Electromagnetism: Force on a parabolic wire in uniform magnetic field

[Aurelius120]

Homework Statement

A conducting wire bent in the form of a parabola $y^2=2x$ carries a current of $2A$. It is placed in a uniform magnetic field of $-4\hat k T$. Find magnetic force on wire in newtons.

Relevant Equations

$$d\vec F=Id\vec l\times \vec B $$
Given correct answer is $32\hat i$

I know the easier method/trick to solve this which doesn't require integration. Since parabola is symmetric about x-axis and direction of current flow is opposite, vertical components of force are cancelled and a net effective length of AB may be considered then $F=2(4)(L_{AB})=32\hat i$

I was trying to reach the same answer via integration. Thus I proceed :
$$y^2=2x \implies A=(2,2) ; B=(2,-2)$$
Net magnetic force on wire is given by :
$$\vec F_{net}=\int ^{2}_{-2} I \left(\frac{x\hat i+y\hat j}{\sqrt{x^2+y^2}}\right)\times \vec B dy$$
Cross multiplying the unit vectors to get direction and integrating with respect to dy:
$$\vec F_{net}=\int ^{2}_{-2} 2 \left(\frac{x\hat i+y\hat j}{\sqrt{x^2+y^2}}\right)\times (-4\hat k) dy$$
From given data and equation of parabola,
$$\vec F_{net}=-8\int ^{2}_{-2} \left(\frac{(y^2/2)\hat i+y\hat j}{\sqrt{(y^4/4)+y^2}}\right)\times (\hat k) dy$$
Using distributive property of cross multiplication,
$$\vec F_{net}=-8\left[\int ^{2}_{-2} \left(\frac{(-y^2)\hat {j}}{\sqrt{y^4+4y^2}}\right)dy+\int^{2}_{-2}\left( \frac{2y \hat i} {\sqrt{(y^4)+4y^2}}\right)dy\right]$$
$$\vec F_{net}=-8\left[\int ^{2}_{-2} \left(\frac{(-y)\hat {j}}{\sqrt{y^2+4}}\right)dy+\int^{2}_{-2}\left( \frac{2 \hat i} {\sqrt{(y^2)+4}}\right)dy\right]$$

The first integral, is of an odd function and yields an even function and the limits make it zero. Only the second remains. Therefore,

$$\vec F_{net}=-8\int^{2}_{-2}\left( \frac{2 \hat i} {\sqrt{(y^2)+4}}\right)dy$$

So here are my questions,

1. How do I proceed further?
2. I tried to compute the integral with help from internet. It gave me the wrong answer. What have I done wrong and how do I get the correct answer using calculus?

 

[Gordianus]

Do you remember what the vector dl stands for?

 

[Orodruin]

In addition to what was said in #2:

I suggest looking at the integral
$$
\int_A^B d\vec l
$$
and thinking about if the path between A and B actually matters.

 

[Aurelius120]

Do you remember what the vector dl stands for?

Small length of wire, correct ?
I took $dy$ as small wire element and then changed $x$ in terms of $y$. So it should be ok?

 

[haruspex]

Net magnetic force on wire is given by :
$$… \left(\frac{x\hat i+y\hat j}{\sqrt{x^2+y^2}}\right)…dy$$

How did you get that from $\vec {dl}$?

 

[Orodruin]

Small length of wire, correct ?

The line element along the wire. It should represent the small change in the position vector along the wire.

I took $dy$ as small wire element and then changed $x$ in terms of $y$. So it should be ok?

No, what you wrote down was not correct. The small change of the position vector when you change y by dy is not given by the expression you wrote down.

 

[Gordianus]

Remember the tangent vector?

 

[Delta2]

The general expression for the infinitesimal element of the position vector along a curve in 3D space is $$d\vec{l}=\vec{i}dx+\vec{j}dy+\vec{k}dz$$ from which you can get the correct dl for your case by observing that this parabola lies exclusively in the xy-plane.

Also you know the equation of the parabola so you can find how dy and dx relate by differentiation.

 

[Orodruin]

The general expression for the infinitesimal element of the position vector along a curve in 3D space is $$d\vec{l}=\vec{i}dx+\vec{j}dy+\vec{k}dz$$ from which you can get the correct dl for your case by observing that this parabola lies exclusively in the xy-plane.

Which of course everybody helping here knew, but wanted the OP to spend some effort to think about.

Also you know the equation of the parabola so you can find how dy and dx relate by differentiation.

The point is, you don’t need to involve any parametrization at all to do the integral here.

 

[Delta2]

The point is, you don’t need to involve any parametrization at all to do the integral here.

Not sure what you mean here. You mean we can do the integral without knowing the equation between dy and dx?

Maybe this can happen here because of the symmetry of the path, but it is not correct in the general case when dx and dy are not independent.

 

[Orodruin]

Not sure what you mean here. You mean we can do the integral without knowing the equation between dy and dx?

Maybe this can happen here because of the symmetry of the path, but it is not correct in the general case when dx and dy are not independent.

It is obviously correct in the general case. The integral is
$$
\int_A^B d\vec l.
$$
The rest of the expression is just constants and vector structure. Please let the OP think about this.

 

[Delta2]

It is obviously correct in the general case. The integral is
$$
\int_A^B d\vec l.
$$
The rest of the expression is just constants and vector structure. Please let the OP think about this.

I have to think it too cause your "obviously" is not so obvious to me. So the integral $\int_A^B d\vec{l}$ along any curve depends only on the edge points A,B of the path of the curve on which we integrate? Not so obvious to me really.

 

[Delta2]

Ok I didn't expect this, the integral $$\int_A^B d\vec{l}$$ is indeed independent of path but the integral $$\int_A^B |d\vec{l|}$$ depends on path. Math can be really unexpected some times.

 

[Aurelius120]

How did you get that from $\vec {dl}$?

When I first read this reply, I thought and started going:
$$\frac{dx}{dy}=y$$
$$dl=\sqrt{(dx)^2+(dy)^2}$$$$\implies \frac{dl}{dy}=\sqrt{\left(\frac{dx}{dy}\right)^2+1}=\sqrt{y^2+1}$$
Then I would have gone:
$$\int....\frac{x\hat i+y \hat j}{\sqrt{x^2+y^2}}....dl$$$$=\int....\frac{x\hat i+y \hat j}{\sqrt{x^2+y^2}}....\frac{dl}{dy} dy$$$$=\int....\frac{x\hat i+y \hat j}{\sqrt{x^2+y^2}}....\sqrt{y^2+1}dy$$

The general expression for the infinitesimal element of the position vector along a curve in 3D space is $$d\vec{l}=\vec{i}dx+\vec{j}dy+\vec{k}dz$$ from which you can get the correct dl for your case by observing that this parabola lies exclusively in the xy-plane.

This comment saved the day.

 

[Orodruin]

Ok I didn't expect this, the integral $$\int_A^B d\vec{l}$$ is indeed independent of path

You did not expect that when you sum up the infinitesimal displacements along a curve you would get the displacement between the end points?

The path independence is also trivially true from Stokes’ theorem.

 

[Delta2]

You did not expect that when you sum up the infinitesimal displacements along a curve you would get the displacement between the end points?

The path independence is also trivially true from Stokes’ theorem.

No I didn't expect it because I know the integral $$\int |d\vec{l}|$$ depends on path so I expected the integral $$\int d\vec{l}$$ to depend too. The raw thought is like I mean the modulus of a vector depends on path hence the vector should depend on path too , but ok as it is usual the case the raw sloppy thoughts driven by intuition are quite often wrong in mathematics where you got to think with razor sharp accuracy and rigor.

 

[Orodruin]

No I didn't expect it because I know the integral $$\int |d\vec{l}|$$ depends on path so I expected the integral $$\int d\vec{l}$$ to depend too. The raw thought is like I mean the modulus of a vector depends on path hence the vector should depend on path too , but ok as it is usual the case the raw sloppy thoughts driven by intuition are quite often wrong in mathematics where you got to think with razor sharp accuracy and rigor.

I disagree. I think it is very intuitive that if you sum up the displacements along a curve, you get the total displacement. If you sum up the sizes of the displacements you get the path length, which is quite clearly path dependent. It is average speed vs average velocity all over again.

 

[Delta2]

You have better intuition than mine :D

 

[Orodruin]

I think you just need to stop and think about what the integrals actually represent. Once you do it is not rocket science.

 

[Aurelius120]

It is obviously correct in the general case. The integral is
$$
\int_A^B d\vec l.
$$
The rest of the expression is just constants and vector structure. Please let the OP think about this.

Does this mean that magnetic force on any weirdly twisted wire in uniform magnetic field can always be reduced to force on a single line of current?

 

[Orodruin]

Does this mean that magnetic force on any weirdly twisted wire in uniform magnetic field can always be reduced to force on a single line of current?

Yes!