Flashcards · 9630 · A Level Topic 10

Magnetic fields (A Level) flashcards

31 cards on A Level Topic 10, subtopics 3.10.1 to 3.10.6. Pick a subtopic or study them all.

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Force on a current-carrying wire in a magnetic field: F = BIL, with the wire at right angles to the field and L the length of wire inside the field.
Magnetic flux density B, and the tesla: B is the force per metre of conductor per ampere of current, with the conductor at right angles to the field. 1 T is the flux density that gives a force of 1 N on 1 m of wire carrying 1 A across the field.
The tesla in base units: 1 T = 1 N A⁻¹ m⁻¹ = 1 kg s⁻² A⁻¹.
Fleming's left-hand rule: Thumb: force (motion). First finger: field (north to south). Second finger: conventional current.
Measuring B with a current balance: Zero the balance, pass current I and record the mass change Δm, so F = Δm g. Plot Δm against I: a straight line through the origin with gradient BL/g, so B = gradient × g/L.

Force on a charged particle moving across a magnetic field: F = BQv, perpendicular to both the field and the velocity. For a negative charge, point the second finger opposite to the velocity.
Radius of the circular path of a charged particle in a uniform magnetic field: BQv = mv²/r, so r = mv/(BQ).
Why does a magnetic field not change a charged particle's speed?: The force is always perpendicular to the velocity, so it does no work and the kinetic energy stays constant; only the direction changes.
Period of a charged particle's orbit in a magnetic field: T = 2πm/(BQ), independent of the speed (frequency f = BQ/(2πm)).
How does a cyclotron accelerate particles?: Two dees in a vacuum with a magnetic field across them. An alternating p.d. across the gap accelerates the particle each crossing; inside a dee the field bends it in a semicircle at constant speed. The radius grows, so the path spirals outwards.
Why can a cyclotron use a fixed-frequency alternating p.d.?: The time for each semicircle, πm/(BQ), does not depend on speed (at speeds well below the speed of light), so the p.d. frequency f = BQ/(2πm) stays in step.

Magnetic flux: Φ = BA, with B perpendicular to the area A. Unit weber, Wb (1 Wb = 1 T m²).
Flux linkage: NΦ for a coil of N turns. Unit Wb (often written Wb-turns).
Flux linkage of a coil at an angle to the field: NΦ = BAN cos θ, where θ is the angle between B and the normal to the coil.
A 30-turn coil, 0.12 m × 0.090 m, normal along a 0.35 T field. Flux linkage?: A = 1.08 × 10⁻² m², Φ = 3.78 × 10⁻³ Wb, NΦ = 30 × 3.78 × 10⁻³ = 0.113 Wb.

Faraday's law: The magnitude of the induced emf equals the rate of change of flux linkage: ε = NΔΦ/Δt.
Lenz's law: The induced emf (and current) acts in a direction to oppose the change in flux linkage that causes it.
emf induced in a straight conductor moving across a field: ε = BLv, with B, L and v mutually perpendicular.
emf of a coil rotating in a uniform field: ε = BANω sin ωt, with peak value BANω (ω = 2πf). The emf is maximum when the plane of the coil is parallel to B (flux linkage zero) and zero when it is perpendicular.
Eddy currents: what they are and how they are reduced: Circulating currents induced in bulk metal when the flux through it changes. They oppose the change, heat the metal and damp motion. Laminating the metal into thin insulated sheets reduces them.
How is the emf found from a graph of flux linkage against time?: The emf is the gradient of the flux linkage–time graph. A flat section gives zero emf however large the flux linkage.

Root mean square (rms) value of an alternating current or p.d.: The steady d.c. value that would give the same mean power in a given resistor.
rms, peak and peak-to-peak for a sinusoidal a.c.: Irms = I₀/√2 and Vrms = V₀/√2. Peak-to-peak = 2 × peak.
Mean power in a resistor on a.c.: P = IrmsVrms = Irms²R = Vrms²/R, using rms values.
Reading an oscilloscope trace: Voltage = vertical divisions × Y-gain (V per division). Period = horizontal divisions for one cycle × time-base (time per division). f = 1/T.
A sinusoidal supply has peak voltage 25 V. rms and peak-to-peak values?: Vrms = 25/√2 = 17.7 V. Peak-to-peak = 50 V.

Transformer equation (ideal): Ns/Np = Vs/Vp. Step-up: Ns > Np. Step-down: Ns < Np.
Transformer efficiency: Efficiency = output power ÷ input power = IsVs/(IpVp).
Causes of transformer energy losses and how each is reduced: Coil resistance (I²R heating): thick copper wire. Eddy currents in the core: laminated core. Hysteresis: soft iron core. Flux leakage: coils wound close together on a closed core.
Np = 500, Ns = 2000, primary 12 V. Secondary voltage?: Vs = 12 × 2000/500 = 48 V.
Why is electricity transmitted at high voltage?: For a given power P = IV, a higher voltage means a smaller current, so the line loss I²R is much smaller. Raising V tenfold cuts the loss by a factor of 100.