Current in wires

1. Feb 28, 2005

waywardtigerlily

Ok, I have been working on this problem for about 2 hours now, and I still can't get the right answer..could someome help me?

two wires carry currents of I=5.12A in opposite directions and are separated by a distance of d0=9.33cm. (the wire on the left has I going up and the one on the right is going down) Calculate the net magnetic field at a point midway between the wires. Use the direction out of the page as the positive direction and into the page as the negative direction in your answer.

B. Calculate teh net magnetic field at point p1- that is 9.32 cm to the right of the wire on the right.

C. calculate the net magnetic field at point p2- that is 19.3 cm to the left of the wire on the left.

To calculate B. I am using:

BR=uoI
2pi (p1/2)

BR= (4pi x 10^-7)(5.12)
(2Pi)(.0466)

= 2.19742 x 10^-5 T

BL=uoI
2pi(p1+do)

BL= (4pi x 10^-7)(5.12)
2Pi(.1865)

Bnet= BR-BL
= 1.65 x 10 ^-5 T

Can anyone see what I am doing wrong? I assume for C you would use the same procedure and for the main question you would just use Bnet= B1+B2
Thanks!

2. Feb 28, 2005

Gamma

You did not write the division sign. Looks like it's a typo.

Any way BR = UoI / 2pi (p1). where p1 = 9.31 cm. Rest of your calculations seems to be ok.

3. Feb 28, 2005

Andrew Mason

I am not sure what you are using for part a. You have to be careful with the directions r. Use:

(1)$$B = \frac{\mu_0I}{2\pi r}$$

For a., r1 = d0/2 and r2=-d0/2, so:

$$B_{mid} = \frac{\mu_0}{\pi d_0}I + \frac{\mu_0}{\pi (-d_0)}(-I) = \frac{2\mu_0}{\pi d_0}I$$

For part b., r1 = d0 and r2 = 2d0 (ie. both positive) . When you work out the fields, you have to subtract them due to the different directions of the currents

For part c., r1 = -(d0 + 10) and r2 = -(2d0 + 10) (ie. both negative) . When you work out the fields, again you have to subtract them due to the different directions of the currents.

AM

Last edited: Feb 28, 2005