Magnetic flux density and force on a conductorAQA A-Level Physics: Revision notes
Section 1
Force on a current-carrying wire
A current-carrying conductor in a magnetic field experiences a force because the moving charges in it experience a force. When the current is at right angles to a uniform field, the force is
F = BIl
where B is the magnetic flux density (T), I the current (A) and l the length of conductor in the field (m). The force is zero if the wire is parallel to the field and is largest when they are at right angles. The force is proportional to each of B, I and l.
Section 2
The direction: Fleming's left-hand rule
The force is perpendicular to both the field and the current. Hold the thumb, first finger and second finger of the left hand at right angles:
- First finger: Field (north to south)
- seCond finger: Conventional current (positive to negative)
- Thumb: Thrust (force/motion)
Reversing the current or the field reverses the force; reversing both leaves it unchanged.
Using the right hand, or using electron flow instead of conventional current, reverses the answer.
Section 3
Defining the tesla
Rearranging F = BIl gives B = F/(Il). The tesla is the flux density of a uniform field that produces a force of 1 newton on a 1 metre length of conductor carrying a current of 1 ampere at right angles to the field.
So 1 T = 1 N A⁻¹ m⁻¹.
Section 4
Worked example
A 0.12 m length of wire carries 3.5 A at right angles to a 0.40 T field.
F = BIl = 0.40 × 3.5 × 0.12 = 0.17 N.
If the wire is to be held in equilibrium against its weight, set BIl = mg and solve for the unknown.
Always convert length to metres and mass to kilograms before substituting.
Section 5
Required practical 10: force on a wire using a top pan balance
Place a magnet assembly (a pair of magnets on an iron yoke) on a top pan balance, with a rigid wire clamped so a known length lies at right angles to the field in the gap and does not touch the magnets. Zero the balance, then pass a current.
The wire is pushed by the field, so by Newton's third law the magnets are pushed with an equal and opposite force and the reading changes by F/g.
- Vary current: use a variable resistor, read the ammeter; reading ∝ I.
- Vary length: use wires or magnet arrangements giving different lengths in the field; reading ∝ l.
- Vary flux density: add more magnets (stack identical pairs); reading ∝ B.
Plot reading against the varied quantity: a straight line through the origin confirms proportionality, and the gradient gives the unknown (for current: gradient = Bl/g). Keep the other two variables constant, and zero the balance each time.
The balance reads mass: multiply the change in reading (kg) by g to get the force in newtons.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Magnetic flux density and force on a conductor
- A technician places a straight horizontal copper rod between the poles of a magnet. The rod carries a conventional current of 6.0 A towards the north. The magnet provides a uniform vertical magnetic field, directed downwards, with flux density 0.24 T over a 0.050 m length of the rod; the rod is at right angles to the field.The current is halved and the length of rod in the field is doubled, with the flux density unchanged. Deduce the new force on the rod and its direction.2 marks
- A student investigates the force on a wire using a top pan balance. A magnet assembly rests on the pan, with a rigid horizontal wire clamped so that 0.040 m of it lies at right angles to the uniform field between the poles, without touching the magnet. The balance is set to zero with no current. When a current of 2.5 A is switched on, the balance reading changes. The flux density between the poles is 0.095 T. Take g = 9.81 N kg⁻¹.The student repeats the experiment for several currents with the length of wire in the field fixed, and plots the balance reading in kilograms against current. State the shape of the graph and explain how the flux density can be found from it.2 marks
- A straight horizontal conductor of length 0.15 m and mass 8.0 g hangs from two very flexible leads in a uniform horizontal magnetic field of flux density 0.060 T, which is at right angles to the conductor. Current in the conductor is arranged so that the magnetic force on it acts vertically upwards. Take g = 9.81 N kg⁻¹.Calculate the current required for the magnetic force on the conductor to balance its weight.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).