Question

One long wire carries current 16.0 A to the left along the x axis. A second long wire carries current 72.0 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?

(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.)

(c) 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)
Amperes law is given by the relation between magnetic field and current as
? B ? ds = µ_o*I

From the given direction and magnitude of the currents, the magnetic field will be zero would be along some line parallel to y < 0.

thus we have

B₁*[2?*(0.280 - y)] = µ_o*(72.0 )

and

B₂*[2?(-y)] = µ_o*(16.0 )

B₁ = B₂

implies that

(72.0 A) / (0.280 m - y) = (16.0 A) / (-y)
or (72.0 A) / (y - 0.280 m) = (16.0 A) / y
(72.0 )(y) = (16.0 )(y - 0.280 m)

56*y = -4.48
y = -0.08 m. at this point, the total magnetic field will be zero.

(b) we have
B₁ *[2?(0.280 - 0.100 m)] = µ_o*(72.0 A)
B₁ = (2 x 10^-7 T.m/A)(72.0 A)/(0.180 m)
B₁ = 800 x 10^-7 = 8 x 10^-5   T

also,

B₂*[2?(0.100 m)] = µ_o*(16.0 A)

B₂ = (2 x 10^-7 T.m/A)(16.0 A)/(0.100 m)
B₂ = 320 x 10^-7 = 3.2 x 10^-5   T

Total magnetic field due to these two is
B = 8.0 x 10^-5 + 3.2 x 10^-5   T
B = 11.2 x 10^-5 T(-z direction)

the magnetic force is given by
F = qv x B

(c) force is related to the electric field E by the relation

F = qE

The force due to the electric field will be equal and opposite to that by the magnetic field.

thus