Forces acting on a coil due to a varying magnetic field

[Zack K]

Homework Statement

A thin rectangular coil lies flat on a low-friction table. A very long straight wire also lies flat on the table, a distance $z$ from the coil. The wire carries a conventional current $I$ to the right as shown, and this current is decreasing: $I=a-bt$, where$t$ is the time in seconds, and $a$ and $b$ are positive constants. The coil has length $L$ and width , where $w\ll z$. It has $N$ turns of wire with total resistance $R$.(I uploaded an image)

What are the initial magnitude and direction of the nonzero net force that is acting on the coil? You can neglect friction. Explain in detail. If you must make simplifying assumptions, state clearly what they are, but bear in mind that the net force is not zero.

Relevant Equations

$\oint {E \cdot d\ell = - \frac{d}{{dt}}} \int {B_n dA}$

$B_{wire}=\frac{\mu_0}{4\pi}\frac{LI}{z\sqrt{z^2+(\frac{L}{2})^2}}\longrightarrow B_{wire}=\frac{\mu_0}{4\pi}\frac{2I}{z}$(for $z\ll L$)

I don't understand why a force would be acting on this rectangular coil at all. The magnetic field of the wire would only induce a force on the coil, if the coil had a current flowing through it. At first I would think that the electric field from the varying magnetic field would induce such current through the wire, but the electric field is going perpendicular to the length of the coil, which would not induce a current. The only thing that seems plausible is that the electric field causes the wire to be polarized along the thickness, but that obviously isn't right.

EDIT: I just realized I'm completely wrong, I think I get what's happening. The magnetic field is pointing into the page at the location of the wire and if the current\magnetic field is decreasing, then the electric field will go clockwise around the coil.

 

[kuruman]

The magnetic field is pointing into the page at the location of the wire and if the current\magnetic field is decreasing, then the electric field will go clockwise around the coil.

I think you got it. Can you finish the problem now?

 

[Zack K]

I think you got it. Can you finish the problem now?

I am actually stuck again,

So you know $F_m=I\Delta\vec l\times \vec B= I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z})$ and $I_{wire}= a-bt$
This force is attractive to the closest length of the coil, and repulsive to the further length of the wire. The magnetic field has no effect on the wires along the width.

But what is $I_{coil}$? I'm assuming it's the induced emf divided by the resistance of the coil, but wouldn't you also get $I_{wire}=\frac{emf}{R}=-\frac{\frac{dB}{dt}}{R}=-\frac{B_{wire}}{R}\frac{dI}{dt}=-\frac{B_{wire}}{R}(a-bt)$

This seems like an odd result to get because I see no reason why the force would be proportional to the square of the magnetic field and and current in a time varying magnetic field.

 

[kuruman]

You are not considering that the magnetic field due to the wire is weaker at the segment of the loop that is at distance $z+w$. Also, although $I_{wire}=\frac{emf}{R}$, it is not true that $emf=\frac{dB}{dt}$. The correct expression for the magnitude of the induced emf is $emf=|\frac{d\Phi_M}{dt}|,$ where $\Phi_M$ is the magnetic flux through the loop.

 

[Zack K]

Also, although Iwire=emfRIwire=emfRI_{wire}=\frac{emf}{R}, it is not true that emf=dBdtemf=dBdtemf=\frac{dB}{dt}. The correct expression for the magnitude of the induced emf is emf=|dΦMdt|,emf=|dΦMdt|,emf=|\frac{d\Phi_M}{dt}|, where ΦMΦM\Phi_M is the magnetic flux through the loop.

Yes that makes sense, that's my bad.

You are not considering that the magnetic field due to the wire is weaker at the segment of the loop that is at distance z+w.

If that's not true then wouldn't the net magnetic force on the coil be 0? (I'm not sure what you mean exactly, but if you're saying that the magnetic field is weaker at z+w, then that is something that I considered.

 

[kuruman]

If that's not true then wouldn't the net magnetic force on the coil be 0? (I'm not sure what you mean exactly, but if you're saying that the magnetic field is weaker at z+w, then that is something that I considered.

What I mean is that actually you didn't consider that the field is weaker at $z+w$. I can see that in your expression

$F_m=I\Delta\vec l\times \vec B= I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z})$

which makes me believe that you treated the force vectors carelessly.
The force on the section of the coil closer to the wire is (assuming the $x$-axis is in the plane of the coil)$$ \vec F_{1}=-\frac{\mu_0}{2\pi}I_{coil} L \left( \frac{I_{wire}}{z} \right)\hat x$$
and the force on the section of the coil farther from the wire is $$ \vec F_{2}=\frac{\mu_0}{2\pi}I_{coil} L \left( \frac{I_{wire}}{z+w}\right)\hat x$$
The net force on the coil is$$\vec F_{net}=\vec F_{1}+\vec F_{2} = \cdots~?$$I suggest that you add the vectors first and only then use the approximation $z>>w$. If you do that before you add the vectors, you will end up with zero which is exactly what the problem is cautioning you against. That's the clean way of doing this instead of tacking a factor of $2$ where it doesn't belong.

 

[Zack K]

What I mean is that actually you didn't consider that the field is weaker at z+wz+wz+w. I can see that in your expression

Yes that's my bad, I do know of this, I was just leaving the general expression for now where $z$ is an arbitrary distance, I probably should have used another variable.

So now I have $I_{wire}=-\frac{d\phi}{dt}=-\frac{d}{dt}(BA)=-\frac{dB}{dt}Lw$

Plugging in, I get $F_{m}=-\frac{dB}{dt}L^2w(\frac{\mu_0}{4\pi}\frac{2(a-bt)}{z})$
If I did this right, however now I have the time varying magnetic field, and so isn't $\frac{dB}{dt}=(\frac{\mu_0}{4\pi}\frac{2(a-bt)}{z})$?

 

[kuruman]

Yes that's my bad, I do know of this, I was just leaving the general expression for now where $z$ is an arbitrary distance, I probably should have used another variable.

So now I have $I_{wire}=-\frac{d\phi}{dt}=-\frac{d}{dt}(BA)=-\frac{dB}{dt}Lw$

You are confusing things. First off $I_{coil}=\frac{1}{R}|\frac{d\Phi}{dt}|$ and $I_{wire}=a-bt$.

Plugging in, I get $F_{m}=-\frac{dB}{dt}L^2w(\frac{\mu_0}{4\pi}\frac{2(a-bt)}{z})$
If I did this right, however now I have the time varying magnetic field, and so isn't $\frac{dB}{dt}=(\frac{\mu_0}{4\pi}\frac{2(a-bt)}{z})$?

Exactly in what expression are you plugging in? To do it right, first find an expression for the net force in terms of $I_{wire}$ and $I_{coil}$ then write down two separate expressions for $I_{wire}$ and $I_{coil}$ and finally put these in your expression for the net force.

 

[Zack K]

So I have

$\vec F_{near}=I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z}) \langle 0,0,1 \rangle$ (attraction)
$\vec F_{far}=I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z+w}) \langle 0,0,-1 \rangle$ (repulsion)

where $I_{wire}=a-bt$

and $I_{coil}= \frac{emf}{R}=N\frac{-\frac{d\phi}{dt}}{R}= N\frac{\frac{-dB}{dt}Lw}{R}$ (for N turns of the coil)

$\vec F_{net}= \vec F_{near} - \vec F_{far}$

I mixed up some terms in the previous comment, but my intent was the same. Though if my expression for my coil is right, is it okay to leave a $-\frac{dB}{dt}$ term? Otherwise, I'm not sure what to break it down to.

 

[kuruman]

You are not paying attention to what you are doing and instead of allowing the algebra to show you the way, you write down things that are not correct.

$\vec F_{near}=I_{coil}L(\frac{\mu_0}{2\pi}\frac{2I_{wire}}{z}) \langle 0,0,1 \rangle$
$\vec F_{far}=I_{coil}L(\frac{\mu_0}{2\pi}\frac{2I_{wire}}{z+w}) \langle 0,0,-1 \rangle$

Where does that $2$ come from? You keep putting it in and it doesn't belong. I gave you the expressions in post #6. In your notation they are
$$ \vec F_{near}=\frac{\mu_0}{2\pi}I_{coil} L \left( \frac{I_{wire}}{z} \right)\langle 0,0,-1\rangle $$
$$ \vec F_{far}=\frac{\mu_0}{2\pi}I_{coil} L \left( \frac{I_{wire}}{z+w}\right)\langle 0,0,1 \rangle$$
Also, an attractive force has a negative sign, in the direction of decreasing $z$. Furthermore, the net force is the sum of the two vectors. This means
$$\vec F_{net}=\vec F_{near}+\vec F_{far} =\frac{\mu_0}{2\pi}I_{coil} L \left( -\frac{I_{wire}}{z} +\frac{I_{wire}}{z+w} \right)\langle 0,0,1 \rangle$$You can take it from here and simplify, but do it right.

... is it okay to leave a $-\frac{dB}{dt}$ term?

It is not OK. You have to find an expression for it, but first you need to figure out where this magnetic field comes from.

 

[Zack K]

Where does that 2 come from?

It comes from the approximation of the length of the wire being much larger than a distance $z$. So I can cancel the $z^2$ term in the denominator

$B_{wire}=\frac{\mu_0}{4\pi}\frac{LI}{z\sqrt{z^2+(\frac{L}{2})^2}}\longrightarrow B_{wire}\approx\frac{\mu_0}{4\pi}\frac{2I}{z}$ (if the distance to the coil $\ll$ the length of the wire)

you need to figure out where this magnetic field comes from.

Doesn't the magnetic field come from the wire?

 

[kuruman]

OK, I see now. Yes, the magnetic field comes from the wire and you know what that is.

 

[Zack K]

I don't know where you got this,
$B_{wire}=\frac{\mu_0}{4\pi}\frac{LI}{z\sqrt{z^2+(\frac{L}{2})^2}}$
but it is not the field generated by a very long wire. See
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

It's the same thing I think.

$B_{wire}=\frac{\mu_{0}I}{2\pi r}=\frac{\mu_0}{4\pi}\frac{2I}{r}$
I just leave the 2 because of $\frac{\mu_0}{4\pi}$ being a defined constant