# Unusual problem from Electromagnetism.

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A wire with resistance per unit length of 32.668 Ω/m is bent at a right angle. The second straight fragment of the same wire (The blue one) moves at the top of the first one at a speed of v = 4.2735 m/s. The whole system is located in a magnetic field with an induction of B = 8.4965 T, perpendicular to the wires (as in the picture). Determine the current flowing through the wires. Ignore the resistance at the points of contact of both wires and also the magnetic field generated by the current itself that is taken into consideration.

Edited by Classical Physicist
corrected
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I have a hard time understanding the latter of it- how come the same wire slide upon itself with a velocity, while being inside the magnetic field B?

The way I see it- these might not actually be 100% same wire, rather two distinct wires, with the same properties, but being separated? -
HOWEVER it wouldn't make sense then, because we need to find the current I, (that is supposedly the same in each "part" of the "wires")…
So that is maybe a wrong assumption to begin with.

So coming back to the idea of the same wire sliding upon itself - we need to take a look at the basic electromagnetic properties in order to push the problem further?

4 minutes ago, Strange said:
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Moved to Homework Help. Please show how far you have got

huh, I was going to just do that right now, because I didn't want to clutter the thread, but hey. Edited by Classical Physicist
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So I think I need to use the faraday's law here to solve it?

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That’s a start. That gives you the voltage. The resistance will change over time as the wire moves. Solve for I.

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My suggestion is to do the geometry first.

Look up the properties of similar triangles if you don't already know them

Use these properties to find out the rate of increase of the area of the triangle and the lengths of the 2 wires that make it up.

Then worry about the induction.

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The way I understand it, the bent wire is immobile versus B while the skewed one, of identical nature, moves and is in contact with the bent one.

d(flux)/dt gives an induced voltage (with a sign) while the summed wire length gives a resistance. Neglecting the self-inductance (="the field produced by the wires themselves") you get a current.

Note that 8T is unusually strong. It take a superconductor or a small duration. 32ohm/m is much even for a resistor wire.

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1 hour ago, Enthalpy said:

The way I understand it, the bent wire is immobile versus B while the skewed one, of identical nature, moves and is in contact with the bent one.

d(flux)/dt gives an induced voltage (with a sign) while the summed wire length gives a resistance. Neglecting the self-inductance (="the field produced by the wires themselves") you get a current.

Note that 8T is unusually strong. It take a superconductor or a small duration. 32ohm/m is much even for a resistor wire.

Yes you're right- that's what it is. two electrical conductors (in a form of wires) with the identical nature.

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33 minutes ago, Classical Physicist said:

Yes you're right- that's what it is. two electrical conductors (in a form of wires) with the identical nature.

So where did this question come from and why are you doing it?

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55 minutes ago, studiot said:

So where did this question come from and why are you doing it?

I found it online maybe ...half a year ago, don't really remember the origin of it, it was from some script or notes from/for students of the first year of some course on physics from A university I don't ever so slightly have an idea of (not sure though)  that I found while looking for something that could help me understand the electromagnetism better? I don't remember, and the problem was handwritten so yeah, I just got down what you see here.

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I had thought this was not homework for a formal course you are doing, online or off.

How are you getting on with the geometry?

When you come to it how do you think the slant angle will affect Flemings generator rule ?

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I just got back from Uni and had not much time to sit down to it. But It also boils down to be quite much easier than it appears to be- For  a while I was thinking about the whole physics that we could use here, but still your problem solving techniques - about the geometry are still a bit "magic" to me, I get them but, intuitively I would have to practice more to see these just like you do,

So the geometry and similarity are well-known to me (or so I suppose), but as you suggested we need to focus on the moving wire that geometrically makes up a triangle with the other wire underneath it. - I sense some differentiation that has to be done with it?

we basically have a function of time (?) that with the upper wire movement makes up for an increasing area of a triangle with angles of 45° 45° and 90°.
which as time passes by --- reaches the point where the sides are just like in a special triangle with angles like that.

Edited by Classical Physicist
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IIRC there are problems like this in Halliday & Resnick (and no doubt in the later versions) with a rectangular cross-section. Making it a triangle just adds a little extra geometry to the mix.

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8 minutes ago, swansont said:

IIRC there are problems like this in Halliday & Resnick (and no doubt in the later versions) with a square cross-section. Making it a triangle just adds a little extra geometry to the mix.

which chapter is it? the 28th - with magnetic fields, right?

Actually the 30th chapter about "Induction and Inductance" is more accurate here, --- you're right the problem 18 from the chapter 30 looks quite similar.

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You will need both lengths and areas

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12 minutes ago, studiot said:

You will need both lengths and areas

I'm confused

The “height” of the triangular area  enclosed by the underneath wire and the one above is the same as the distance traveled in time v: d = vt, also the base of the triangle like that equals to λ
(as in picture)

so we get
1/2*
λ*vt=A

Do I understand correctly that λ is the length of the wires? - in which case the bent one has  λ as a sum of these "sides",so that each side is λ/2?

however geometric look at the picture denies that- so the bent part would be 2λ, 1
λ for each side, just like the moving wire above, with l=1 λ.

38 minutes ago, studiot said:

You will need both lengths and areas

I still don't get which exact triangles should I take into consideration,  I mean I understand these ratios, but which  triangles  with an angle of 45, 45 and 90 from the problem  do I need, exactly?

Never Thought that I would have to ask you about such trivial stuff, but hey I am very tired right now… However that is still not an excuse.

Edited by Classical Physicist
math
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Moderator Note

As my assumption that this was homework appears to be wrong, I have moved this back to the original forum.

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On 10/6/2019 at 10:18 PM, Classical Physicist said:

So I think I need to use the faraday's law here to solve it?

d(flux)/dt gives an induced voltage (with a sign)

So putting this into mathematical form

$d\Phi = - BdA$

divide both sides by dt

$\frac{{d\Phi }}{{dt}} = - B\frac{{dA}}{{dt}}$

Where A is the area enclosed by the (triangular loop)

$E = - \frac{{d\Phi }}{{dt}}$

Where is is the voltage generated in a wire at right angles to the field

Now draw some positions of the blue wire in at varous times.
Also extend the velocity vector of the blue wire in both directions.

What do you notice about the triangular areas so formed?

Can you connect this to the above in terms of the velocity of the blue wire?

The point of the similar triangles is that you don't need any actual values of area, ratios will do.

One further note.

E is generated only in the blue wire because it is the only wire moving.

The black wire is stationary in the field so no voltage is generated in this one.

The circulating current in the loop is entirely due to the motion of the blue wire.

One final twist - is the blue the blue wire motion at right angles to the field, in accordance with Flemings rule?

Edited by studiot
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Love it, Thanks again, I'll post  the further answer shortly,   …. so how can I  insert  the maths in here? beause I tried with LateX syntax but it didn't output this nice formatting that you post here?

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2 minutes ago, Classical Physicist said:

Love it, Thanks again, I'll post  the further answer shortly,   …. so how can I  insert  the maths in here? beause I tried with LateX syntax but it didn't output this nice formatting that you post here?

Build your formulae in one of these and copy/paste into SF.

or

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15 hours ago, Classical Physicist said:

which chapter is it? the 28th - with magnetic fields, right?

Actually the 30th chapter about "Induction and Inductance" is more accurate here, --- you're right the problem 18 from the chapter 30 looks quite similar.

I don't have the book in front of me. I remember seeing the problem in the book, ~40 years ago when I took the class, and ~30 years ago, when I was a TA for the class.

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On 10/7/2019 at 11:57 PM, studiot said:

d(flux)/dt gives an induced voltage (with a sign)

So putting this into mathematical form

dΦ=BdA

divide both sides by dt

Where A is the area enclosed by the (triangular loop)

E=dΦdt

Where is is the voltage generated in a wire at right angles to the field

Now draw some positions of the blue wire in at varous times.
Also extend the velocity vector of the blue wire in both directions.

What do you notice about the triangular areas so formed?

Can you connect this to the above in terms of the velocity of the blue wire?

The point of the similar triangles is that you don't need any actual values of area, ratios will do.

One further note.

E is generated only in the blue wire because it is the only wire moving.

The black wire is stationary in the field so no voltage is generated in this one.

The circulating current in the loop is entirely due to the motion of the blue wire.

One final twist - is the blue the blue wire motion at right angles to the field, in accordance with Flemings rule?

oh I forgot about it so I got this: is this correct?

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w/q = p/I is a long way round to get to Ohm's law, but yes ok to here

$E = IR = B\frac{{dA}}{{dt}}$

However you will not get the right answer if you just follow the example in Res & H . Theirs is simpler.

Remember that R depends upon the length of the perimeter of the triangle and this is increasing with time and the motion of the blue wire.

Backalong I suggested to get the geometry done first.

This was because in R&H the blue wire has a constant length whereas here it is increasing.

In fact all linear dimesions increase in the asame ratio, which is why I wanted you to draw the line velocity vector through the origin at 45o.
So you could see for youself that the length of the base of the triangle (ie the blue line) increases linearly with time by the area of the triangle increase as the square if this since it is the product of two dimensions, both subject to the same increase in time. The geometry of similar triangles, as mentioned, confirms this by another way.

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