Moving Conductors in a Magnetic Field
- Similar to a coil or a solenoid, a straight conducting rod moving through a magnetic field will also have an e.m.f induced in it
- This is only when the conductor moves perpendicular to the magnetic field lines where it cuts through the magnetic flux lines
Conducting rod moving perpendicular to a magnetic field directed into the page
- The conducting rod has a length L and moves through a uniform magnetic field with flux density B at a constant velocity v
- The distance travelled by the conductor is:
s = vΔt
- Where:
- s = distance travelled (m)
- v = velocity (m s-1)
- Δt = time interval (s)
- Therefore, the area A of the magnetic flux that it cuts through is:
A = LvΔt
- The total flux the conductor cuts through is:
ΔΦ = BA = BLvΔt
- Faraday's law gives the e.m.f induced:
- Where for the conductor, N = 1
- Substituting the change in magnetic flux ΔΦ into the e.m.f equation:
- Therefore, the induced e.m.f in the conductor as it moves through the magnetic field is:
ε = BLv
- This equation shows that the e.m.f induced increases if:
- A longer conductor is in the field
- The magnetic field strength is larger
- The conductor cuts through the field lines faster
Worked example
An aeroplane with a wingspan of 34.5 m flies at a speed 253 m s-1 perpendicular to the Earth's magnetic field. The Earth's magnetic field at the aeroplane's location has a strength of 0.06 mT.
Calculate the induced e.m.f between the wing tips.
Step 1: Write down the known quantities
- Length, L = 34.5 m
- Velocity, v = 253 m s-1
- Magnetic field strength, B = 0.06 mT = 0.06 × 10-3 T
Step 2: Write down the e.m.f equation
ε = BLv
Step 3: Substitute in the values
ε = (0.06 × 10-3) × 34.5 × 253 = 0.52371 = 0.52 V
Examiner Tip
Although calculations about a straight conductor moving through a magnetic field are common, the exam questions can be in other contexts.For example, the wing of an aeroplane moving through the Earth's magnetic field like in the worked example. If this is the case, treat the situation like the straight conductor and use the same equations