1
$\begingroup$

Today we started learning about the electromagnetic induction. Out teacher gave us the following explanation:

Suppose we have a conduction frame inside a magnetic field $\vec{B}$ going towards the page. That frame has a stationary edge (also a conductor), which we will call $AB$. $AB$ is pushed right with a starting velocity $\vec{v}$. Lets say $AB$ moves at a constant speed. As a result of the edge's movement, Lorentz force acts on it's inside charges. It's direction on the positive charges is up (towrads $A$), and on the negative charges, down (towards $B$). This seperation creates an electric potencial $V_{AB}$, which will cause an electric force $\vec{F_E}$ to act on the edge's charges as well. The situation will balance when:

$$F_B = F_E$$.

Thus

$$qvB=qE$$

We know that $E=\frac{V_{AB}}{d}$, where $d$ is the length of $AB$. Thus:

$$qvB=q\frac{V_{AB}}{d}$$

Thus

$$V_{AB}=vBd$$

This represents the voltage created between the two sides of the edge, which moves inside a magnetic filed. It will, obviously, create a current in the circuit, as follow:

$$I=\frac{V_{AB}}{R}=\frac{vBd}{R}\quad (1)$$

Where $R$ is the resistance of the circuit.

From the last equation, we can learn that a longer edge $d$, will produce a larger current $I$. So far so good.

But, if we dig down in the definition of the resistance of a wire, we will remember that:

$$R = \rho\frac{d}{A} \quad (2)$$

Where $d$ is the length of the conductor, $A$ is the cross-sectional area of the conductor and $\rho$ is the electrical resistivity. The last formula states that there is a direct relation between $R$ and $d$. If we combine $(1)$ and $(2)$, we produce the follow:

$$I=\frac{vBd}{\rho\frac{d}{A}}=\frac{vBA}{\rho}$$

Wait - the last formula shows us that the current $I$ has no relation to the length of the wire $d$ - but how is that possible? Moving a million metre wire and a one meter wire inside a magnetic field will produce the same current? My teacher said that isn't true, but we had no time to find the error, if such one trully exist.

Thanks in advance :)

P.S - i'm just a highschool student, so please use math as simple as possible to answer. Thanks.

$\endgroup$
  • $\begingroup$ The length enters two ways: in the size of the motional EMF and in the resistance of the wire. Now ask yourself how each of these things scales with length and how current scales with the resistance and EMF. $\endgroup$ – dmckee --- ex-moderator kitten Apr 26 '15 at 17:09
1
$\begingroup$

You need to specify what is kept constant in your problem.

What you want to compare is two systems:

  • 1) one with potential difference $V_{AB}$, length $d_1$ and $R_1 = \rho d_1/A$

  • 2) another with the same potential difference $V_{AB}$, length $d_2$ and $R_2 = \rho d_2/A$

In such a scenario, you will find that the intensity will of course depend on the length of the circuit.

Now, in the particular scenario you have in mind where one drags a frame in a magnetic field, that's different because the induced potential difference depends on the size of the frame. The thing that you keep constant is actually the magnetic field imposed to the system and not the potential difference.

Moreover, the current created by moving a conducting material in a magnetic field is only reacting to local things, which is directly the electric field that is a potential gradient and not an actual potential difference; so if the conditions are locally exactly the same for every bit of the wire, then the length doesn't matter.

Finally, a crucial point in this thought experiment is that the magnetic field is uniform. If you have a two hundred meters wire or a one centimetre one, the actual energy to spend to sustain a uniform magnetic field is not at all the same.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I dont understand. Which two systems are you talking about? There is one frame, in which we move an edge onto it. We dont drag the frame in the magnetic field, but rather an edge on top of the frame.And yes, I forgot to mention that the magnetic field is uniform. $\endgroup$ – Eminem Apr 26 '15 at 18:52
1
$\begingroup$

Right, but you are not getting something for nothing. What about the energy losses? These scale as $I^2 R$ and so does depend on the length (and area) of the wire.

$$ W = I^2 R = \frac{v^2 B^2 A^2}{\rho^2} \frac{\rho d}{A} = \frac{v^2 B^2 A d}{\rho}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ True. But if the net force is 0, then there is no work being done, then there is no power as well, right? $\endgroup$ – Eminem Apr 26 '15 at 19:02
  • 2
    $\begingroup$ @Eminem How do you keep the velocity of the wire constant? $\endgroup$ – Rob Jeffries Apr 26 '15 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.