Magnetic Field Homework Question

[wr1015]

Two long, straight wires are oriented perpendicular to the computer screen, as shown in Figure 22-43, in which L = 6.0 cm. The current in one wire is I1 = 3.7 A, pointing into the screen, and the current in the other wire is I2 = 4.0 A, pointing out of the screen. Find the magnitude and direction of the net magnetic field at point P.


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ok so first off i used pythagorean theorem to find the straight line distance from point to the wire coming out of the page and used that as $$r_{2}$$. then, i used the formual for magnetic force to find the force on point P from the two wires $$B = \mu_{0}I_{1}/2 \pi r_{1}$$ and $$B= \mu_{0}I_{2}/2 \pi r_{2}$$ and added the results. The answer I'm getting is wrong.. any suggestions?? i know the current in each of the wires are going in opposite directions so the forces repel but how does that relate to a point??

 

[wr1015]

for a more detailed description of what i did:

$$B= (4 \pi x 10^-7)(3.7A)/(2 \pi (.06m) + (4 \pi x 10^-7)(4.0A)/(2 \pi (.0848m))$$ which gave $$2.1767x10^-5 T$$

 

[Meir Achuz]

You have to add the two B's vectorially. B1 is horizontal.
B2 makes and angle of 45 degrees.

 

[wr1015]

You have to add the two B's vectorially. B1 is horizontal.
B2 makes and angle of 45 degrees.

like this??:

$$I_{1}_{y}$$= (4$$\pi$$x 10$$^-7$$)(3.7)/(2$$\pi$$(.06)) which gives 1.233x10$$^-5$$

$$I_{1}_{x}$$= 0

then, $$I_{2}_{y}$$= (4$$\pi$$x 10 $$^-7$$)(4.0)/(2$$\pi$$(.0848)(sin 45$$^0$$)

$$I_{2}_{x}$$= (4$$\pi$$^-7[/tex](4.0)/(2$$\pi$$(.0848)(cos 45$$^0$$)

then use pythagorean theorem again to find the total force??

 

[Meir Achuz]

I_1y=0. I_1x=what you have for I_1y.
I_2x and I_2y are both negative.
Then use Phyth for magniktude of B.
You should know the right hand rule to give you the correct direction of
B from each wire.