The picture below shows a cross-sectional view through three long straight curre

Question

The picture below shows a cross-sectional view through three long straight current-carrying wires, which are each carrying current perpendicular to the page (or the screen). In this problem, use these values: the distance a60.0 cm the current I 6.50 amps. 31 31 -a +a +2a +3a (a) Calculate the net force per unit length acting on the wire that carries a current of 31 out of the page, at x = +2a. Use a plus sign if the force is directed to the right, and a minus sign if the force is directed to the left. 0.0000034 X N/m (b) Calculate the net magnetic field at x-+a. Use a plus sign if the field is directed up, and a minus sign if the field is directed down. 0.0000021X T (c) Calculate the magnitude of the net magnetic field at at a distance of a directly above the wire located at x = 0 0 (d) In which region, on the x-axis, is there a point a finite distance from the wires at which the net magnetic field is zero? Select all the correct answers to the left of x -2a in between x =-2a and x = 0 in between x = 0 and x = +2a to the right of x = +2a (e) Find the distance from x 0.19 0 to the point closest to x = 0 where the net magnetic field is zero cm

Explanation / Answer

part(a)

force exerted by I1 on I3

F13x = uo*I1*I3/(2*pi*4a) = 4*pi*10^-7*3*6.5*3*6.5/(2*pi*4*0.6) = 3.17*10^-5 N/m

force exerted by I2 on I3

F23x = -uo*I2*I3/(2*pi*2a) = -4*pi*10^-7*6.5*3*6.5/(2*pi*2*0.6) = -2.11*10^-5 N/m

Fnet = F13x + F23x = 1.06*10^-5 N/m

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part(b)

B1y = -uo*I1/(2*pi*3a) = -4*pi*10^-7*3*6.5/(2*pi*3*0.6) = -2.17*10^-6 T

B2y = uo*I2/(2*pi*a) = 4*pi*10^-7*6.5/(2*pi*0.6) = 2.17*10^-6 T

B3y = -uo*I3/(2*pi*a) = -4*pi*10^-7*3*6.5/(2*pi*0.6) = -6.5*10^-6 T

Bnet = B1y + B2y + B3y = -6.5*10^-6 T

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part(c)

tantheta = a/2a = 1/2

theta = 26.5

r = sqrt(a^2+(2a)^2) = a*sqrt5 = 134 cm = 1.34 m

B1x = uo*I1*sintheta/(2*pi*r)

B1y = -uo*I1*costheta/(2*pi*r)

B3x = -uo*I3*sintheta/(2*pi*r)

B3y = -uo*I3*costheta/(2*pi*r)

B2x = uo*I2/(2*pi*a)

B2y = 0

Bx = B1x + B2x + B3x = uo*I2/(2*pi*a) = 4*pi*10^-7*6.5/(2*pi*0.6) = 2.17*10^-6 T

By = B1y + B2y + B3y = -2*uo*3I*costheta/(2*pi*r) = -2*4*pi*10^-7*3*6.5*cos26.5/(2*pi*1.34) = -5.21*10^-6 T

B = sqrt(B^2+By^2) = 5.64*10^-6 T

==================================

d)

to the left of   = -2a

in between = 0 and = +2a

-----------------------------------

(e)

at point x the net B = 0

B1y = -uo*I1/(2*pi*(2a+x))

B2y = uo*I2/(2*pi*x)

B3y = -uo*I3/(2*pi*(2a-x))

B1y + B2y + B3y = 0

-I1/(2a+x) + I2/x - I3/(2a-x) = 0

-3I/(2a+x) + I/x - 3I/(2a-x) = 0

-3/(2a+x) + 1/x - 3/(2a-x) = 0

-3/(1200+x) + 1/x - 3/(120-x) = 0

x = 28.4 cm <<<---ANSWER

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