# Homework Help: Net magnetic field in wires

1. Feb 27, 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!

Last edited: Feb 27, 2005
2. Feb 28, 2005

### vincentchan

just wanna net u no, your notasion is hard to folow, plis don't write it dat way next time...
$$B_{right} = \frac{\mu_0 I}{2 \pi p_1}$$
and
$$B_{left} = \frac{\mu_0 I}{2 \pi (p_1+d_0)}$$

$$B_{total} = B_{right} - B_{left}$$
since $B_{right}$ and $B_{left}$ have different direction

your $B_{right}$ have a factor of one half at the bottom for some reason... is that because that is the formula for part a ??????
and yes, in general, the total B field is the sum of its component