Magnetic Field at the Center of a Wire Loop

[cse63146]

Homework Statement

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.

Homework Equations

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.

 

[Reshma]

Homework Statement

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.

Homework Equations

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.

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

 

[Doc Al]

Biot-Savart law

I know this equation:

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

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.

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

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]

Isnt the equation I posted the Biot-Savart law?

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 = \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?

 

[Doc Al]

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}

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.

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

Not relevant; No circular motion here.

 

[cse63146]

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

Would it be

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

and then integrate that?

 

[Doc Al]

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!)

 

[cse63146]

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

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

 

[Doc Al]

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

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. )

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

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.