Magnetic fields and forces flashcards

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Magnetic field
A region in which a moving charge, a current-carrying conductor or a magnetic material experiences a force.
Magnetic flux density B
The force per unit current per unit length on a conductor placed at right angles to the field. Unit: tesla, 1 T = 1 N A⁻¹ m⁻¹.
Magnetic flux φ
φ = BA, the flux density × the area at right angles to the field. Unit: weber, 1 Wb = 1 T m².
Flux linkage
Nφ: the flux through a coil multiplied by its number of turns. Unit: Wb turns.
The motor effect
A current-carrying conductor in a magnetic field (not parallel to it) experiences a force at right angles to both the current and the field.
Fleming's left-hand rule
First finger = Field (N to S), seCond finger = Current (conventional), thuMb = Motion (force); all at right angles.
Right-hand grip rule
Thumb along the conventional current; the curled fingers give the direction of the magnetic field lines (or, for a coil, fingers with the current and thumb towards the N end).
Equation for magnetic flux through an area
φ = BA (A at right angles to B). (No — learn it)
Equation for the flux linkage of a coil
Nφ = NBA. (No — learn it)
Flux through a tilted coil
φ = BA cos θ, θ = angle between the field and the normal to the coil. (No — learn it)
Force on a current-carrying conductor
F = BIl sin θ, θ = angle between the wire and the field; l = length in the field. (Yes)
Force on a moving charge
F = Bqv sin θ, θ = angle between the velocity and the field. (Yes)
Radius of a charged particle's path in a uniform field
r = p/BQ = mv/BQ. (Yes)
Deriving the radius of the circular path
Bqv = mv²/r (magnetic force = centripetal force), so r = mv/BQ. (No — derive it)
Field pattern of a long straight wire
Concentric circles centred on the wire, further apart with distance; direction from the right-hand grip rule.
Field pattern of a solenoid
Inside: parallel, equally spaced lines (uniform field). Outside: like a bar magnet, lines leaving the N end and looping to the S end.
Field pattern of a flat circular coil
Lines loop round each side of the coil; near the centre they run almost straight along the axis.
Combined (catapult) field of a wire between two poles
The fields add on one side of the wire (crowded lines) and oppose on the other (sparse lines, neutral point); the wire is pushed towards the weak side.
Current balance graph
Change in balance reading Δm against current I: straight line through the origin with gradient Bl/g.
Symbols × and • for fields and currents
× = into the page (tail of an arrow); • = out of the page (point of an arrow).
Why does a magnetic field not change a charged particle's speed?
The force is always perpendicular to the velocity, so no work is done and the kinetic energy stays constant.
Why does a charged particle move in a circle in a uniform magnetic field?
The force has constant size and is always perpendicular to the velocity, so it acts as a centripetal force.
Current balance: why does the reading increase when the wire is pushed up?
Newton's third law: the wire pushes the magnet down with an equal and opposite force.
Why do two parallel wires with currents in the same direction attract?
Each wire is a current in the other's field; the grip rule then Fleming's left-hand rule give a force towards the other wire.
Why does a loudspeaker cone vibrate on an alternating current?
The current reverses, so the force on the coil in the magnet's field reverses with the same frequency.
What does a detector track that spirals inwards tell you?
r = p/BQ: the radius shrinks because the particle loses momentum (energy); it travels towards the centre of the spiral.
Square coil, side 4.0 cm, 250 turns, face-on to 0.15 T. Flux linkage?
A = 1.6 × 10⁻³ m²; φ = 2.4 × 10⁻⁴ Wb; Nφ = 0.060 Wb turns.
0.30 m of wire, 4.0 A, at 35° to a 0.080 T field. Force?
F = BIl sin θ = 0.080 × 4.0 × 0.30 × sin 35° = 0.055 N.
An 18 g rod with 0.12 m in a 0.30 T field hangs on flexible leads. Current for zero tension?
BIl = mg → I = (0.018 × 9.81) ÷ (0.30 × 0.12) = 4.9 A.
Electron at 3.0 × 10⁷ m s⁻¹ perpendicular to 2.0 mT. Force?
F = Bqv = 2.0 × 10⁻³ × 1.60 × 10⁻¹⁹ × 3.0 × 10⁷ = 9.6 × 10⁻¹⁵ N.
Proton at 2.2 × 10⁶ m s⁻¹ perpendicular to 0.25 T. Radius of path?
r = mv/BQ = (1.67 × 10⁻²⁷ × 2.2 × 10⁶) ÷ (0.25 × 1.60 × 10⁻¹⁹) = 0.092 m.
Current balance: gradient of Δm against I = 0.325 g A⁻¹, l = 5.0 cm. B?
Bl = 0.325 × 10⁻³ × 9.81 = 3.19 × 10⁻³ N A⁻¹; B = 3.19 × 10⁻³ ÷ 0.050 = 0.064 T.
Crossed fields: E = 6.0 × 10⁴ V m⁻¹, B = 0.050 T. Speed passing undeflected?
EQ = BQv → v = E/B = 1.2 × 10⁶ m s⁻¹ (independent of charge and mass).
Direction of the force on an electron
Point the second finger opposite to the electron's velocity (conventional current), so the force is opposite to that on a proton.
Which angle goes in F = BIl sin θ and φ = BA cos θ?
Force: angle between the wire (or velocity) and the field. Flux: angle between the field and the normal to the coil.
Using a current balance reading in F = BIl
Convert the change in reading from g to kg and multiply by 9.81 to get the force in N.
Two particles with the same kinetic energy in the same field
Use p = √(2mEₖ): r ∝ √m/Q, not m/Q.

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