Magnetic Field Acceleration

1. May 19, 2015

roam

1. The problem statement, all variables and given/known data
A square aluminium loop, of side a and uniform cross-section falls in a vertical plane under gravity, as shown in the diagram. A uniform, horizontal magnetic field B, indicated as the shaded area, points into the page.

(a) If the loop starts when its lower edge is coincident with the lower edge of the magnetic field, how long does it take to clear the field?

(b) We were careful to avoid starting the loop from above the shaded area. Why?

2. Relevant equations

Attached

3. The attempt at a solution

In the attachment I have already worked out the equation for acceleration, velocity, and terminal velocity.

(a) If I understand correctly, the distance to be traveled in order to clear the magnetic field is $a$ (side length of the square loop).

So since the loop is fully immersed in the magnetic field, it would just accelerate at the local gravitational acceleration g (the flux through the loop would not change if it is fully inside). So

$\frac{dv}{dt} = g - \frac{B^2l^2 v}{mR} = g$​

I have used the equation $s=\frac{1}{2} at^2$ to work out the time:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(a)}{g}}$​

Is this right?

(b) I have no idea. It think the loop would simply start to fall freely with acceleration g, then its acceleration would change when it comes to the upward force exerted by the magnetic field. So why does the question avoid starting the loop from above the shaded area?

Any explanation here would be greatly appreciated.

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2. May 19, 2015

BvU

the acceleration is no longer equal to g once it is moving. So you can't use the simple SUVAT $s=\frac{1}{2} at^2$

3. May 19, 2015

roam

Then what other equation do I have to use? I'm a bit confused, I thought it should be falling with the local gravitational acceleration g while transitioning out of the magnetic field (as the flux through the loop is not changing).

4. May 19, 2015

BvU

I'm not with you: if it starts, the flux is Ba2 and by the time it has fallen over a distance a, the flux is 0. How can you say the flux isn't changing ?

Your $\frac{dv}{dt} = g - \frac{B^2l^2 v}{mR} = g$ is a differential equation for $v$. $v = {ds\over dt}$, so you get an expression for $s(t)$ and solve $s(t) = a$. RIght ?

5. May 19, 2015

roam

Thank you for the hint, however when I solve the DE:

$\frac{dv}{dt} = g - \frac{B^2l^2 v}{mR} = g \implies v=\frac{ds}{dt}= gt+c_1 \implies s = \frac{gt^2}{2} + c_1 t + c_2$

How do I solve for the two constants? I mean, in a situation where I need to solve this numerically?

6. May 19, 2015

BvU

$v$ in the second term is a function of t. ${dv\over dt} = g$ only at t=0. Your $v$ does not satisfy the differential equation.

Integration constants follow from initial conditions: v(0) = 0 and s(0) = 0.

7. May 19, 2015

roam

So if I understand correctly, do you mean I should solve $\frac{dv}{dt} = g- \frac{B^2l^2 v}{mR}$ for finding an expression for s(t)? (I mean, instead of the other expression with g alone)

I thought from Lenz's law if the loop is falling the magnetic flux through it will drop so the induced current will act to increase this flux. But since the loop is moving inside the field the whole time, flux should not change. That is why I wrote dv/dt=g, ignoring the other term.

8. May 19, 2015

roam

Should the answer not be the same as for the free fall case (if we cut the loop, breaking the circuit, for instance)?

The $B^2l^2v/mR$ term is really due to the interaction with the magnetic field and the upward magnetic force, therefore it should be set to zero when the loop has no interaction with the field. In the free fall case for example we obtain $v(t) = gt + c_1 \implies s(t)=gt^2/2 + c_1 t + c_2$. I'm not sure how the initial condition helps us to solve the constants as we have two unknowns.

9. May 19, 2015

BvU

If your expression were a solution, which it is not, then $\ \ v(0) = 0\ \ \Rightarrow c_1 = 0\ \$ and $\ \ s(0) = 0 \Rightarrow \ \ c_2 = 0 \ \$.

Now, you have a differential equation to solve. Something of the form$${dy\over dx} = g - \alpha \; y(x)$$where$$\alpha = B^2l^2/mR$$
Ever done before ?

10. May 19, 2015

BvU

This I don't understand. At t=0 the lower edge of the loop is at the edge of the B field area. if s = a/2, the magnetic field is B only in half the area of the loop and if s = a the loop is completely outside, so flux = 0

11. May 19, 2015

roam

You are absolutely right. I thought it was moving from the top, I completely I misunderstood the problem, my bad.

I tried to find the velocity of the loop as a function of time:

$\frac{dv}{g-\frac{B^2l^2 v}{mR}} = dt \implies - \frac{mR}{B^2l^2} ln (g- \frac{B^2l^2 v}{mR}) = t + c$
Letting $a = B^2l^2/mR$ we obtain:

$\therefore g- av = e^{-a(t+c)} = e^{-at} e^{-ac} = A e^{-at}$​

Using initial condition v(0)=0 we obtain the constant A=g. Therefore:

$v(t)=\frac{ds}{dt}=\frac{g}{a} (1- e^{-a t})$

(The g/a part is just the terminal velocity, a numerical value we have already calculated)

So basically, I need to use $s(t)=\frac{g}{a}(t+\frac{e^{-at}}{a} + c_1)$ (with $s(0)=0 \implies c_1=-1/a$) in order to find the time it takes to clear the field by substituting the length of the edge of the square?

Edit: If the length of the side of the loop is $x$, then

$s(t)=\frac{g}{a} (t+ \frac{e^{-at}}{a} - \frac{1}{a})=x \implies t +\frac{e^{-at}}{a} = \frac{ax}{g}+ \frac{1}{a}$

How can I solve that?

Last edited: May 19, 2015
12. May 19, 2015

roam

The expression I got for s(t) is not solvable. What's wrong here?

Also, for part (b), what would happen if the loop was started from above the shaded area (magnetic field)? Is that unsolvable because we don't know the initial downward velocity of it, or something else? Any explanation would be appreciated.

Last edited: May 19, 2015
13. May 19, 2015

BvU

I don't know of a standard way to solve this kind of equation (other than numerically).
If you have a numerical value for $v_t$ you also have one for a. Perhaps $a t$ is small enough to use a power series approximation for the exponent ?

 Griffiths 3rd ed. exercise 7.11 makes life easier for itself by asking when v is 90% of vt -- so a simpler equation results. (Solution here)

Last edited: May 19, 2015
14. May 17, 2016

Bucho

Almost a year later and I have this exact same problem and am stuck in the same spot.

Did you ever work your way past that point @roam?

15. May 17, 2016

Bucho

I should add that I took length of side to be s = a (rather than x as roam did).
I also had used opposite signs for my z-axis values from roam (i.e., I took +ve z direction to be upward, so Fmag was +ve and Fg -ve) so have an exponential with a positive argument.
And I had called my notational simplifying constant k = B2A2/(mR) rather than a, since a is given as the length of the sides of the square loop.

So the equation I ended up with for s(t) = a was:

a = (g/k)t - (g/k2)ekt + g/k2

t - (1/k)ekt = k(a + g/k2) / g

It's no prettier or more advanced than roam's, but that's where I'm at.

16. May 17, 2016

nrqed

This can only be done numerically (or graphically). Do you have numerical values for all the quantities? You can also try to Taylor expand $e^{kt}$ to second order, solve for t and check if the expansion was valid (i.e. if indeed $kt \ll 1$).

17. May 17, 2016

Bucho

Thanks @nrqed, I did have numerical values and eventually solved it numerically.