9. 0/3 points | Previous Answers SerPSE9 30.P.014.MI. My Notes Ask Your Teacher

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9. 0/3 points | Previous Answers SerPSE9 30.P.014.MI. My Notes Ask Your Teacher One long wire carries current 20.0 A to the left along the x axis. A second long wire carries current 68.0 A to the right along the line (-0.280 m, z-0) (a) Where in the plane of the two wires is the total magnetic field equal to zero? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m (b A particle with a charge of 2.00 c is moving with a velocity of 150 Mm sai ng the line y-010 2 0 Calculate the veto 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. N/C Need Help?Read ItMaster Submit Answer Save Progress Practice Another Version

Explanation / Answer

a)For magnetic field to be zero at any point

B1 = B2

u0 I1/2 pi r1 = u0 I2/ pi r2

68/(0.28 - y) = 20/y

68y = 5.6 - 20 y

88y = 5.6 => y = 0.064 m

Hence, y = 0.064 m = 6.4 x 10-2 m

b)We need the magnitude of the B field where the charge is moving

B = u0 I2/ 2 pi r2

B2 = 4 pi x 10^-7 x 68/2 pi x (0.28 - 0.1) = 7.55 x 10^-5 T

B1 = 4 pi x 10^-7 x 20/2 pi x (0.1) = 4 x 10^-5 T

Bnet = B1 + B2 = (7.55 + 4) x 10^-5 = -1.155 x 10^-4 T

F = q v B

F = 2 x 10^-6 ( 150 i x 1.155 x 10^-4 k ) = -3.46 x 10^-8 j

Hence, F = 0i - 3.46 x 10^-8 j + 0k

c)We know that F = q E

E = F/q

E = - 3.46 x 10^-8 j/2 x 10^-6 = 1.73 x 10^-2 j

Hence, E = -1.73 x 10^-2 N/C

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