How Does the Biot-Savart Law Apply to Forces Between Wire Segments in a Loop?

[Scarlitt14]

1. A circular loop of wire of radius R = 5.14 m lies in the xy-plane, centered about the origin. The loop is carrying a current of I = 6.91 A flowing in counterclockwise direction. Consider two l = 1.52 mm segments of the loop: one centered about the positive x-axis, the other centered about the positive y-axis. Hint: Use Biot-Savart law. The permeability of free space is 1.25664 × 10−6 Tm/A. What is the magnitude of the force the first exerts on the second? Answer in units of N.

Attached is a copy of the illustration for the problem.

2. dB = Uo*i*ds×r/(4*pi*r^2)


3. To be honest, I'm not even really sure where to start with this problem! Any assistance would be appreciated! Thank you!

 

[gneill]

Start by finding the magnetic field at the location of the second segment due to the first segment. This will employ Biot-Savart. Note that the lengths of the segments are so small that you can probably forgo integrating, just use the segment length as the magnitude of ds, and the dB will become the B.

In your line 2, you're crossing ds with the radius vector rather than a unit vector in the direction of r, so you'll need another magnitude of r in the denominator. Have a look here:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/biosav.html

 

[Scarlitt14]

Ok, so assuming that ds can be viewed as just s since the segment is so small and dB becomes just B; I got:

B=($$\mu$$₀Is)/(4$$\pi$$r$$^{3}$$)
Where r=$$\sqrt{2R^{2}}$$=$$\sqrt{2}$$ R

And then to get the force the equation F=I$$\vec{l}$$x$$\vec{B}$$

In this case l is the s mentioned above and that gives:

F={$$\mu$$₀I²s²}/{4$$\pi$$r³}

Does that seem right? I'm not really sure if I'm on the right track or not. Thanks SO much for your help!

 

[gneill]

At least one of the cross products will involve a sin(θ) term, and I think you might have "lost" an r in the numerator (which will cancel with one of the r's in the denominator). Also keep in mind that the force will have a direction, which you'll need to work out from the cross product operations.

 

[Scarlitt14]

Since the angle is 90 between the two segments sin(90)=1 so I dropped that out, and the problem also asked for magnitude so I wasn't worried about the resultant vector from the cross product. You said that I lost and r somewhere? Does that mean that my final equation should end up as:

F = ($$\mu$$₀I²s²)/(4$$\pi$$r²)

I tried that and I didn't get the right answer.

 

[bobasp1]

This might help you, same question

Field at the point on the y-axis due to the current element on the x-axis is
dB = μο/4π*i*ds*sinθ/r^2. Here θ is 45 degrees. Direction of the field is perpendicular to the current element on the y-axis. Hence the force on the current element is
df = dB*i*ds.

https://www.physicsforums.com/showthread.php?t=323251

 

[gneill]

Going back to the original Biot-Savart law,

$$\vec{dB} = \frac{\mu_0 I \vec{ds} \times \vec{I_r}}{4 \pi r^2}$$

where I_r is a unit vector in the direction of the r vector. If you choose to use the r vector itself rather than the (unitless) unit vector, it becomes:

$$\vec{B} = \frac{\mu_0 I \vec{ds} \times \vec{r}}{4 \pi r^3}$$

Note that in this case the r vector makes an angle of 45° with the ds vector; the ds vector points in the +y direction, while the r vector points from (R,0) to (0,R).

So in our case this becomes:

$$\vec{B} = \frac{\mu_0 I \; ds \; r sin(45°)}{4 \pi r^3} = \frac{\mu_0 I \; ds \; \sqrt{2}/2}{4 \pi r^2}$$

Plugging in values, the magnetic field at the first segment due to the second segment is

B = 1.406 x 10^-11 T pointing out of the page (in the +z direction).

When you then use F = I L x B to find the force, then the IL and B vectors are at right angles and the sine of the angle for that cross product is 1.