Induction of emf in moving conductors in magnetic fields

[tzxn3]

In preparation for an upcoming physics exam I have been going through some past paper questions. This question is from the multiple choice section of the paper, so while I can check the answer there is no source from which I can see it being worked through.

Homework Statement

An aircraft of wing span 60m flies horizontally at a speed of 150 ms^-1. If the vertical component of the Earth's magnetic field is 1.0 * 10^-5 T, what emf is induced across the wing tips of the plane?

radius of Earth = 6.37 * 10^6 m

Homework Equations

F = Bqv

F = BIlsin(\theta)

\Phi = BAcos(\theta)

\epsilon = N \frac{Δ\Phi}{Δt}

\epsilon = BANωsin(ωt)

The Attempt at a Solution

I know the length of the conductor, its velocity, and the relevant magnetic flux density.

If F = Bqv and F = BIl, then \frac{Il}{qv} = 1

as q = It, \frac{l}{tv} = 1, so t = \frac{l}{v}

substituting in the values given for v and l gives t as 0.4s

This seems intuitively wrong, but I can't think of anything else to do.

I also know the radius of the Earth, so I can find an angular velocity for the aircraft. This comes out as about 2.35 * 10^{-5} s^{-1}.

I'm not sure what to do about the area term in the equations.

Anyone care to give some advice?

 

[maCrobo]

Hi!
You can deal with this kind of problems in two ways: using lorentz' Law or Faraday's Law. For Faraday's Law you should keep in mind that you need a closed circuit, so even if you don't practically have one you can imagine a closed path that suit your calculations.

Just to follow you reasoning: first draw the wings from the above and use them as one of the 4 edges of a rectangle so to have a rectangle with 2 edges of length 60m (the length of the other 2 edges will change with time because of the airplane motion). So you have your path, a constant magnetic field and enough information to describe how the area inside the rectangle changes with time.

 

[Spinnor]

In equilibrium,

F = qvB - qE = 0 ?

E = V/L where L is the length of the wing and V is the potential difference between the wing tips?

 

[technician]

Microbo has the right idea
Induced emf = rate of change of magnetic flux
You need to find the area swept through per second

 

[maCrobo]

Well I wrote down how to solve it, because I understand it may be difficult to see how the picture should be done if you have never thought about it before, but I haven't posted it because I want to focus on lorentz' law as I found it easier to explain what physically happen. Then you should check that is the same by using Faraday's Law.

As for the lorentz' force. Think in this way: you have a bar of length 60m full of charges; the bar is moving with a speed v, so charges move in that direction and you have F=qv x B (vector form). You know that the electromotive force is nothing than a voltage that makes electrons move: how do you evaluate the voltage when you have the force?

 

[tzxn3]

Sorry, I think all this is slightly over my head.

Lorentz's law is F = Bqv - qE, so how do I find E when I don't know q?

Is E = V / L correct? Not sure I've seen that before.

Using Faraday's law relies on knowing the area cutting the flux, right? How do I calculate that area when I only have a length?

What am I trying to model the airplane as, anyway? A big square or a bar of conducting material?

 

[maCrobo]

First: on the airplane there's no external E applied so F reduces to F= qvB. That's to say if you have a certain charge that moves in a magnetic field there will be a force applied on it due to its motion in the magnetic field.

Second: Fx=W=- delta U; U=Vq. So (focusing just on magnitudes) Fx=(delta V)q, but F=Eq. So, Eqx=(delta V)q, then you have an equation that doesn't care about q: Ex=deltaV.

Then you should apply all that to your problem. First you should verify the direction of this force and you'll see it is parallel to the length of the wings; then keep in mind that charges move just inside the metallic material (the wings in this case). So you have the length x. Then you find deltaV that is exactly the same of the electromotive force.

 

[Spinnor]

See page 8 of,

http://200.105.152.242/olimpiada/file.php/1/LIBROS_OLIMPIADAS/FISICA SERWAY/31-FaradaysLaw.pdf

 

[tzxn3]

Oh, brilliant, thanks a lot everyone!

Not sure how I was supposed to calculate it within the confines of the syllabus (my teacher's good but I don't get a lot of time with him, while the textbooks I have are frankly rubbish), but at least I've learned something new.

Just worked through it:

x = 60m, v = 150ms^-1, B = 1.0 * 10^-5 T

Force, velocity and length are all perpendicular to the field, so no need for resolving vectors, thankfully.

F = qvB, E = F / q
∴ E = vB = 150ms^-1 * 1.0 * 10^-5 T = 1.5 * 10^-3 V m^-1

W = Fx = Eqx = -ΔU = -ΔVq
∴ Ex = ΔV = ε (direction doesn't matter)
∴ ε = 1.5 * 10^-3 V m^-1 * 60m = 0.090 V

This was the correct answer to the question, if I recall correctly. Thanks again for all your help.