Moving frame in a magnetic field

1. Aug 16, 2015

kaspis245

1. The problem statement, all variables and given/known data
A quadratic frame which has an edge of $a=10 cm$, was put inside a magnetic field $B=0.01T$. The frame's opposite vertices were being pulled with a uniform speed $v=1mm/s$ until the frame became flat. Find the charge that passed through the frame. The plane of the frame remained perpendicular to the magnetic field all the time, the resistance of the frame equals $R=5Ω$.

2. Relevant equations
$ε=vLB$
$I=\frac{q}{t}$
$ε=IR$

3. The attempt at a solution
Here's my try:

Clearly the answer I get is incorrect, because $v$ does not appear in the final formula. How should I approach this problem?

2. Aug 16, 2015

Staff: Mentor

I wouldn't v to appear in the final formula. Closing it faster increases the current but decreases time in the same way. Also, I don't think there is a way to get the units right with speed in the final answer.
You can insert the known value for the angle to simplify the final answer a bit.

How did you find I? I would expect that you have to calculate some derivative for it.
The vertical velocity indicated in the sketch starts with v, but it changes over time.

3. Aug 16, 2015

rude man

In line with mfb I not only agree that total charge Q is not a function of v but that that fact can be used to get the answer in a jiffy!

4. Aug 17, 2015

kaspis245

I must use $v$ since it is given in my problem. There has to be some use of it.

I used this formula to find I:
$I=\frac{ε}{\frac{R}{4}}$

I see that $v$ direction changes over time so this means that $ε$ also changes. I don't know which velocity I should be interested in: vertical or horizontal?

I've denoted in my drawing that the top vertex of the triangle will be moving downwards with a uniform velocity $v$. Are you saying that this velocity will change over time?

Last edited: Aug 17, 2015
5. Aug 17, 2015

BvU

It indeed has a function in this exercise: to make you understand it doesn't play a role. To distract you, so to say.

Note that the exercise doesn't ask for the current, but only for the charge. Now $q = \int i\; dt$ and $i$ does depend on v, but the integral only on the integral of v.

6. Aug 17, 2015

Staff: Mentor

The horizontal speed is constant but the vertical one is not. It is possible to find the current as function of time and integrate it, but it is not necessary. This will become easier to see once you find the right formula for the current. Think about the area...

7. Aug 17, 2015

rude man

The instructor ORDERED you to use it? Merkwuerdig, as they say in Germany and in the movie Dr. Strangelove ...
Current and v are not parameters to be concerned with. Just Farady's law, integrated.
BvU's post #5 is right-on!

8. Aug 18, 2015

kaspis245

Ah, math is my weak spot. Please check if this is correct:

$ε=\frac{dΦ}{dt}$
$Φ=\int B⋅dA=B⋅\int dA=B⋅\int_{π/4}^{π/2} \frac{(acosθ)^2}{2}=-\frac{Ba^2}{2}$
$q=\frac{16Φ}{R}=\frac{8Ba^2}{R}$

9. Aug 18, 2015

rude man

good.
bad. What is θ? There's no need for any θ. And where is the differential? Every integral has to have a differential.
I can only repeat my previous hint: integrate faraday's law: emf = -dΦ/dt.
You should seriously review your basic calculus. You're too far into physics not to have the necessary math tools at hand.

10. Aug 18, 2015

kaspis245

Ok, let's do it in small steps. Is it correct:

$ε=-\frac{dΦ}{dt}=-B\frac{dA}{dt}$

11. Aug 18, 2015

Staff: Mentor

That is a good start.

12. Aug 19, 2015

kaspis245

Then, I need to find $\frac{dA}{dt}$.

$x=\frac{a}{\sqrt{2}}$
From the triangle, I can say that it's area $A=\frac{(x-vt)(x+vt)}{2}=\frac{x^2-(vt)^2}{2}=\frac{a^2-2(vt)^2}{4}$.

Then, $\frac{dA}{dt}=\frac{d}{dt}[\frac{a^2-2(vt)^2}{4}]=-v^2t$

Therefore, $ε=Bv^2t$

$\frac{q}{t}=\frac{4Bv^2t}{\frac{R}{4}}=\frac{16Bv^2t}{R}$

$q=\frac{16Bv^2t^2}{R}$

$t=\frac{\frac{a}{\sqrt{2}}}{v}$

$q=\frac{8Ba^2}{R}$

13. Aug 19, 2015

Vibhor

Hello kaspis245

I might be wrong but I think your analysis is not correct . The relation $x=\frac{a}{\sqrt{2}}$ doesn't seem reasonable.

I believe the answer should be $q=\frac{Ba^2}{R}$

14. Aug 19, 2015

Staff: Mentor

That is possible, but not necessary. It is better to integrate over time as early as possible.

In the sketch you fix the two shorter sides to have the same length. This is true in the initial setup but won't stay true. This is also the main reason your earlier analysis didn't give the right result.

15. Aug 22, 2015

Vibhor

@rude man ,@mfb

Do you agree that $q=\frac{Ba^2}{R}$ should be the answer ?

16. Aug 22, 2015

Staff: Mentor

It is kaspis245's homework, he has to find the answer.