0
$\begingroup$

I've got a basic question on the method of computing the average induction voltage. Here is the task:

A conductor loop with the area $A_1 = 50\text{cm}^2$ in a magnetic field with the flux density of $B_1 = 0.2 \text{T}$ is downsized to $A_2 = 5\text{cm}^2$ and at the same time the flux density is reduced to $B_2 = 0.1\text{T}$. This all happens in $\Delta t = 0.1\text{s}$. Compute the average voltage in that loop.

My teacher and I have come up with two different way of computing it. I want to know which way is the right one.

Way 1:

$$U_{\text{ind}} = -\frac{\Delta \Phi}{\Delta t} = -\frac{\Phi_2-\Phi_1}{\Delta t} = -\frac{A_2B_2-A_1B_1}{\Delta t} = -0.0095 \text{V}$$

Way 2:

$$U_{ind} = -\frac{d\Phi}{dt} = -\left(\frac{dA}{dt}B+\frac{dB}{dt}A\right) = -\left(\frac{\Delta A}{\Delta t}B+\frac{\Delta B}{\Delta t}A\right) = -\left(\frac{A_2-A_1}{\Delta t}B_1+\frac{B_2-B_1}{\Delta t}A_1\right) = 0.014 \text{V}$$

Obviously these are not equal. My teacher said that the second one is correct. If he is right, why is the first way wrong?

$\endgroup$
1
$\begingroup$

Let's do this the slow and careful way.

Assuming that $B(t) = B_1 + b\cdot t$ and $A(t) = A_1 + a\cdot t$, then the instantaneous voltage change is

$$V = \frac{d}{dt}((B_1 + bt)(A_1 + at))\\ = aB_1 + bA_1 + 2abt$$

If we integrate over $t$, then divide by the interval, we get

$$V_{av} = aB_1 + bA_1 + ab\Delta t$$

You teacher is using $B_1$ and $A_1$ as though they last for the entire duration of the experiment - so his approach calculates the voltage at the start of the period, not the average over the entire experiment. The difference is that last term...

Putting it in the same notation as your original,

$$a = \frac{A_2-A_1}{\Delta t}\\ b = \frac{B_2-B_1}{\Delta t}$$

From which it follows that

$$V_{av} = \frac{A_2-A_1}{\Delta t}B_1+\frac{B_2-B_1}{\Delta t}A_1 + \frac{A_2-A_1}{\Delta t}\frac{B_2-B_1}{\Delta t}\Delta t$$

$$v_{av}\Delta t = A_2 B_1 -A_1 B_1 + B_2 A_1 - B_1 A_1 + A_2 B_2 -A_1 B_2 -A_2 B_1 + A_1 B_1\\ =A_2B_2 - A_1B_1$$

So in fact, it seems that you were right and your teacher was wrong.

$\endgroup$
  • $\begingroup$ $A$ and $B$ change in the same time. But if you consider them as functions of time (say $A(t)$ and $B(t)$) then wouldn't the average voltage just be $U_{av} = \frac{1}{0.1\text{s}}\int_{0\text{s}}^{0.1\text{s}}{U(t)}{dt} = \frac{1}{0.1\text{s}}\int_{0\text{s}}^{0.1\text{s}}{-\frac{d\Phi}{dt}}{dt}$ which is just the same as Way 1? $\endgroup$ – Andrei Kh Mar 12 '15 at 20:36
  • $\begingroup$ Yes - see my update. $\endgroup$ – Floris Mar 12 '15 at 20:40

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.