One very long wire carries current 40.0 A to the left along the x axis. A second
ID: 1451249 • Letter: O
Question
One very long wire carries current 40.0 A to the left along the x axis. A second very long wire carries current 80.0 A to the right along the line (y = 0.280 m, z = 0). Where in the plane of the two wires is the total magnetic field equal to zero? m (along the y axis) A particle with a charge of -2.00 A mu C is moving with a velocity of 150t Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. ( i + j + k)N 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/CExplanation / Answer
Given data:
I1=40A
I2=80A
(y, z)= (0.280, 0)
(a) Where in the y plane of the two wires is the total magnetic field equal to zero?
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.
from the Amperes law,
(Bt)[2(0.280 m - y)] = (µo)(80 A) [y < 0 means 0.280 m - y > 0.280 m]
(Bb)[2(-y)] = (µo)(40 A)
according to the given condition
(Bt) = (Bb)
(80) / (0.280 - y) = (40) / (-y)
y=-0.28 m
B) Calculate the vector magnetic force acting on the particle
Find the magnetic field where the charge is moving.
(Bt)[2(0.280 - 0.100 m)] = (µo)(80 A)
(Bt) = (2*10^-7 T-m/A)(80 A)/(0.180 m)
(Bt) = 8.88*10^-5 T
(Bb)[2(0.100 m)] = (µo)(40 A)
(Bb) = (2*10^-7T-m/A)(40 A)/(0.100 m)
(Bb) = 8*10^-5 T
B = 8.88*10^-5+8*10^-5
B = (16.88*10^-5T) (-z direction)
B= (16.88*10^-5)(- k)
v=150 i+0j+0k
F = qv x B
F=(-2*10^-6)((150 i+0j+0k) x (0i+0j+ (16.88*10^-5)(- k))
F=(-2*10^-6) (0i+2532*10^-5 j+0k)
F=( 0 i - 5064*10^-9 j +0 k )N
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 i - 5064*10^-9 j +0 k ) = (-2*10^-6) E
E= (0i+2532*10^-5 j+0k) N/C
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