Magnetic Field at the Center of a Wire Loop

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:

[tex]\frac{(\mu_0)I}{2(\pi)r}[/tex]

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

Thank You.

Reshma

Integrate the magnetic field around the circular path of radius r.
[tex]\oint \vec B \cdot d\vec r = ?[/tex]

Doc Al

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.)

cse63146

Isnt the equation I posted the Biot-Savart law?

Doc Al

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

cse63146

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

B = [tex]\frac{\mu_0}{4\pi}[/tex] [tex]\frac{q(v X r}{r^2}[/tex]

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

Doc Al

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

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.

cse63146

Would it be

[tex]\vec{B} = \frac{\mu_0 I d}{4 \pi r^2}[/tex]

and then integrate that?

Doc Al

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

[tex]d\vec{B} = \frac{\mu_0 I}{4 \pi r^2}\;d\ell[/tex]

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

cse63146

is the [tex]d \ell[/tex] distance*length or the derivative of length.

Then I would [tex]\oint \vec{B} dr[/tex] like Reshma said?

Doc Al

Neither. [tex]d \ell[/tex] 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.