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One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y 5 0.280 m, z 5 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 22.00 mC is moving with a velocity of 150i ^ Mm/s along the line (y 5 0.100 m, z 5 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.

Respuesta :

Hashirriaz830

Answer:

y = -0.42 m

1.16*10^-4 T(-k)

-1.73*10^4 N/C (j)

Explanation:

(a) Above the pair of wires, The field out of the page of the 50 A current will be stronger than the (—k) field of the 30 A current (k).  

Between the wires, both produce fields into the page.  

below the wires, y = -  | y |

B = u_o*I/2Ï€r (-k) + u_o*I/2Ï€r (k)

0 = u_o/2πr[50/  | y |+0.28 (-k) + 30/| y | (k) ]

50 | y | = 30(| y | + 0.28)

| y | = -y

-50 y = 30*(0.28 - y)

y = -0.42 m

b)  B = u_o*I/2πr (-k) + u_o*I/2πr (k)

B = 4Ï€*10^-7/2Ï€[ 50/0.28 -1 (-k) +30/1(-k) ]

  = 1.16*10^-4 T(-k)

F = qv*B

F = (-2*10^-6)*(150*10^6(i) )(1.16*10^-4(-k))

F = 3.47*10^-2 N(-j)

c) F_e = qE

      E = F_e/q

E = 3.47*10^-2/-2*10-6

  = -1.73*10^4 N/C (j)

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