I am having trouble finding the equation for the magnetic field inside of a current loop, can anyone help me?

- Caboose

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- anonymous

I'm hoping radar will answer it, because I'm curious to see what the loop is and what the answer turns out to be. I thought magnetic fields go clockwise, if the current is going left to right, but I'm not sure and don't have my books around me.

- anonymous

If radar doesn't want it, I'll take it.

- anonymous

so as long somebody does, I'm dying to find out, I'm sure Caboose is too,

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## More answers

- radar

Go ahead lolisan, I don't know what the desired unit is, gilberts oersteds etc.

- anonymous

The Biot-Savart Law gives the differential for the magnetic field to be\[dB=\frac{\mu_0IdL \times r}{4\pi R^2}\]where dB, dL and r are vector quantities. The symmetry allows us to write the vector cross product component as\[dL \times r = dLr \sin \theta=dL r\]since the angle between the position vector and current element is always 90 degrees. You can then write\[B=\frac{\mu_0I}{4\pi R^2} \int\limits_{L}dL=\frac{\mu_0I}{4 \pi R^2}.2 \pi R=\frac{\mu_0I}{2R}\]

- anonymous

It's taken that the point at which the magnetic field is desired is in the centre of the loop.

- anonymous

The 'r' above is a unit vector in the direction from the current loop to the point where the magnetic field is to be calculated (here, the centre).

- radar

For daomowon for the direction of the field take (not literally) the right hand with your thumb pointing in the direction of conventional current your circled fingers will be in the direction of the magnetic field of course if it is alternating current the magnetic field is alternating in direction also.

- anonymous

in ac it alternates circularly right?

- anonymous

Yeah, the field points from the centre of the loop along the normal to the plane of the loop.

- radar

conventional is in the opposite direction of electron flow i.e. conventional flows from positive to negative, while electron flow (and a lot of people call this current) is from negative to positive.

- anonymous

Yes.

- radar

in ac it periodically reverses

- anonymous

thanks for the insights fellas, I can't wait to get into it fully

- Caboose

What if the current loop is centered in the x-y axis, and the electrical field is to be calculated along the x-axis, from the center to the loop. Essentially I need to find the electrical field at any point on the x-axis, inside the loop.

- anonymous

You said magnetic field.

- Caboose

sorry, magnetic field

- Caboose

my mistake.

- anonymous

You use symmetry considerations to kill off vector components.

- radar

I went basic radar in 1970 and it was covered, but memory fades.

- Caboose

Am I interpreting this correctly? Because the points are on the x-axis, the magnetic field vector will only have a z component.

- Caboose

- anonymous

I can take you through it, but I'm going to need a medal.

- radar

If the wire carrying current was aligned with the x axis picture the magnetic field expanding in circules vertical to the x axis

- Caboose

I'd really appreciate it, thanks. Eventually I am looking to make a plot of the Magnetic Field against the distance from the center, stopping at the loop. I assumed that the vectors would only have a Z component, and the magnitude would decrease the farther away you move from the center. I have to write the program in MatLab, which isnt much of a problem, I am just having trouble finding the equation of the magnetic field, at each point on the x-axis.

- radar

If the conventional current was flowing in a direction from right to left the magnetic lines of forece would be coming out of the page in -y side and rotate over and enter the page on the +y side.

- anonymous

Do you have a picture of what this setup looks like?

- Caboose

Yea, give me a sec.

- anonymous

It's for your reference for when I start going through it.

- anonymous

For your reference, I have a loop sitting in the y-z plane, and the x-axis is perpendicular to this plane...okay?

- anonymous

brb...

- Caboose

This is how I pictured it, flat on the x-y plane, with the Z axis being vertical.

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- anonymous

Pick a point on the loop (probably on y-axis for ease) and have your current vector, IdL, moving anti-clockwise. Draw a vector, r, from the point on the loop to an arbitrary point on z and label the angle between r and the z-axis to be theta. Update your picture so I know we're on the same page.

- Caboose

Like this?

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- anonymous

Sorry...

- anonymous

Yeah, like that.

- Caboose

Okay, whats next?

- anonymous

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- anonymous

This is - and now for the explanation.

- anonymous

The current element along the loop, IdL, is tangent to the loop and perpendicular to the vector, r (by geometry of the problem).

- anonymous

The magnetic field, dB, according to the Biot-Savart Law is perpendicular to both IdL and r.

- anonymous

The magnitude of the magnetic field is\[|dB|=\frac{\mu_0}{4\pi}\frac{I|dL \times r|}{r^2}=\frac{\mu_0}{4\pi}\frac{IdL}{r^2}=\frac{\mu_0}{4\pi}\frac{IdL}{z^2+R^2}\]where R is the radius on the loop.

- anonymous

\[r^2 = z^2+R^2\]by Pythagoras' Theorem, and\[|dL \times r|=dL\]since dL and r are perpendicular to each other and r is a unit vector.

- anonymous

Are you okay so far?

- Caboose

Yes I follow so far.

- anonymous

Now, because of the symmetry, when you move the vector element of current along the loop and draw the corresponding position vector r, and *then* draw in the component of the magnetic field, dB, you should see that the component vectors of dB (i.e. dB_z, dB_y) are such that dB_y's cancel off pair-wise (opposite on the loop) whereas the dB_z's keep adding.

- Caboose

That makes sense

- anonymous

From the diagram, \[dB_z=dB \sin \theta\]but\[\sin \theta = \frac{R}{\sqrt{z^2+R^2}}\]so that\[dB_z=dB \frac{R}{\sqrt{z^2+R^2}}\]

- anonymous

Hence combining everything,\[dB_z=\frac{\mu_0I R}{4\pi (z^2+R^2)^{3/2}}dL \rightarrow B_z=\frac{\mu_0I R}{4\pi (z^2+R^2)^{3/2}} \int\limits_{L}dL \rightarrow B_z=\frac{\mu_0I R}{4\pi (z^2+R^2)^{3/2}}.2\pi R\]that is\[B_z=\frac{\mu_0I R^2}{2 (z^2+R^2)^{3/2}}\]along the z-axis of the loop.

- anonymous

Hello?

- Caboose

okay i follow you there, now what if I want to find the vector at this point? In the blue

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- anonymous

I can possibly look at it later, but I really have to go right now. Just try to set the problem up as per above and look for any symmetry that may be of use :)

- Caboose

Thank you so much!

- Mendicant_Bias

I don't understand how the integration of dL yielded a 2*pi*R.Could somebody explain this if possible? @radar

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