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.

Question

- 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.

Answer 1

(a)
Magnetic field of the long current carrying wire according to Biot-Savert 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 y-axis).
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)

Answer 2

thank you Elena!

Answer 3

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))
= 1.2E-4 (-k) Teslas

Then you take the cross product and eventually get F = -.035j N