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) = (-12.5 cm, 0) carries current I_1 = 3.6 A in the negative z-direction. The wire at (x,y) = (12.5 cm, 0) carries current I_2 = 0.9 A in the positive z-direction. The wire at (x,y) = (0, 21.7 cm) carries current I_3 = 6.6 A in the positive z-direction. What is B_x(0,0), the x-component of the magnetic field produced by these three wires at the origin? What is B_y(0,0), the y-component of the magnetic field produced by these three wires at the origin? What is F_x(1), the x-component of the force exerted on a one meter length of the wire carrying current I_1? What is F_y(1), the y-component of the force exerted on a one meter length of the wire carrying current I_1?

Explanation / Answer

The direction of the magnetic field by a current carrying wire is given by the right-hand rule, that is if we point our right thumb in the direction of the current , the curl of the fingers gives us the magnetic field. using this we find that at the origin, the magnetic field due to wire 1 is in -ve y direction, and due to wire 2 is also in the - ve y-direction and due to wire 3 it is in +ve x direction.

The magnitude of B at a distance r from the wire, according to the Ampere's law is given by B = oI / 2r

1.) So, the x-component of magnetic field at origin is only by wire 3

= oI3 / 2r3

r3 = height of equilateral triangle with side 25 cm = 25 x Sin 60 = 21.65063509 cm

therefore x-component of B at Origin = oI3 / 2r3 = 4 x 10-7 x 6.6 / 2 ( 0.2165063509) = 6.096818844 x 10-6 T

2.) Magnetic field due to wire 2 at origin is = oI2 / 2r2 = 4 x 10-7 x 0.9 / 2 ( 0.125) = 1.44 x 10-6 T

Magnetic field due to wire 1 at origin is = oI1 / 2r1 = 4 x 10-7 x 3.6 / 2 ( 0.125) = 5.76x 10-6 T

Both are in the - ve y-direction. So, the net y-component at origin is = 7.2 x 10-6 T

their sum in -ve y-direction.

3.) Current is the same direciton attract and those in the opposite direction repel. So, the force interaction between wires 1 and 3 is repulsive ( wire 1 is pushed away from wire 3) and that between wires 1 and 2 is also repulsive. So, wire 2 also pushes wire 1 away (in the -ve x direction)

Force per unit length between two wires carrying currents Ia and Ib is given by Fab = oIaIb / 2d

So, F12 = oI1I2 / 2d = 2 x 10-7 x 3.6x 0.9 / 0.25 = 2.592 x 10-6 N

since this is a replusive force it's direction is exactly away from wire 2 on wire 1 that is in the -ve x-direction.

similarly,

F13 = oI1I3 / 2d = 2 x 10-7 x 3.6x 6.6 / 0.25 =19.008 x 10-6 N

Since this repulsive force it's direction will exactly away from wire 3 on wire 1, that is making an angle of 60 degrees in the anti-clockwise direction with the -ve x-axis.

So, its x-component will be F13Cos 60 in the -ve x-direction

19.008x 10-6 x Cos 60 = 9.504 x 10-6 N

So, the net Fx(1) = - 2.592 x 10-6 N -   9.504 x 10-6 N = - 12.096 N i

-ve sign indicating the -ve x -direction.

4.) y - component will be only the y component of F13 , because F12 is completely in the -ve x direction.

y component of F13 = - F13Sin 60 j

= - 19.008 x 10-6 x Sin 60 = - 1.646141088 x 10-5j

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