Three infinite straight wires are fixed in place and aligned parallel to the z-a

Question

Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) = (-21 cm, 0) carries current I1 = 3.8 A in the negative z-direction. The wire at (x,y) = (21 cm, 0) carries current I2 = 0.8 A in the positive z-direction. The wire at (x,y) = (0, 36.4 cm) carries current I3 = 5.5 A in the positive z-direction.1)

1)What is Bx(0,0), the x-component of the magnetic field produced by these three wires at the origin?

T

2)

What is By(0,0), the y-component of the magnetic field produced by these three wires at the origin?

T

3)

What is Fx(1), the x-component of the force exerted on a one meter length of the wire carrying current I1?

N

4)

What is Fy(1), the y-component of the force exerted on a one meter length of the wire carrying current I1?

N

5)

What is Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current I2?

N

Explanation / Answer

magnetic field by an wire carrying current I at a distance of d is given by

magnitude=mu*I/(2*pi*d)

direction is given by right hand thumb rule

hence magnetic field at origin due to wire carrying I1:

distance d=0.21 m

magnetic field magnitude=4*pi*10^(-7)*3.8/(2*pi*0.21)=3.619*10^(-6) T

direction is along -ve y axis.

hence magnetic field at origin due to wire carrying I2:

distance d=0.21 m

magnetic field magnitude=4*pi*10^(-7)*0.8/(2*pi*0.21)=0.7619*10^(-6) T

direction is along -ve y axis.

hence magnetic field at origin due to wire carrying I3:

distance d=0.364 m

magnetic field magnitude=4*pi*10^(-7)*5.5/(2*pi*0.364)=3.022*10^(-6) T

direction is along +ve x axis.

part 1:

x component of magnetic field at origin=3.022*10^(-6) T

part 2:

y component of magnetic field at origin=3.619*10^(-6)+0.7619*10^(-6)=4.3809*10^(-6) T

part 3:

force per unit length between two parallel wires is given by

F=mu*I1*I2/(2*pi*d)

force is attractive nature if currents are in same direction and force is repulsive in nature if currents are in opposite direction.

force per unit length on wire carrying I1 due to I2:

distance between I1 and I2=0.42 m

as they carry current in opposite direction, force is repulsive in nature

hence force direction is -ve x axis.

force magnitude=4*pi*10^(-7)*3.8*0.8/(2*pi*0.42)=1.4476*10^(-6) N/m

force per unit length on wire carrying I1 due to I3:

as they cary current in opposite direction, force is repulsive in nature.

vector along the force direction=(-0.21,0)-(0,0.364)=(-0.21,-0.364)

distance=sqrt(0.21^2+0.364^2)=0.42023 m

unit vector=vector/magnitude=(-0.49973,-0.86619)

force magnitude=4*pi*10^(-7)*3.8*5.5/(2*pi*0.42023)=9.9469*10^(-6) N

in vector notation,

force=9.9469*10^(-6)*(-0.49973,-0.86619)=(-4.9708*10^(-6),-8.6159*10^(-6))

hence force along x direction=-1.4476*10^(-6)-4.9708*10^(-6)=-6.4184*10^(-6) N

part 4:

force along y direction=-8.6159*10^(-6) N

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