Determining which wing tip is positive in an aeroplane's induced EMF

[kingstar]

Homework Statement

An aeroplane in horizontal flight has a wing span of 50 m and is traveling due north at a speed of
500 kmph. Calculate the emf induced between the wing tips of the aeroplane.
The value of the vertical component of the Earth’s magnetic field is 5.5 x 10^-5 T.

What additional information would be needed to determine whether the left wing tip was positive
or negative?

Homework Equations

ε = BvL

The Attempt at a Solution

500 kmph = 500,000/3600 = 138.89 ms^-1

ε = 5.5 x 10^-5 T x 50m x 138.89 ms^-1
ε = 0.38194V

I hope that's right? lol

Anyways I'm confused about the second part 'What additional information would be needed to determine whether the left wing tip was positive or negative?'

 

[technician]

do you know the right hand rule, used to determine the direction of any induced current (or emf)

 

[kingstar]

Yeah, but I don't see how this would help determine if the left wing tip was positive or negative :S

 

[gneill]

Yeah, but I don't see how this would help determine if the left wing tip was positive or negative :S

Consider a typical conduction electron in the metal of the plane; it's being carried through the magnetic field along with the plane. So, electron moving through magnetic field...

 

[technician]

you are given the vertical component of the Earth's field...Do you know its direction?

 

[rude man]

To determine emf direction, think of a unit positive charge along the wing. The force on it is F = q v x B so you need to know direction of v and B to determine the direction of F on the charge.

The emf is simply the force on a unit charge times length L, or F*L which is work done on the unit charge. So for example if the force is from left wing to right wing then the left wing is - and the right wing is +. In vector notation, emf = ( v x B ) * L. So L defines the direction (and sign) of the emf. In the above example, L points to the right wing and so emf is +.