Question

One long wire carries current 12.0 A to the left along the x axis. A second long wire carries current 80.0 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?
y = ________m

(b) A particle with a charge of −2.00 µC is moving with a velocity of 150î Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (Ignore relativistic effects.)
= ___________N

(c) A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.
= __________N/C

Answer 1

(a)
By Ampere's law,
∫ B ▪ ds = (µo)I

Given the directions and magnitudes of the currents, the only region in the xy-plane where the magnetic field would be zero would be along some line parallel to the currents with y < 0.

(Bt)[2π(0.280 m - y)] = (µo)(80.0 A) [y < 0 means 0.280 m - y > 0.280 m]

(Bb)[2π(-y)] = (µo)(12.0A)

(Bt) = (Bb)
(80.0 A) / (0.280 m - y) = (12.0 A) / (-y)
(80.0 A) / (y - 0.280 m) = (12.0 A) / y
(80.0 A)(y) = (12.0A)(y - 0.280 m)
72y = -2.80
y = -3.88 e-2 m

(b)
Find the magnetic field where the charge is moving.
(Bt)[2π(0.280 - 0.100 m)] = (µo)(80.0A)
(Bt) = (2e-7 T-m/A)(80.0A)/(0.180 m)
(Bt) = 8.88e-5 T

(Bb)[2π(0.100 m)] = (µo)(12.0A)
(Bb) = (2e-7 T-m/A)(12 h hi A)/(0.100 m)
(Bb) = 2.000e-5 T

B = 5.778e-5 + 2.000e-5 T
B = (7.778e-5 T) (-z direction)

F = qv x B
F = (-2.00e-6 C)<150 m/s, 0, 0> x <0, 0, -7.778e-5 T>
F = <0, (-2.00e-6 C)[0 - (150 m/s)(-7.778e-5 T)], 0>
F = <0, -2.33e-8 N, 0>

(c)
F = qE

The force due to the electric field will be equal in magnitude and opposite in direction to the force due to the magnetic field.
<0, +2.33e-8 N, 0> = (-2.00e-6 C)(E)
E = <0, -1.17e-2 N/C, 0>