. The zero magnetic-work result does not imply the total Lorentz force always does zero work.
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Question
A charge of −3 μC moves at 100 m/s along +x in a 0.20 T field along +y. Find its magnetic force.
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Force on a Current-Carrying Conductor
Magnetic fields exert forces not just on individual charges, but also on wires carrying electric currents.Predict: what happens if you reverse both current and field?
I = 4 A, l = 0.50 m, B = 0.20 T. |F| = IlB = 0.40 N; it points up. Reverse either I or B to reverse F; reverse both to restore it. The length vector follows conventional current. B is the external field, not the wire’s own field.
- Force on a straight conductor
- In a uniform external field, F⃗ = I(l⃗ × B⃗); l⃗ points along conventional current.
The magnetic forces on charge carriers transfer to the wire. For a straight segment of length l carrying current I in a uniform external field, force magnitude is IlB sinθ.
A perpendicular wire has maximum force IlB; a parallel wire has zero magnetic force. Reverse current or field to reverse the force; reverse both to restore its direction.
The tesla is equivalent to N A⁻¹ m⁻¹. A perpendicular 1 m conductor carrying 1 A experiences 1 N in a uniform 1 T external field.
Go deeper: Reason it through
Use the external field, not the wire’s own field. For a curved wire or nonuniform field, add elemental forces dF⃗ = I(dl⃗ × B⃗). A suspension calculation also needs a force direction that can oppose gravity.
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Question
An original 0.10 kg horizontal wire of length 0.50 m carries 5.0 A. What perpendicular field magnitude would balance its weight? Use g = 9.8 m/s².
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Motion in a Magnetic Field
The trajectory of a charged particle in a uniform magnetic field depends on the angle at which it enters the field.Predict: will a stronger magnetic field change speed, radius, or both?
A non-relativistic proton has v_perp = 1.0 × 10⁵ m/s in B = 0.02 T. r = mv_perp/(eB) = 5.22 cm; T = 2πm/(eB) = 3.280 μs. B is into the page; the velocity arrow stays tangent and the force points towards the centre. Speed stays constant. Advance one eighth of a revolution at a time. These formulas assume a uniform static B, E = 0 and speeds well below c.
- Helical pitch
- The distance along the field in one revolution: p = |vparallel|T.
In a uniform static magnetic field with no electric field, perpendicular motion becomes circular. |q|v⊥B supplies centripetal force, giving r = mv⊥/(|q|B). Charge sign changes the turning sense, not the radius.
For nonrelativistic motion, T = 2πm/(|q|B) and f = |q|B/(2πm). The period is independent of speed in this model; relativistic mass-energy effects invalidate that simple conclusion at high speeds.
A parallel velocity component is unchanged by the magnetic force. Combining it with the perpendicular circle gives a helix, with pitch |vparallel|T. Purely parallel motion is straight.
Go deeper: Reason it through
A larger perpendicular speed increases the radius but leaves the nonrelativistic period unchanged. Increasing only parallel speed increases pitch without changing radius. Cyclotron acceleration requires an electric field to supply energy; the magnetic field alone cannot accelerate the particle’s speed.
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Question
In the same field, a nonrelativistic particle’s perpendicular speed triples while its parallel speed stays fixed. How do radius, period and pitch change?
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The Biot-Savart Law
The Biot-Savart law provides a mathematical method to calculate the magnetic field produced by a current element.Predict: how does one element’s field contribution change when distance doubles?
A short element carries 5 A, has dl = 1 cm and is 20 cm from P. At θ = 60°, |dB| ≈ μ₀I dl sinθ/(4πr²) = 108.3 nT, out of the page. This is one element’s contribution; integrate vector contributions for the whole wire. The infinitesimal-element approximation treats r and θ as constant across dl.
- Biot–Savart contribution
- The magnetic field due to an infinitesimal steady-current element: dB⃗ = (μ₀/4π)I(dl⃗ × r⃗)/r³.
The position vector r⃗ runs from the current element to the observation point. The magnitude contribution is (μ₀/4π)I dl sinθ/r², where θ is between dl⃗ and r⃗.
Use the cross product for direction, then integrate contributions as vectors over the complete current path. An element’s contribution vanishes on its own line; that does not mean every other element contributes zero.
For school-level calculations, μ₀ ≈ 4π × 10⁻⁷ H/m. This is an excellent approximation, not an exact defining value under the current SI.
Go deeper: Reason it through
Unlike the electric field of a point charge, this contribution is perpendicular to the plane of dl⃗ and r⃗. The formula here assumes steady currents. A short finite element can approximate an infinitesimal element only when distance and angle change little across its length.
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Question
For a small steady-current element, what happens to its field contribution if distance doubles while I, dl and θ stay fixed?
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Magnetic Field of a Circular Loop
Using the Biot-Savart law, we can determine the magnetic field on the axis of a circular current-carrying loop.Predict: where along the axis will the field be strongest?
For N = 10 turns, |I| = 1 A, R = 0.10 m and x/R = 0.0, |B_axis| = μ₀NIR²/[2(x²+R²)³ᐟ²] = 62.83 μT. Symmetric transverse contributions cancel; axial components add rightwards. The graph is |B|/B₀, where B₀ = μ₀NI/(2R). It is an axial result, not a formula for every point around a loop.
- Axial loop field
- On the axis of an N-turn circular coil: B = μ₀NIR²/[2(R² + x²)³ᐟ²].
Symmetrically opposite elements cancel their transverse field components, while axial components add. The result is an axial field; it does not describe every off-axis point.
At the centre, x = 0 and B = μ₀NI/(2R). Curl your right-hand fingers along conventional current; your thumb gives the axial field direction.
Far along the axis, |x| ≫ R, the magnitude approaches μ₀NIR²/(2|x|³). Reversing current reverses the field.
Go deeper: Reason it through
The field falls smoothly on either side of the centre. Multiplying by N assumes closely spaced turns with approximately the same radius and axis. The far-field expression is an approximation and cannot be substituted at the centre.
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Question
An original 20-turn air-core coil has radius 0.050 m and carries 0.50 A. Estimate its central field.
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Ampere's Circuital Law
Ampere's circuital law relates the integrated magnetic field around a closed loop to the electric current passing through the loop.Predict: why does the field rise inside the wire but fall outside?
A long wire has a = 0.10 m and uniform total I = 2 A. At r/a = 1.00, I_enclosed = 2.00 A and |B| = 4.00 μT. Inside, B ∝ r; outside, B ∝ 1/r. The graph joins continuously at a. Symmetry makes B tangential and constant on each circular Amperian loop.
- Amperian loop
- An imaginary closed path used to evaluate ∮B⃗ · dl⃗; it is not a physical wire.
For steady currents, ∮B⃗ · dl⃗ = μ₀Ienclosed. Count current with signs consistent with the loop orientation and its right-hand surface normal.
For a long straight wire, circular symmetry makes B tangent and constant on a circle of radius r, so B(2πr) = μ₀Ienclosed. Outside the wire, B = μ₀I/(2πr).
Inside a wire of radius a with uniform current density, Ienclosed = I r²/a². Therefore B = μ₀Ir/(2πa²): it rises linearly from zero and joins the outside result continuously at r = a.
Go deeper: Reason it through
The integral equation does not make B constant on an arbitrary path; symmetry is what allows B to leave the integral. Currents outside the surface can affect local B without contributing to Ienclosed. Time-varying electric fields require the Maxwell displacement-current extension.
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Question
An anticlockwise loop encloses 4 A out of the page and 1.5 A into the page. An extra 3 A wire lies outside. What is the signed enclosed current?
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The Solenoid
A solenoid is a long wire wound into a tight helix, used to generate a uniform magnetic field.Predict: what changes when turn density doubles at the same current?
With n = 500 turns/m and I = 2 A, the long-solenoid interior estimate is |B| ≈ μ₀nI = 1.257 mT. A finite coil has returning external field lines and fringing at its ends; outside B is not exactly zero. The drawn turns are illustrative, not a count of all turns. Reversing I reverses B.
- Turn density
- Number of turns per unit length, n = N/L, measured in m⁻¹.
A solenoid is a wire wound into a helix. A long closely wound air-core solenoid has an approximately uniform field in its interior away from the ends: B ≈ μ₀nI.
A rectangular Amperian loop includes an interior segment parallel to B. In the long-solenoid approximation, the exterior field is neglected and the transverse sides contribute zero, giving Bh = μ₀(nh)I.
A finite solenoid has fringing near the ends and returning field outside. Reverse current to reverse the field. A magnetic core changes the field response; the simple air-core formula cannot describe every core.
Go deeper: Reason it through
Uniform interior field and negligible exterior field are idealisations, not literal properties of every coil. Near either end of a finite coil the field is weaker and less uniform. Draw external return lines to preserve the continuous magnetic-field pattern.
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Question
An original air-core solenoid has 600 turns over 0.30 m and carries 0.80 A. Estimate the central field using the long-solenoid approximation.
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Force Between Parallel Currents
One long wire produces magnetic field at the other; that field exerts force on the second wire’s current. Same-direction parallel currents attract, and opposite-direction currents repel.Predict: what happens to the forces when one current reverses?
Two long wires carry |I₁| = 4 A and |I₂| = 6 A, with d = 0.20 m. |F|/L = μ₀|I₁I₂|/(2πd) = 24.0 μN/m. Same-direction currents attract; forces are equal and opposite. The wire-force ampere definition in the textbook is historical: today SI fixes the elementary charge. Here μ₀ ≈ 4π × 10⁻⁷ H/m is an approximation.
- Parallel-current force
- For long wires in vacuum, |F|/L = μ₀|I₁I₂|/(2πd).
One long wire produces magnetic field at the other; that field exerts force on the second wire’s current. Same-direction parallel currents attract, and opposite-direction currents repel.
The force magnitude per unit length is μ₀|I₁I₂|/(2πd). The two forces are equal and opposite in this ideal arrangement; doubling separation halves their magnitude.
The textbook’s two-wire force definition of the ampere is historical. Since 20 May 2019, the SI ampere is defined by fixing the elementary charge at exactly 1.602176634 × 10⁻¹⁹ C. Use μ₀ ≈ 4π × 10⁻⁷ H/m for these classroom calculations.
Go deeper: Reason it through
The force result remains useful after the unit definition changes. It assumes wires long compared with their separation and neglects end effects. The current rule differs from attraction/repulsion of stationary electric charges because force depends on both the field source current and the second current’s direction.
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Question
Two original long wires carry 3 A and 5 A in opposite directions, 0.10 m apart. Estimate their force per unit length and state its sense.
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Torque on a Current Loop
A current-carrying loop in a uniform magnetic field experiences a torque that tends to rotate it.Predict: which orientation gives maximum turning effect?
N = 20, I = 0.50 A, A = 0.010 m² and B = 0.20 T give m = NIA = 0.100 A·m². θ = 60° is measured from m (the area normal), not the coil’s plane. τ = mB sinθ = 0.0173 N·m; net force = 0 in uniform B. The side view shows two opposite forces: their lever arm vanishes at 0° and 180°. Aligned equilibrium is stable; anti-aligned is unstable.
- Magnetic dipole moment
- For a small N-turn plane coil, m⃗ = NIA n̂; n̂ follows the right-hand current rule.
A loop in a uniform magnetic field has zero net force but can have a force couple. Its torque is τ⃗ = m⃗ × B⃗, with magnitude NIAB sinθ.
The angle θ is measured between the area normal and B, not between the coil’s plane and B. Torque is maximum at 90° and zero at 0° or 180°.
Aligned m⃗ and B⃗ give stable equilibrium; anti-alignment gives unstable equilibrium. Far from a small loop on its axis, B ≈ (μ₀/4π)(2m/|x|³), with the appropriate axial direction.
Go deeper: Reason it through
Zero torque alone does not establish stability: test what a small rotation does. In a uniform field, opposite forces can cancel while their turning moments add. A nonuniform field can also exert net force on a dipole; that is a different condition.
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Question
An original 10-turn coil carries 0.40 A, has area 0.020 m² and sits in 0.30 T. Find maximum torque and the angle to its normal.
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The Moving Coil Galvanometer
The moving coil galvanometer (MCG) is an instrument used to detect and measure small electric currents.Predict: why must a shunt be parallel, while a voltmeter resistor is in series?
For N = 50, A = 10⁻³ m², B = 0.20 T and k = 10⁻⁶ N·m/rad, I = 20 μA gives φ = NABI/k = 0.20 rad = 11.46°. A radial field keeps the magnetic torque NIAB; the spring provides opposite kφ. This is the settled deflection, not a simulation of damping.
- Current sensitivity
- Equilibrium deflection per current: φ/I = NAB/k, where k is restoring torque per radian.
| Conversion Type | Component Added | Connection | Resulting Resistance |
|---|---|---|---|
| Ammeter | Low Resistance (Shunt) | Parallel | Very Low |
| Voltmeter | High Resistance | Series | Very High |
A moving coil lies in a radial magnetic field. Magnetic torque NIAB balances restoring torque kφ, so φ = (NAB/k)I. The linear relation describes settled deflection.
Current sensitivity is NAB/k. Voltage sensitivity is NAB/(kRg), where Rg is coil resistance. Adding turns can increase current sensitivity while also increasing resistance, so voltage sensitivity need not increase in the same proportion.
For an ammeter, add a low-resistance parallel shunt: Rs = IgRg/(Irange − Ig). Put the whole ammeter in series with the measured circuit. For a voltmeter, add a high series resistance: Rs = Vrange/Ig − Rg. Put the whole voltmeter across the measured component.
Go deeper: Reason it through
The radial field keeps the magnetic torque proportional to current throughout the useful deflection range. Meter loading still matters: an ammeter should add little series resistance, while a voltmeter should draw little current. Neither ideal zero ammeter resistance nor infinite voltmeter resistance is achieved by an ordinary meter.
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Question
An original galvanometer has Rg = 80 Ω and full-scale Ig = 100 μA. Find the shunt for a 20 mA ammeter and the added series resistance for a 4 V voltmeter.
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