# Magnetic flux problem

1. Jan 24, 2014

### VishalChauhan

1. The problem statement, all variables and given/known data
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).

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

3. 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)

2. Jan 24, 2014

### 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.]

#### Attached Files:

• ###### Square Current Loop.png
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Last edited: Jan 24, 2014
3. Jan 24, 2014

### 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.

4. Jan 24, 2014

### 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.

#### Attached Files:

• ###### Square Current Loop 2.png
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5. Jan 24, 2014

### 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?

6. Jan 24, 2014

### TSny

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?

7. Jan 24, 2014

### 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

8. Jan 24, 2014

### 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.

9. Jan 24, 2014

### 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.

10. Jan 24, 2014

### TSny

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

11. Jan 25, 2014

### 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

12. Jan 25, 2014

### 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.

13. Jan 26, 2014

### VishalChauhan

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?

14. Jan 26, 2014

### TSny

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

15. Jan 26, 2014

### 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.

16. Jan 26, 2014

### TSny

I don't get 0 at y = infinity.

17. Jan 26, 2014

### VishalChauhan

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

18. Jan 26, 2014

### TSny

Yes.

19. Jan 27, 2014

### VishalChauhan

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

20. Jan 27, 2014

### VishalChauhan

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