Physics  Electromagnetism
posted by Robert .
This is the last question for a final exam preparation sheet and I can't figure out the answer. Any help would be greatly appreciated. Thank you.
 One very long wire carries current 10.0 A to the left along the x axis. A second very long wire carries current 75.0 A to the right along the line
(y = 0.280 m, z = 0).
(a) Where in the plane of the two wires is the total magnetic field equal to zero? (along the y axis)
(b) A particle with a charge of 2.00 µC is moving with a velocity of 150 Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle.
(c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.

(a)
Magnetic field of the long current carrying wire according to BiotSavert Law is
B=μₒI/2•π•a
The desired point is separated by distance “x” from the 1st current and by distance (d+x) from the 2nd current (along yaxis).
μₒI1/2•π•x=μₒI2/2•π•(d+x),
10•(0.28+x)=75•x
2.8+10x=75 x.
65x=2.8
x=0.431 m.
(b) Magnetic force (Lorentz force) is
F=qvB sinα.
Since the motion of the particle is along the straight line, then F=0
(c) Electric field has to be applied along the direction of the particle motion (E↑↑ῡ, E↑↓ῡ) 
thank you Elena!

For people of the future, Elena's answer for part B is wrong. It's better to use the equation
F = q*v(cross)B
where B is the sum of the B field created by the 50 A wire and the 30 A wire.
> 50A

 .18m

> v = 150E6 m/s

 .1m
< 30A
B_tot = mu/2pi * (50/.18 (k) + 30/.1 (k))
= 1.2E4 (k) Teslas
Then you take the cross product and eventually get F = .035j N
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