One long wire carries current 26.0 A to the left along the x axis. A second long

Question

One long wire carries current 26.0 A to the left along the x axis. A second long wire carries current 54.0 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.)

Vector F = _______________ N

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

Vector E = ___________ N/C

Explanation / Answer

a] By Ampere's law,

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)[2pi(0.280 m - y)] = (uo)(54.0 A) [y < 0 means 0.280 m - y > 0.280 m]

(Bb)[2pi(-y)] = (uo)(26 A)

(Bt) = (Bb)

(54 A) / (0.280 m - y) = (26 A) / (-y)

(54 A) / (y - 0.280 m) = (26 A) / y

(54 A)(y) = (26 A)*(y - 0.280 m)

(54 - 26)y = -26*0.280

y = -26*0.280/(54 - 26)

= -0.26 m

b] The magnetic field where the charge is moving.

(Bt)[2pi(0.280 - 0.100 m)] = (uo)(54 A)

(Bt) = (2e-7 T-m/A)(54 A)/(0.180 m)

(Bt) = 6*10^-5 T

(Bb)[2pi(0.100 m)] = (uo)(26 A)

(Bb) = (2e-7 T-m/A)(26 A)/(0.100 m)

(Bb) = 5.2*10^-5 T

B = 6 e-5 + 5.2 e-5 T

B = 11.2*10^-5 T (-z direction)

F = qv * B

F = (-2.00e-6 C)<150 m/s, 0, 0> x <0, 0, -11.2*10^-5T>

F = <0, (-2.00*10^-6 C)*[0 - (150 m/s)*(-11.2*10^-5 T)], 0>

F = <0, -3.36*10^-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, +3.36*10^-8 N, 0> = (-2*10^-6 C)(E)

E = <0, -1.68*10^-2 N/C, 0>

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