Calculating Magnetic Field of a Conducting Loop Using Biot-Savart's Law

[formulajoe]

A square conducting loop of side 2a lies in the z=0 plane and carries a current in the counterclockwise direction. Show that at the center of the loop
H = sqrrt2*I/pi*a in the z direction.

I am stuck in this problem. Heres what I've got. I placed the center of the loop at the origin. I am using Biot-Savarts Law. I found an R vector to be sqrt x^2 + y2 in the rho direction and cos x/sqrt x^2 + y2 in the phi direction. Heres where I am stuck. When I go to do the integral, my dl depends on the side I am integrating. So if I am my dl is dx, i don't know how to cross that why R because R has the unit vectors rho and phi.

 

[Gamma]

I didn't quiet understand your approach. Find the magnetic field due to one wire of length 2a at a distance a from it's center. Contribution from each wire is same. So B = 4 * B(A).

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| a
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From the center of the wire at a distance l choose current element dl.
Distance between dl and A is r.

Using biot law,

B(A) = \frac{\mu_0}{4\pi}\int_{-a}^{a} \frac{I dl X n}{r^2}

B(A) = \frac{\mu_0 I}{4\pi}\int_{-a}^{a} \frac{dl sin \theta}{a^2+l^2}

sin \theta = a / \sqrt (a^2 + l^2)

Do the integration.

B(A) = \frac{\mu_0 I}{4\pi a}* \sqrt 2


H = 4* H(A) = \frac{I}{\pi a} * \sqrt 2

 

[formulajoe]

where does the sin theta come from in that second step

 

[Andrew Mason]

A square conducting loop of side 2a lies in the z=0 plane and carries a current in the counterclockwise direction. Show that at the center of the loop
H = sqrrt2*I/pi*a in the z direction.

I am stuck in this problem. Heres what I've got. I placed the center of the loop at the origin. I am using Biot-Savarts Law. I found an R vector to be sqrt x^2 + y2 in the rho direction and cos x/sqrt x^2 + y2 in the phi direction. Heres where I am stuck. When I go to do the integral, my dl depends on the side I am integrating. So if I am my dl is dx, i don't know how to cross that why R because R has the unit vectors rho and phi.

You only have to do one side because they all add together. Call them sides 1, 2, 3, 4.

B_1 = \mu_0I \int \frac{d\vec L \times \vec r}{4\pi r^2} = \frac{\mu_0I}{4\pi} \int \frac{rdlsin\theta}{r^2}

In polar coordinates:

dL = cos\theta d\theta
r = a/cos\theta

So:
B_1 = \frac{\mu_0I}{4\pi} \int_{-\pi/4}^{\pi/4} \frac{(a/cos\theta)cos\theta d\theta sin\theta}{(a/cos\theta)^2}

B_1 = \frac{\mu_0I}{4\pi} \int_{-\pi/4}^{\pi/4} \frac{asin\theta cos^2\theta}{a^2}d\theta


Since B = 4B_1:

B = \frac{\mu_0I}{\pi a} \int_{-\pi/4}^{\pi/4} sin\theta cos^2\theta d\theta

You have to use some trigonometric identities to integrate that. Better check my math too.

AM

 

[Gamma]

In my first integral, n is the unit vector along the direction of vector r. sin \theta comes in when you take the cross product of dl X n. sin \theta is the angle between r and dl.