Two very long parallel wires are separated by a distance d = 8.4 cm and each car

Question

Two very long parallel wires are separated by a distance  d= 8.4 cm and each carry a current  I = 16.8 A currents in the z-direction.

Part A:

Find the vector magnetic field due to the two wires at the point P which is positioned a distance L= 13.2cm from one wire and a distance w=  12.1 cm from the other. See figure(Figure 1) . Express the field direction as an angle relative to the positive x-axis.

[Hint: The relevant angles may be found using the Law of Cosines ]

|B| = _______________T

= ___________   relative to the positive x-axis

Explanation / Answer

current in a wire i=16.8 A

magnetic field B=(uo*i)/(2pi*r)

at a distance of w=12.1 cm

B1=(uo*i)/(2pi*w)

B1=(4pi*10^-7*16.8)/(2pi*12.1*10^-2)

B1=2.78*10^-5 T

and

at a distance of L=13.2 cm

B2=(uo*i)/(2pi*L)

B2=(4pi*10^-7*16.8)/(2pi*13.2*10^-2)

B2=2.54*10^-5 T

angle between these two field is theta,

by using cosine angle,

d^2=L^2+w^2-2L*W*cos(theta)

==

8.4^2=13.2^2+12.1^2-2*13.2*12.1*cos(theta)

==>

cos(theta)=((13.2^2+12.1^2)-(8.4^2))/(2*13.2*12.1)

===>

theta=38.47 degrees

now,

B=(B1*cos(38.47)-B2*cos(90-38.47))i+(B1*sin(38.47)-B2*sin(90-38.47))j

B=(2.78*10^-5*cos(38.47)-2.45*10^-5*cos(90-38.47))i+(2.78*10^-5*sin(38.47)+2.45*10^-5*sin(90-38.47))j

the magnetic field B=(0.65*10^-5)i+(3.65*10^-5)j

and

angle theta =38.47

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