# Falling loop in magnetic field

1. Homework Statement

This is a question regarding Problem 7.11 in Griffiths Electromagnetism book, which considers a falling square loop of aluminum through a uniform magnetic field pointing into the page with only part of the loop in the field at t=0 (see picture below).

To solve this, normally one calculates that there is an upward magnetic force of IlB, where I is the current, l is the length of the top portion of the wire, and B is the constant magnetic field. To get I, you'd calculate the emf ξ using the flux rule and then divide by the resistance of the aluminum, R.

However, this is the way to solve this problem using the flux rule to get the emf. I'd like to get the emf by calculating the work done per unit charge (as Griffiths does previously in the chapter).

2. Homework Equations
F
mag = q(v x B)

3. The Attempt at a Solution
As the loop starts to fall, its velocity is (initially) all in the negative z-hat direction so the magnetic force pushes charges to the right, creating a current (u in the diagram). However, consequently, the charges now take on a diagonal path (w). So the magnetic force is then = quB(z-hat) + qvB(x-hat) (blue in the diagram). Now I can get the correct emf if I assume u is constant, because with qu (or really λu) being the current I, the total work done here is IlB (the top part is l and the sides don't contribute).

My question is why can I assume u is constant? If u is the current per electron, doesn't u increase since there is a magnetic force of qvB(x-hat) on a given electron, hence increasing it's speed?

Everywhere I've looked people solve this problem using the flux rule, but I haven't seen it done this way.

Related Introductory Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
Hello. Welcome to PF!
My question is why can I assume u is constant? If u is the current per electron, doesn't u increase since there is a magnetic force of qvB(x-hat) on a given electron, hence increasing it's speed?
Just after dropping the loop, u does increase. Are you asking why u remains constant once the loop attains terminal velocity? Does the electrical resistance R of the loop play a role here?

Last edited:
Hello. Welcome to PF!

Just after dropping the loop, u does increase. Are you asking why u remains constant once the loop attains terminal velocity? Does the electrical resistance R of the loop play a role here?
Thank you! Been a long time viewer, but finally signed up. In terms of my question, let me run through my reasoning below (and this is an adaptation of what Griffiths does on the example of someone pulling a rectangular loop partially situated in a uniform magnetic field pointing into the page to the right):

Using my original drawing there is a component of magnetic force upward quB. So the gravitational force downward includes a "component" that cancels this upward force out. This gravitational pulling force is then quB(-z-hat). The work that this does per charge, then, is ∫uB(-z-hat)⋅w = ∫uBvdt. This is the emf. Now this integral is supposed to equal Bvl, but u and v both functions of t, so I don't see how that integral works out?

TSny
Homework Helper
Gold Member
Using my original drawing there is a component of magnetic force upward quB.
Yes
So the gravitational force downward includes a "component" that cancels this upward force out.
I don't feel comfortable with this statement. Are you referring to the gravitational force on the single charge q? Besides the gravitational force on a single charge carrier q, there will also be forces on q from the other particles making up the system. (These forces make sure q stays within the aluminum material.) The gravitational force on q would be negligible if q represents an electron. The vertical forces on q do not cancel out exactly while the loop is still accelerating (before "reaching" the terminal velocity). However, if q is a single electron, then the mass is essentially negligible and the forces on q can be taken to essentially add to zero even while the loop is still accelerating downward.

Once terminal speed is essentially attained, the upward component of the magnetic force on all of the charge carriers will cancel the weight of the entire aluminum loop (rather than cancelling just the weight of the charge carriers). The weight of the loop is essentially determined by the weight of the "background" aluminum ions.
∫uB(-z-hat)⋅w = ∫uBvdt. This is the emf.
Did you mean to include dt in the integral on the left?
Now this integral is supposed to equal Bvl, but u and v both functions of t, so I don't see how that integral works out?
First, you might consider the case where terminal speed has essentially been attained. So, u and v are constant. You then need to consider the appropriate time interval for the integration with respect to t.

I don't feel comfortable with this statement. Are you referring to the gravitational force on the single charge q?
I'm saying that the magnetic force pushes the charge up (magnitude is quB(z-hat)), but gravity counteracts this. A portion of the magnitude of the gravitational force then counteracts this. So mg > quB. In that sense I'm translating the problem of someone pulling a rectangular loop to the right through a uniform mangetic field. Here's how Griffiths writes it for that problem:

So what I'm referring to is what he's referring to as fpull, which, when dotted with the infinitesimal displacement of the charge (wdt) and then integrated around the loop gives the emf.

Did you mean to include dt in the integral on the left?
First, you might consider the case where terminal speed has essentially been attained. So, u and v are constant. You then need to consider the appropriate time interval for the integration with respect to t.
As I alluded to above, the infinitesimal displacement vector dl is (in both my and Griffith's diagrams) wdt, since w is the resultant velocity of the current in the wire u plus the downward velocity of the loop v. So what I'm curious about is how ∫uB(-z-hat)⋅w (which= ∫uBvdt) = vBl?

Or if you'd rather, in the analagous example I pasted above, how does he go from
to
? My question pertains to both cases.

#### Attachments

• 116.8 KB Views: 310
TSny
Homework Helper
Gold Member
As I alluded to above, the infinitesimal displacement vector dl is (in both my and Griffith's diagrams) wdt, since w is the resultant velocity of the current in the wire u plus the downward velocity of the loop v. So what I'm curious about is how ∫uB(-z-hat)⋅w (which= ∫uBvdt) = vBl?
I still think you're missing a factor of dt in the integrand of ∫uB(-z-hat)⋅w.

Or if you'd rather, in the analagous example I pasted above, how does he go from View attachment 106695 to View attachment 106696? My question pertains to both cases.
In the integral $\int \vec{f}_{\rm pull} \cdot \vec{dl}$, how would your write the integrand in terms of $B$, $u$, $dl$ and $\theta$?

I still think you're missing a factor of dt in the integrand of ∫uB(-z-hat)⋅w.
Yup, you're right.

In the integral $\int \vec{f}_{\rm pull} \cdot \vec{dl}$, how would your write the integrand in terms of $B$, $u$, $dl$ and $\theta$?
Ok so here (referring to Griffiths example), the integral is ∫uBdlsin(θ). Now I see that insofar as u and θ are constant at each point on the line of integration, this integral becomes uB[h/cos(θ)]sin(θ) = uBh(v/u) = Bhv = emf.

So as the guy pulls the loop to the right at constant velocity v, the magnetic force has a constant upward component on the charge, qvB, and (ignoring gravity) by Newton's Second Law, the acceleration in the (z-hat) direction is qvB/m. So the upward component of the charge's velocity, u(t), = qvBt/m(z-hat) = qBx(t)/m(z-hat). The horizontal component is of course just constant v.

Does that look right so far?

TSny
Homework Helper
Gold Member
That looks ok except that in applying the 2nd law to get the acceleration in the $\hat{z}$ direction there would be an additional force due to the electrical resistance. Griffiths does not include this retarding force in his picture, presumably because it is not needed in order to derive the expression for the emf.

That looks ok except that in applying the 2nd law to get the acceleration in the $\hat{z}$ direction there would be an additional force due to the electrical resistance. Griffiths does not include this retarding force in his picture, presumably because it is not needed in order to derive the expression for the emf.
True. So if we ignore the resistance, then, we have $\int \vec f_{pull} \cdot \vec {dl}$ = $\int u(t)B \hat x \cdot \vec {dl}$. Now $\vec {dl} = u(t)dt \hat z + vdt \hat x$ and so we get $\int u(t)B \hat x \cdot vdt\hat x$ = $vB\int u(t)dt$ = $vBh$ = $ξ$, since $u(t)$ is the vertical velocity. Ok, this I can see. However, this worked out because $v$ is constant. But if $v$ is not constant, it seems we're stuck at $B\int u(t)v(t)dt$ =? $ξ$, which seems to be the case for the original falling loop analogue.

TSny
Homework Helper
Gold Member
True. So if we ignore the resistance, then, we have $\int \vec f_{pull} \cdot \vec {dl}$ = $\int u(t)B \hat x \cdot \vec {dl}$. Now $\vec {dl} = u(t)dt \hat z + vdt \hat x$ and so we get $\int u(t)B \hat x \cdot vdt\hat x$ = $vB\int u(t)dt$ = $vBh$ = $ξ$, since $u(t)$ is the vertical velocity. Ok, this I can see. However, this worked out because $v$ is constant. But if $v$ is not constant, it seems we're stuck at $B\int u(t)v(t)dt$ =? $ξ$, which seems to be the case for the original falling loop analogue.
The emf $\xi(t_0)$at a given time $t_0$ should depend only on the conditions within the circuit at the instant $t_0$. In particular, the emf at $t_0$ should depend only on the values of $u$ and $v$ at $t_0$. The emf at $t_0$ is the work done per unit charge that it would take to move charge a distance $h$ through the wire assuming conditions at $t_0$. So, when considering the emf at $t_0$, I believe that $\int \vec f_{pull} \cdot \vec {dl}$ should be evaluated treating $u$ and $v$ as having fixed values $u(t_0)$ and $v(t_0)$. Then your derivation would yield $\int \vec f_{pull} \cdot \vec {dl} = v(t_0)Bh = \xi(t_0)$.

rude man
Homework Helper
Gold Member
I won't try to improve on the comments tsny has made. Somewhat complicated subject. I would just add the following:

You are wise to seek an alternative means to the flux rule for computing induced emf in the presence of moving media. Authors such as Prof. Skilling of Stanford have pointed out the dangers of this. If you're interested I can send you an example I've worked out that gives the wrong answer if emf = -dφ/dt is used.

The safe way to handle induced emf's in moving media is via the "Blv" law which gives the emf induced in a moving conductor of length l in a B field moving at velocity v. In its most general differential form it's d(emf) = (B x dl)⋅v. This equation can readily be derived from the Lorentz equation F = qv x B (fast-and-loose we have F = qvB but F = qE so E = vB and emf = El = Blv). In your case the only segment constituting l above is the horizontal one moving thru the B field. There is an electric field thruout the loop due to the current but the only B-induced one is in that horizontal segment which is the E above.

The emf $\xi(t_0)$at a given time $t_0$ should depend only on the conditions within the circuit at the instant $t_0$. In particular, the emf at $t_0$ should depend only on the values of $u$ and $v$ at $t_0$. The emf at $t_0$ is the work done per unit charge that it would take to move charge a distance $h$ through the wire assuming conditions at $t_0$. So, when considering the emf at $t_0$, I believe that $\int \vec f_{pull} \cdot \vec {dl}$ should be evaluated treating $u$ and $v$ as having fixed values $u(t_0)$ and $v(t_0)$. Then your derivation would yield $\int \vec f_{pull} \cdot \vec {dl} = v(t_0)Bh = \xi(t_0)$.
I see your point. So, if we recall, there are two ways to calculate emf -- 1) the "snapshot" view (as Griffiths calls it) which only looks at one instance of time and 2) the work per unit charge. I think combined with rude man's response (below yours), it seems to be the case that option 2) is a bit tenuous, or perhaps should be clarified. For instance, in the falling loop example, the effect of the loop falling under influence of gravity (as opposed to the horizontal loop problem, wherein there is no force impeding the guy from pulling the loop to the right at constant velocity v) kinda serves as an impediment to otherwise normal flow (like an implied resistance or inductance of sorts) and that shouldn't be included in the calculation of ideal emf, which assumes no resistance to current. Perhaps..

Interesting and challenging discussion, thanks for the feedback!

I won't try to improve on the comments tsny has made. Somewhat complicated subject. I would just add the following:

You are wise to seek an alternative means to the flux rule for computing induced emf in the presence of moving media. Authors such as Prof. Skilling of Stanford have pointed out the dangers of this. If you're interested I can send you an example I've worked out that gives the wrong answer if emf = -dφ/dt is used.

The safe way to handle induced emf's in moving media is via the "Blv" law which gives the emf induced in a moving conductor of length l in a B field moving at velocity v. In its most general differential form it's d(emf) = (B x dl)⋅v. This equation can readily be derived from the Lorentz equation F = qv x B (fast-and-loose we have F = qvB but F = qE so E = vB and emf = El = Blv). In your case the only segment constituting l above is the horizontal one moving thru the B field. There is an electric field thruout the loop due to the current but the only B-induced one is in that horizontal segment which is the E above.
Hmm, that's interesting, I'd be curious to see that example if you don't mind sending it to me.

As for the "Blv" law, as I was mentioning in my previous reply to tsny, it seems like the calculation of emf as the work done per unit charge perhaps needs some further clarification, whereas the alternative methodology isn't subject to the same issue since it doesn't consider time in it's calculation. That is, in calculating work done per unit charge of the falling loop, as the charge is pushed around the loop, the "background environment" changes, and so it appears that calculating emf this way is introducing "environmental" effects to something that should be calculated in a sort of steady state environment. That's sort of my stab at it at least

rude man
Homework Helper
Gold Member
Hmm, that's interesting, I'd be curious to see that example if you don't mind sending it to me.
Here it is as a.doc attachment for you and all. Sorry I don't provide a drawing; I'm lousy at using my powerpoint software. Perhaps someone else can. I tried hard to depict the situation verbally as best I could.

The fact is that Maxwell's ∇ x E = - ∂B/∂t does not hold for moving media, but must then be modified to ∇ x E = - ∂B/∂t + ∇ x (v x B). The v x B term can be shown to provide the Blv law, and if you work the attachment you can see that there is no flux change over time inside the loop, and you can show that, no matter how slowly or rapidly the B field inside the loop tapers off in the y direction from B to zero, the last term provides the correct answer as does the Blv law. With the Blv law the thing to keep in mind is the cutting of flux lines as the medium l moves with velocity v thru the B field.

#### Attachments

• 28.7 KB Views: 99