Force on a Current F = BIL, Circular Motion of Charges F = Bqv, Hall Probe & Cyclotrons
A magnetic field of flux density B (measured in Tesla, T = N A⁻¹ m⁻¹ = Wb m⁻²) exerts a magnetic force on any moving electric charge or current-carrying conductor that has a component perpendicular to the magnetic field lines. Using Fleming's Left-Hand Rule (where the Second finger always points in the direction of conventional current, the motion of positive charge, or opposite to electron velocity), the magnetic force on a wire of length L is F = BIL sin θ, while the force on a single moving charge q is F = Bqv sin θ. Because the magnetic force F = Bqv is always strictly perpendicular (at 90°) to the particle's instantaneous velocity v, a uniform magnetic field does zero work on a charged particle (speed v and kinetic energy stay constant) and acts purely as a centripetal force (Bqv = mv² / r), bending the particle into a circle of radius r = mv / (Bq) with a constant frequency f = Bq / (2πm) exploited in particle cyclotrons.
- Magnetic Flux Density (B), Fleming's Left-Hand Rule & F = BIL sin θ
- Magnetic flux density B (in Tesla, T) is defined by the force per unit length per unit current on a straight wire placed perpendicular (at 90°) to a uniform magnetic field: F = BIL sin θ, where θ is the angle between the wire and the magnetic field lines (F = BIL when θ = 90°; F = 0 when the wire is parallel to the field lines, θ = 0°). Fleming's Left-Hand Rule: Thumb = Force / Motion (F); First finger = Magnetic Field (B, North to South); Second finger = Conventional Current (I, positive to negative).
- Current Balance Experiment to Measure Magnetic Flux Density B
- A horizontal copper wire of known length L inside the magnetic field is clamped above a U-shaped magnet resting on a top-pan digital balance (zeroed with I = 0). When current I flows through the wire, a horizontal magnetic force acts vertically upwards on the wire, and by Newton's Third Law an equal and opposite downward force (F = mg = BIL) pushes down on the magnet, increasing the balance reading by mass m. Varying I and plotting F = mg (y-axis) against I (x-axis) gives a straight line through the origin of gradient = BL, so B = gradient / L.
- Force on a Moving Charge (F = Bqv) & Circular Trajectory (r = mv / Bq)
- For a particle of charge q and mass m moving at velocity v at angle θ to a magnetic field B: F = Bqv sin θ (for a negative electron, point the Second finger of Fleming's Left-Hand Rule opposite to the electron's direction of travel). When θ = 90°, the magnetic force F = Bqv is always perpendicular to velocity v, so: 1. Zero work is done by the magnetic field (W = Fd cos 90° = 0), so speed v and kinetic energy ½mv² remain constant. 2. The magnetic force acts as the centripetal force: Bqv = mv² / r → Radius of circular path r = mv / (Bq) = p / (Bq).
Electric Field Simulator & Interactive NPN Transistor Builder
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1. Coulomb’s Law Point-Charge Calculation
Formula: F = k · |Q₁ · Q₂| / r² where k = 1/(4πε₀) = 8.99 × 10⁹ N·m²·C⁻²
3.60 N
Attractive (towards each other)7.19e+6 N/C
At r/2 = 0.10 m2. Uniform Parallel-Plate Field Calculator
Formula: E = V / d (units: V m⁻¹ or N C⁻¹)
Key Concepts & Exam Definitions (6)
- Magnetic Flux Density (B), Fleming's Left-Hand Rule & F = BIL sin θ
- Magnetic flux density B (in Tesla, T) is defined by the force per unit length per unit current on a straight wire placed perpendicular (at 90°) to a uniform magnetic field: F = BIL sin θ, where θ is the angle between the wire and the magnetic field lines (F = BIL when θ = 90°; F = 0 when the wire is parallel to the field lines, θ = 0°). Fleming's Left-Hand Rule: Thumb = Force / Motion (F); First finger = Magnetic Field (B, North to South); Second finger = Conventional Current (I, positive to negative).
- Current Balance Experiment to Measure Magnetic Flux Density B
- A horizontal copper wire of known length L inside the magnetic field is clamped above a U-shaped magnet resting on a top-pan digital balance (zeroed with I = 0). When current I flows through the wire, a horizontal magnetic force acts vertically upwards on the wire, and by Newton's Third Law an equal and opposite downward force (F = mg = BIL) pushes down on the magnet, increasing the balance reading by mass m. Varying I and plotting F = mg (y-axis) against I (x-axis) gives a straight line through the origin of gradient = BL, so B = gradient / L.
- Force on a Moving Charge (F = Bqv) & Circular Trajectory (r = mv / Bq)
- For a particle of charge q and mass m moving at velocity v at angle θ to a magnetic field B: F = Bqv sin θ (for a negative electron, point the Second finger of Fleming's Left-Hand Rule opposite to the electron's direction of travel). When θ = 90°, the magnetic force F = Bqv is always perpendicular to velocity v, so: 1. Zero work is done by the magnetic field (W = Fd cos 90° = 0), so speed v and kinetic energy ½mv² remain constant. 2. The magnetic force acts as the centripetal force: Bqv = mv² / r → Radius of circular path r = mv / (Bq) = p / (Bq).
- Time Period & Cyclotron Frequency: T = 2πm / (Bq) and f = Bq / (2πm)
- Substituting circular speed v = 2πr / T into r = mv / (Bq) cancels both r and v completely: Time period for one full revolution T = 2πm / (Bq), and Cyclotron Frequency f = 1 / T = Bq / (2πm). Key exam fact: As long as the particle is non-relativistic (m is constant), the time period T and frequency f of circular motion in a uniform magnetic field are completely independent of the particle's speed v and radius r! A faster particle simply travels around a proportionally larger circle in the exact same time.
- The cyclotron particle accelerator
- Consists of two hollow semi-circular metal electrodes ('dees') inside a vacuum chamber, with a uniform magnetic field B applied perpendicular to the dees and a high-frequency alternating voltage (AC) applied across the narrow gap between the dees: 1. Inside the metal dees (which act as Faraday cages, E = 0), the uniform magnetic field B applies a perpendicular force F = Bqv that bends the protons/ions in a semi-circle at constant speed (does not accelerate/speed up the particles). 2. Every time the particles cross the narrow gap between the dees, the electric field E accelerates them (increasing their kinetic energy by ΔE_k = qV and increasing their radius r = mv / Bq). 3. Because time spent in one dee (½T = πm / Bq) is independent of speed v and radius r, the AC supply can stay at a fixed frequency f = Bq / (2πm) so the gap polarity reverses every half-period.
- Crossed Electric & Magnetic Fields (Velocity Selector v = E / B & Mass Spectrometer)
- In a velocity selector (Wien filter), a uniform electric field E (between parallel plates, E = V / d) and a uniform magnetic field B are applied at 90° to each other so that the electric force (F_E = qE) and magnetic force (F_B = Bqv) act in exact opposite directions on an ion beam: 1. Only ions with speed v where F_B = F_E pass straight through undeviated: Bqv = qE → v = E / B (independent of mass m and charge q!). 2. The velocity-selected ions (all having the exact same speed v = E/B) then enter a uniform magnetic field B₂ alone, where they follow semi-circular paths of radius r = mv / (B₂q) = mE / (B₁B₂q), separating isotopes strictly in proportion to their mass-to-charge ratio (m / q).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Why does a uniform magnetic field never change the speed or kinetic energy of a moving charged particle?
Derive the expressions for the radius r and time period T of a particle of mass m and charge q moving at speed v perpendicular to a uniform magnetic field B.
Explain the distinct roles of the magnetic field and the alternating electric field in a cyclotron.
Why can the alternating voltage applied across the dees of a cyclotron have a constant frequency even as the protons speed up and spiral outward?
In a velocity selector with perpendicular electric field E and magnetic field B, show that only particles with speed v = E / B pass through undeviated.
In a bubble chamber photograph with a magnetic field into the page, why does a charged particle leave an inward-spiralling track, and how can you tell its charge sign?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A straight horizontal copper rod of length 0.250 m and mass 18.0 g is suspended in mid-air inside a uniform horizontal magnetic field of flux density 0.480 T that acts perpendicular to the rod. Calculate the magnitude of the electric current required to keep the rod suspended in equilibrium without falling, state how Fleming's Left-Hand Rule determines the current direction, and explain what happens if the magnetic field is tilted so it makes an angle of 30.0° to the rod. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. For the rod to remain suspended in vertical equilibrium, the upward magnetic force must equal the downward weight of the rod: BIL = mg [1 mark]. 2. Rearrange for current I and substitute values (m = 0.0180 kg, g = 9.81 m s⁻²): I = mg ÷ (BL) = (0.0180 × 9.81) ÷ (0.480 × 0.250) = 0.1766 ÷ 0.120 = 1.47 A [1 mark]. 3. Using Fleming's Left-Hand Rule: point the Thumb vertically upwards (force F opposing gravity) and the First finger in the direction of the magnetic field B (North to South); the Second finger then points in the required direction of conventional current I [1 mark]. 4. If the magnetic field is tilted to 30.0° to the rod, the upward magnetic force decreases to F = BIL sin(30.0°) = 0.5 BIL [1 mark]. 5. Since the upward magnetic force is now half the weight of the rod, there is a net downward force of ½mg and the rod accelerates downwards at ½g = 4.91 m s⁻² (unless the current is doubled to 2.94 A) [1 mark].
- Q2
An electron (mass m = 9.11 × 10⁻³¹ kg, charge e = -1.60 × 10⁻¹⁹ C) is accelerated from rest in a vacuum through a potential difference of 2.50 kV and then enters a uniform magnetic field of flux density 8.00 mT at right angles to the field lines. Calculate: (a) the speed of the electron as it enters the magnetic field, (b) the radius of its circular path in the magnetic field, and (c) the time taken for the electron to complete one semi-circle in the field. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Equate work done by the accelerating p.d. (eV) to kinetic energy (½mv²): v = √(2eV ÷ m) = √(2 × 1.60 × 10⁻¹⁹ × 2500 ÷ 9.11 × 10⁻³¹) [1 mark]. 2. Calculate speed v: v = √(8.782 × 10¹⁴) = 2.96 × 10⁷ m s⁻¹ [1 mark]. 3. (b) Equate magnetic force to centripetal force (Bqv = mv² / r) to find radius r = mv ÷ (Be): r = (9.11 × 10⁻³¹ × 2.963 × 10⁷) ÷ (8.00 × 10⁻³ × 1.60 × 10⁻¹⁹) [1 mark]. 4. Calculate radius r: r = (2.699 × 10⁻²³) ÷ (1.28 × 10⁻²¹) = 0.0211 m = 21.1 mm [1 mark]. 5. (c) Time for one semi-circle t = ½T = πm ÷ (Be) (or πr ÷ v): t = (π × 9.11 × 10⁻³¹) ÷ (1.28 × 10⁻²¹) = 2.24 × 10⁻⁹ s (2.24 ns) [1 mark].
- Q3
A medical cyclotron accelerates protons (mass m = 1.67 × 10⁻²⁷ kg, charge q = +1.60 × 10⁻¹⁹ C) using a uniform magnetic field of flux density 1.50 T and dees of maximum radius 0.420 m. Calculate the frequency of the alternating voltage supply required and the maximum kinetic energy of the emerging protons in MeV. Explain why a fixed-frequency cyclotron cannot accelerate electrons or protons to speeds approaching the speed of light. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate the required AC frequency (equal to the proton cyclotron frequency): f = Bq ÷ (2πm) = (1.50 × 1.60 × 10⁻¹⁹) ÷ (2π × 1.67 × 10⁻²⁷) = 2.29 × 10⁷ Hz (22.9 MHz) [1 mark]. 2. From r_max = mv_max / (Bq), find maximum speed v_max = B q r_max ÷ m = (1.50 × 1.60 × 10⁻¹⁹ × 0.420) ÷ (1.67 × 10⁻²⁷) = 6.036 × 10⁷ m s⁻¹ [1 mark]. 3. Calculate maximum kinetic energy in Joules: E_k = ½m(v_max)² = 0.5 × 1.67 × 10⁻²⁷ × (6.036 × 10⁷)² = 3.042 × 10⁻¹² J [1 mark]. 4. Convert E_k to MeV (÷ 1.60 × 10⁻¹³ J MeV⁻¹): E_k = (3.042 × 10⁻¹²) ÷ (1.60 × 10⁻¹³) = 19.0 MeV [1 mark]. 5. At relativistic speeds approaching c, the relativistic mass (m) of the particle increases significantly, increasing the orbital period T = 2πm / (Bq) so the particles fall out of synchronisation (out of phase) with the fixed-frequency AC voltage across the gap [1 mark].
- Q4
Singly charged neon-20 ions (²⁰Ne⁺) and neon-22 ions (²²Ne⁺) pass through a velocity selector with parallel plates separated by 15.0 mm at a potential difference of 450 V and a perpendicular magnetic field of 0.120 T. The undeviated ions then enter a deflection chamber with a uniform magnetic field of 0.250 T and strike a photographic plate after completing a semi-circle (diameter 2r). Calculate the speed of the selected ions and the distance separating the ²⁰Ne⁺ and ²²Ne⁺ strike marks on the photographic plate. (1 u = 1.66 × 10⁻²⁷ kg). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Electric field strength in the velocity selector: E = V ÷ d = 450 V ÷ 0.0150 m = 3.00 × 10⁴ V m⁻¹ [1 mark]. 2. Speed of selected ions: v = E ÷ B₁ = (3.00 × 10⁴) ÷ 0.120 = 2.50 × 10⁵ m s⁻¹ [1 mark]. 3. In the deflection chamber (B₂ = 0.250 T), each ion travels a semi-circle of diameter D = 2r = 2mv ÷ (B₂q) [1 mark]. 4. Difference in mass Δm between ²²Ne⁺ and ²⁰Ne⁺ is 2 u = 2 × 1.66 × 10⁻²⁷ = 3.32 × 10⁻²⁷ kg [1 mark]. 5. Separation of strike marks ΔD = 2(Δm)v ÷ (B₂q) = (2 × 3.32 × 10⁻²⁷ × 2.50 × 10⁵) ÷ (0.250 × 1.60 × 10⁻¹⁹) = 0.0415 m = 41.5 mm [1 mark].
Common Exam Questions (4)
How do I avoid getting the direction backwards with Fleming's Left-Hand Rule when the particle is an electron or beta-minus particle?
Remember that the Second finger ('Current') always represents the direction of conventional current (flow of positive charge). Because an electron is negatively charged, if an electron is moving to the right, you must point your Second finger to the left (opposite to the electron's velocity).
How does a Hall Probe measure magnetic flux density B, and why must the flat face of the semiconductor slice be held perpendicular (at 90°) to the magnetic field lines?
When current flows through a thin semiconductor wafer placed perpendicular to a magnetic field B, the magnetic force F = Bqv deflects charge carriers to one side of the wafer until the transverse electric force (qE_H = qV_H / d) balances Bqv, producing a Hall Voltage V_H ∝ B sin θ. Holding the flat face at 90° to the field lines ensures sin(90°) = 1, giving the maximum reading V_H.
Why do two parallel wires carrying currents in the same direction attract each other, whereas currents in opposite directions repel?
By the Right-Hand Grip Rule, Wire 1 creates concentric magnetic field lines at the position of Wire 2. Applying Fleming's Left-Hand Rule to the current in Wire 2 inside Wire 1's magnetic field shows that parallel currents experience an inward attractive force toward each other, whereas antiparallel currents experience an outward repulsive force.
What is the relationship between the radius of curvature r of an alpha particle (⁴₂α²⁺) and a proton (¹₁p⁺) entering the same magnetic field B at the same speed v?
An alpha particle has 4 times the mass (4m) and 2 times the charge (2q) of a proton. Substituting into r = mv / (Bq) gives r_α = (4m)v / (B(2q)) = (4/2) × (mv / Bq) = 2 r_p (the alpha particle follows a circle of twice the radius of the proton).
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