# Magnetic Field at the center of a square

1. Jun 19, 2013

### aChordate

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

2. Relevant equations

not sure

3. The attempt at a solution

I am not sure how to start this problem.

2. Jun 19, 2013

### rude man

3. Jun 19, 2013

### Goddar

You've probably seen how to obtain the field of a single line, whether through the Biot-Savart law or, more simply, through Ampere's law. Then just remember that the fields of individual components of a configuration add-up...

4. Jun 20, 2013

### aChordate

So, would I use Ampere's Law for static magnetic fields?

ƩB||Δl=μ0I ?

5. Jun 20, 2013

### rude man

Yes, since the wires are "long", you can use Ampere's law individually for each wire and then add the results vectorially.

6. Jun 20, 2013

### aChordate

Ok, so I used ƩB||Δl=μ0I

and got:

ƩB||=3.8x10^-5
ƩB||=4.3x10^-5
ƩB||=5.0x10^-5
ƩB||=3.0x10^-5

For x,
(3.8x10^-5)+(-5.0x10^-5)=-1.2x10^-5

For y,
(-4.3x10^-5)+(-3.0x10^-5)=-7.3x10^-5

I have a feeling I didn't do this correctly...

7. Jun 20, 2013

### rude man

Let's take one wire at a time, and show all your work.
For the wire carrying the 1.2A, write ampere's law and determine B at the center of the square.

8. Jun 20, 2013

### aChordate

Ok, so I used ƩB||Δl=μ0I

and got:

ƩB||=[(4∏*10^-7T*m/A)(1.5A)]/0.05m
ƩB||=3.8x10^-5

ƩB||=[(4∏*10^-7T*m/A)(1.7A)]/0.05m
ƩB||=4.3x10^-5

ƩB||=[(4∏*10^-7T*m/A)(2.0A)]/0.05m
ƩB||=5.0x10^-5

ƩB||=[(4∏*10^-7T*m/A)(1.2A)]/0.05m
ƩB||=3.0x10^-5

9. Jun 20, 2013

### aChordate

Ampere's Law:
ƩBΔl=μ0I

ƩB(0.025m)=(4∏*10^-7)(1.5)

ƩB=7.54x10^-5

10. Jun 20, 2013

### rude man

Ampere's law refers to the integration of a path around the current-carrying wire, i.e in this case you want the circumference, not the radius.

11. Jun 20, 2013

### aChordate

ƩB(0.050m)=(4∏*10^-7)(1.5A)
ƩB=3.77x10^-5

ƩB(0.050m)=(4∏*10^-7)(1.7A)
ƩB=4.27x10^-5

ƩB(0.050m)=(4∏*10^-7)(2.0A)
ƩB=5.03x10^-5

ƩB(0.050m)=(4∏*10^-7)(1.2A)
ƩB=3.02x10^-5

12. Jun 20, 2013

### aChordate

Do I add these all together?

13. Jun 20, 2013

### rude man

They were wrong before & are still wrong.
In your post #9 you seemed to realize that the radius is 0.025m, not 0.05m. Now you're back to 0.05m?

And you still haven't understood that the path of integration is not along the radius but the circumference.