1
$\begingroup$

I'm reviewing test questions from an old exam, but getting a little confused.

enter image description here

a) I thought one of the locations of 0 magnetic field would be right where the middle wire is, since the fields from wires to the left and to the right of it cancel there, but I'm not sure if that's true. I also found the other two locations as follows:

$ B_1 = \frac{\mu_0 I}{2 \pi s}, \:\:\: B_2 = \frac{\mu_0 I}{2 \pi (a - s)}, \:\:\: B_3 = \frac{\mu_0 I}{2 \pi (2a - s)} \:\:\:$

$ B_1 = B_2 + B_3 $

$ \frac{1}{s} = \frac{1}{a-s} + \frac{1}{2a - s} $

$ s = \frac{a (\sqrt{3}+3)}{3}, \frac{-a (\sqrt{3}-3)}{3} $

Which would put the points of 0 magnetic field in the middle-ish (a bit further from the middle wire) between wires 1 and 2, and 2 and 3, which seems reasonable.

c) I'm not sure how to find the frequency of oscillation here. I know that $ F = \int I \times B \: dl $ and we need to put it in the form $ F =-kx$ but I don't really know how to do that. Any hints?

Same for part d.


Is it something like this?

enter image description here

$ B_1 \cos \theta + B_2 \cos \theta = (B_1 +B_2) \frac{\epsilon}{\sqrt{a^2 + \epsilon^2}} = \frac{\mu_0 I \epsilon}{\pi (a^2 + \epsilon^2)} (-\hat{x})$

$\endgroup$
  • $\begingroup$ I know that in part (c) the wire will be pulled down and oscillate before coming to rest right between the other two wires, and in part (d) I think it will just move all the way to the wire closest to it and stay there. I don't think there's any oscillation in the second case since the wire will stop as soon as it strikes the other wire, but I'm not sure. $\endgroup$ – Raksha Jul 29 '15 at 19:15
  • 1
    $\begingroup$ * tap tap * ... is this thing on? +) $\endgroup$ – Raksha Jul 29 '15 at 21:12
1
$\begingroup$

(a) You're about right on the calculation for the two points on $(-a, 0)$ and $(0, a)$ but the $2a-s$ thing looks weird. Is $s = x + a$ or something?

On $(0, a)$ it should be $1/x + 1/(x + a) + 1/(x - a) = 0$ or $$(x + a)(x - a) + x (x - a) + x (x + a) = 0$$ simplifying to $3 x^2 = a^2$ and giving $x = \pm a \sqrt{1/3}$, which appears to be what you got by a more complicated route. Lesson: use B2 as the origin, not B1.

As for x=0 being a zero, well, I guess that's almost true. It's not well-defined because the field of B2 is singular there and we're asking about the field B1 + B2 + B3.

(c) Basically, yes: at a position $(0, y)$ the field is $$\frac{\mu_0 I}{2\pi\sqrt{a^2 + y^2}} \left(\left[\begin{array}{c}-\sin\theta\\\cos\theta\end{array}\right] + \left[\begin{array}{c}-\sin\theta\\-\cos\theta\end{array}\right]\right)$$where we define $\theta$ by the triangle $y = a \tan\theta.$ Hence you get a $-2 \sin \theta = \frac{-2y}{\sqrt{a^2 + y^2}}$ by simple geometry on that triangle. For small displacements $a^2 + y^2 \approx a^2$ and you get:$$\vec B = \frac{-\mu_0 I y}{\pi~a^2}~\hat x$$and the only "mistake" I can see is that you're not replacing $a^2 + \epsilon^2$ with $a^2$.

If the infinitesimal force on a line segment $d\ell$ is $dF_y = (\vec I \times \vec B)_y~d\ell$ and its mass is $dm = \lambda ~d\ell$ then its Newton's equation of motion is just $$dm~\ddot y = dF_y$$hence when we cancel out $d\ell$ we get $$\lambda ~\ddot y = -(\mu_0 I^2 / \pi a^2)~y$$ which is a harmonic oscillator and you've been to the dentist a thousand times -- you know the drill, $\sqrt{k/m}$ and all that.

$\endgroup$
  • $\begingroup$ hmmmm .... looking at this again after a while, how do you know that $ \theta $ is defined at $ y=a tan \theta $ ? (Actually did you mean $ a=y tan \theta $?) I feel like the way I drew it is misleading. I don't think that it's the case that the horizontal leg of that triangle (made up by axes and B2) is actually $ a $ ... it's smaller. But then how do we define $ \theta? $ $\endgroup$ – Raksha Dec 28 '15 at 20:04
  • $\begingroup$ Also, I think it's $ dq =λ dℓ $ not $ dm=λ dℓ $ $\endgroup$ – Raksha Dec 28 '15 at 20:17
  • $\begingroup$ (2) I mean, you can think it's $dq=\lambda~d\ell$ all you want, but if $q$ is not some mass measurement then you're wrong. (1) You can define your $\theta$ however you want, but I defined my $\theta$ how I wanted, so that $\theta=0$ when $y=0$. There were therefore two unit vectors, one for each of the field sources: the one from the left goes (for small $\theta$) mostly towards $+\hat y$ and a little towards $-\hat x$ therefore it is $[-\sin\theta,\cos\theta];$ the one from the right goes mostly towards $-\hat y$ and a little towards $-\hat x$ therefore it is $[-\sin\theta,-\cos\theta].$ $\endgroup$ – CR Drost Dec 28 '15 at 20:48
  • $\begingroup$ oooh.... similar triangles........ $\endgroup$ – Raksha Dec 30 '15 at 21:26
2
$\begingroup$

A few hints:

The field at the middle wire is "indeterminate" since there is a singularity due to the current in the middle wire. If you sketch the field as a function of $x$ you would get something like this:

enter image description here

(this is the plot of $\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}$ courtesy of Wolfram Alpha)

The zeros in the field are easily seen as occurring roughly at ±0.6 a - consistent with your calculation.

The magnetic field lines will look something like this:

enter image description here

where the 'x' marks the point where the field is null.

Now for the oscillation, you need to consider the B field that is experienced by the wire in the middle. As you already surmised, the sum of the fields of the two wires on either side is zero - but you need to ask yourself what is the field when you displace a small distance in x.

One way to do that is to take the derivative of the sum of the two fields - or you can just write down the expression for the field due to the wires at ±a at a position $dx$. This would be of the form

$$B \propto \frac{1}{a+dx}-\frac{1}{a-dx}$$

Doing a Taylor expansion (and figuring out the constant of proportionality) you will see that the field is indeed proportional to the displacement in x. This means that the central wire will experience a force that is proportional to the displacement. Convince yourself that the force is restoring - that is, points back to the middle - and you will have all you need to calculate the frequency.

You can then repeat the same procedure for the vertical field; but eyeballing it, I have the feeling that there is no change of direction of the field and therefore no oscillation. But I'll leave it up to you to take it from here - that's the policy on this site for homework-like questions.

Feel free to ask more questions if you don't make progress with these hints.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.