# Magnetic Field at the Center of a Wire Loop

1. Mar 7, 2008

### cse63146

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

A piece of wire is bent to form a circle with radius r. It has a steady current I flowing through it in a counterclockwise direction as seen from the top (looking in the negative z direction).

What is B_z(0), the z component of B at the center (i.e., x = y = z = 0) of the loop?

Express your answer in terms of I, r, and constants like mu_0 and pi.

2. Relevant equations

3. The attempt at a solution

I know this equation:

$$\frac{(\mu_0)I}{2(\pi)r}$$

but there is a hint that says I need to find the Integrand.

Thank You.

2. Mar 7, 2008

### Reshma

Integrate the magnetic field around the circular path of radius r.
$$\oint \vec B \cdot d\vec r = ?$$

3. Mar 7, 2008

### Staff: Mentor

Biot-Savart law

That's the magnetic field from an infinite straight current-carrying wire.

Look up the Biot-Savart law. That will give you the field from a current element.

Right. Once you have the field from a current element, you'll need to integrate around the entire loop. (Since you are only asked to find the field at the center of the loop--as opposed to some arbitrary location--the integral will turn out to be quite doable.)

4. Mar 7, 2008

### cse63146

Isnt the equation I posted the Biot-Savart law?

Last edited: Mar 7, 2008
5. Mar 7, 2008

### Staff: Mentor

No. As I said, the equation you posted is the field from a long current-carrying wire. Look up the Biot-Savart law.

6. Mar 7, 2008

### cse63146

Sorry, about that, I was looking at the wrong equation in my book.

B = $$\frac{\mu_0}{4\pi}$$ $$\frac{q(v X r}{r^2}$$

since its circular motion B = $$\frac{qmv}{r}$$ <=Would I need to ingetrate this equation?

7. Mar 7, 2008

### Staff: Mentor

The one you want is in terms of current:
$$d\vec{B} = \frac{\mu_0 I d\vec{\ell}\times \hat{r}}{4 \pi r^2}$$

Figure out what that is for a point in the center of the loop, then integrate around the loop.

Not relevant; No circular motion here.

8. Mar 8, 2008

### cse63146

Would it be

$$\vec{B} = \frac{\mu_0 I d}{4 \pi r^2}$$

and then integrate that?

9. Mar 9, 2008

### Staff: Mentor

Almost. After taking care of the vector product, it would be:

$$d\vec{B} = \frac{\mu_0 I}{4 \pi r^2}\;d\ell$$

Integrate that around the loop. (It's easy!)

10. Mar 9, 2008

### cse63146

is the $$d \ell$$ distance*length or the derivative of length.

Then I would $$\oint \vec{B} dr$$ like Reshma said?

Last edited: Mar 9, 2008
11. Mar 9, 2008

### Staff: Mentor

Neither. $$d \ell$$ is an element of length around the circumference of the circle. (That should tip you off as to what the integral is. )

No. Integrate the expression I gave in the last post, which is the field at the center due to a small element of the current, over the complete loop.