One very long wire carries current 25.0 A to the left along the x axis. A second

Question

One very long wire carries current 25.0 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? m (along the y axis)

(b) A particle with a charge of -2.00 µC is moving with a velocity of 150i Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. ( _i + _j + _k) N

(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. ( _i + _j + _k) N/C

Explanation / Answer

I = 25 A to the left (along -x-axis)
i = 75 A to the right (along +x - axis)
d = 0.28 b/w the wires
B due to long wire = 2kI/r

a. the B is additive for the region between the wires, so B != 0 between the wires
Now, the top wire has more current, so at any point above the top wire, B top > B bottom, so no point with net 0 B

at distance r from the bottom wire, B due to top wire = 2ki/(r+0.28)
B due to bottom wire = 2kI/(r)
Net B = 2ki/(r+0.28) - 2KI/(r) = 0
=> i/(r+0.28) - I/(r) = 0
=> 75/(r+0.28) - 25/(r) = 0
=> 75r - 21 - 25r = 0
=> 50r = 21
=> r = 0.42 m from the bottom wire (0,-0.42,0) m

b. q = -2*10^-6 C
v = 150i
B ( at y = 0.1) = 2k[I/0.1 + i/(0.28-0.1)] = 2*10^(-7)[25/0.1 + 75/0.18] k = 1.33*10^(-4) k T
Force on charge, F = qvxB = -2*10^-6 * 150i x 1.33*10^-4 k = 39.9*10^-9 j N

c. Force = 39.9*10^-9 k N
E applied should cancel this force
so qE = -39.9*10^ -9 = -2*10^-9 E
E = 19.95 k N/C

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