Finding Magnetic Flux Through a Wire Frame

[VishalChauhan]

Homework Statement

A square wire frame,in the x-y plane, carrying a current 'I' and with side length 'A' is kept at a distance 'A' from the y-axis . Find the magnetic flux through the other half of the plane(not containing the square loop).

Homework Equations

The magnetic field due to line charge is μi/4πr(sinø1+sinø2)
The magnetic field due to a dipole is μM/2πl3

The Attempt at a Solution

I tried to find the magnetic field at a given distance from from the loop,but it was too clumsy.Symmetry considerations won't work(though i guess its the only way to solve this problem)

 

[TSny]

Hello, VishalChauhan.

Symmetry considerations will be helpful. Which sides of the loop will produce zero net flux through the half plane?

For the other sides, it looks like you will need to work through the integrations. Can you show us more detail of your attempt to set up the magnetic field at a point in the half plane?

[EDIT: I've intepreted the problem as shown in the attached figure.]

 

[VishalChauhan]

Hello TSny
The total magnetic flux due the horizontal wires (parallel to the x-axis will be zero).As far as the vertical wires are concerned,the magnetic field at a perpendicular distance x will be μi/4πx(sinθ1+sinθ2).

This is where i am stuck. I could take a vertical line of thickness dx as my differential area element,calculate the flux through it and integrate for the rest of the half-plane , but i can not understand how to go about it.

 

[TSny]

OK, you have the right idea. So if you pick a little patch of area dxdy in the half plane at coordinates (x,y) you need to express the magnetic field at that patch in terms of x, y, and A.

So in the expression $\frac{\mu_0 I}{4 \pi d}(\sin\theta_1 - \sin \theta_2)$ try to express $d, \sin \theta_1$, and $\sin\theta_2$ in terms of x, y, and A.

 

[VishalChauhan]

d will be x+a, tanø will be a/2(x+a)[(x+a)^2+y(y-a/2)] {with minus a for the other angle).But converting this into sin form will be very clumsy.Maybe we can use trig identities to convert the expression into tan form?

 

[TSny]

tanø will be a/2(x+a)[(x+a)^2+y(y-a/2)]

I don't see how you're getting that expression. Can you describe the angle ø₁ as it would be draw in the figure for side #1 of the loop and also describe the right triangle that you are using to express tanø₁?

 

[VishalChauhan]

I'm sorry, i jumbled up a bit in the angles.
Sinθ1 will be (a/2-y)/l, where l^2=(x+a)^2 + (y-a/2)^2.similarly for sinθ2

 

[TSny]

OK. I get (y - a/2)/l instead of (a/2 - y)/l. But that might just be a difference in how we choose the sign of the angle.

Now you can set up an expression for the flux through the patch of area. Then you can decide which will be easier: to integrate over x first or to integrate over y first.

 

[VishalChauhan]

I made the required integral expression and integrated with y first. what i get is (some factor of x+a)*log[r^2+(y+a/2)^2]-similar exp with y-a/2. But putting limits as -∞ to +∞ leaves me with a zero.

 

[TSny]

I don't get a log function for the y integration. Can you write out the integrand for y integration?

 

[VishalChauhan]

Hello again TSny.
You're right. There is no logarithm.I integrated with l^2 by mistake. The actual integarted expression is some factor of (x+a)*[L1-L2] , where (L1)^2 is (x+a)^2+(y+a/2)^2 and L2 is a similar expression with y-a/2

 

[TSny]

When you say "some factor of (x+a)", do you mean some factor of 1/(x+a) ? Otherwise, I agree with your result so far. Next thing is to evaluate at the y-limits of integration.

 

[VishalChauhan]

When you say "some factor of (x+a)", do you mean some factor of 1/(x+a) ? Otherwise, I agree with your result so far. Next thing is to evaluate at the y-limits of integration.

Yes, i mean 1/(x+a).
But now if i were to put the y limits as -∞ to +∞, would i not end up with a zero?

 

[TSny]

But now if i were to put the y limits as -∞ to +∞, would i not end up with a zero?

No. You have to be careful evaluating at -∞. I find it easier to integrate from 0 to ∞ and double the result.

 

[VishalChauhan]

I'm sorry, but i don't follow you. Even if i were to integrate from 0 to infinity, the integrated expression is 0 at infinity and zero again at y=0.

 

[TSny]

I don't get 0 at y = infinity.

 

[VishalChauhan]

I thing i figured out my mistake.
The expression i get is some constants*1/(x+a) dx.

 

[TSny]

Yes.

 

[VishalChauhan]

So on integrating wrt x i get log(x+a), with limits as 0 to infinity

 

[VishalChauhan]

If i were to put infinity in this expression, would the expression not become infinity?

 

[TSny]

Before evaluating at x = infinity, combine with the expression from the opposite side of the loop.

 

[VishalChauhan]

I get my final ans as μialog2/8∏

 

[TSny]

That's close to what I get. Did you remember to double you answers when integrating from 0 to infinity instead of -infinity to infinity?

 

[VishalChauhan]

Oops! You are right, I didn't. So the final ans should be μialog2/(4∏).

 

[TSny]

I integrated twice from 0 to infinity, so I get 2π in the denominator. I haven't checked my work too carefully, so I could be off by a factor of 2 somewhere else.

 

[VishalChauhan]

I am not so sure, but maybe you used μi/2πr(sinø1+sinø2) in the formula instead of μi/4πr(sinø1+sinø2).(I make this mistake often due to similarity with the simpler μi/2πr formula)
Or then, I may have missed something myself.
Anyway, I learned a few cool tricks from this question.Thanks for sticking with me and helping me out.
(I'll try and solve this again thoroughly and see if I can figure out anything else.)