Q: A length of wire is formed into a closed circuit with radii a and b, as shown in the figure, and carries a current I.

(i) What is the magnitude and direction of the magnetic field, B at point P which is located at the center of both circular regions?

(ii) Determine the magnetic dipole moment of this circuit.

Solution:

Attack this problem as you would a large pizza: divide and conquer. The first goal is to find the magnetic field at the center due to the current running through the entire wire. It will be an easier job to break the loop into 4 segments, calculate B individually for each wire segment, and add them up.

Then, the magnetic field due to the whole loop is just the sum of the parts:

Since we'll be adding magnetic fields along strange and unusual paths, the equation to use is the Biot-Savart Law:

dl A little piece of wire. dB A little piece of magnetic field cased by the current going through dl r The distance from dl to the point P. The direction of a unit vector which points from P to dl. i The current which flows through the wire. 0, , 4 Some silly constants that you can look up in the back of your textbook.

WAIT! Do you recall seeing a different version of the Biot-Savart Law? There is a similar and common way to write the same equation. If you're interested, check out the following note about it.

A lot of effort can be saved by look at the cross product. The cross product is easy in cases where the two items being crossed are parallel to each other (or anti-parallel). In that case, the cross product is 0, and there's nothing easier to deal with than 0 (just ask those cosmologists about singularities).

dl is a vector which points in the direction of the current along any tiny wire segment. is just a direction to cross dl with. It points radially from point P to the segment dl. Take a look at how these vectors relate to each other at various points along the wire:

Now, before you go screaming for the hills, take a good look at the diagram to the right. Along the straight segments, (2) and (4), and dl point along the same line, so dl x = 0, and there is no contribution to the B-field from those wire segments.

Are we going to be that lucky along the circular wire segments? 'Fraid not. But there still is good news. At every point along the curved wires, dl is perpendicular to . You see, the magnitude of the cross product is :

(Keep in mind, is just a unit vector - it has a magnitude of 1.)

The fact that the two vectors are perpendicular means that = 90^o everywhere along the circular wire! (If you're not jumping for joy, perhaps you have forgotten the little known fact that sin(90^o) = 1). The cross product reduces down to dl.

Now we can get busy! We can now do some calculating! Along wire segment (1):

	Since the radius is equal to b everywhere in the circular segment, we can just stick that value in.
	To get the total B-field contribution, we need to integrate over the semicircle. We can pull all constant values outside of the integral sign. Then we add up all of the tiny length segments. The result is the length of a semi-circle (half of 2b). Things cancel out rather nicely and we're left with a nice, neat result for the magnitude of the B-field caused by the upper circular segment.

Likewise, with the bottom semi-circle:

Things seem okay... BUT what's a magnetism question without the right hand rule? Placing your thumb in the direction of the current, the fingers curl into the page inside of the loop. The direction of the magnetic field inside of the loop due to segment (1) is into the page.

For loop segment (3) the fingers also curl into the page inside of the loop. This means that the magnetic fields created by both the top and bottom curved segments add together to create a field into the page.

Now onto the second part... The magnetic dipole moment is regular product of the current and the area inside the loop.

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