Moving frame in a magnetic field

[kaspis245]

Homework Statement

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 $R=5Ω$.

Homework Equations

$ε=vLB$
$I=\frac{q}{t}$
$ε=IR$

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?

 

[mfb]

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.

 

[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!

 

[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?

The vertical velocity indicated in the sketch starts with v, but it changes over time.

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?

 

[BvU]

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

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.

 

[mfb]

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...

 

[rude man]

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

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

 

[kaspis245]

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

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

 

[rude man]

The instructor ORDERED you to use it? Merkwuerdig, as they say in Germany and in the movie Dr. Strangelove ...
This problem is easily solvable in your head.
Current and v are not parameters to be concerned with. Just Farady's law, integrated.

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

$ε=\frac{dΦ}{dt}$

good.

$Φ=\int B⋅dA=B⋅\int dA=B⋅\int_{π/4}^{π/2} \frac{(acosθ)^2}{2}=-\frac{Ba^2}{2}$

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.

 

[kaspis245]

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

$ε=-\frac{dΦ}{dt}=-B\frac{dA}{dt}$

 

[mfb]

That is a good start.

 

[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, $ε=Bv^2t$

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

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

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

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

 

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

 

[mfb]

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

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.

 

[Vibhor]

@rude man ,@mfb

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

 

[mfb]

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