Question

Two long thin parallel wires 13.0cmapart carry 35 A

currents in the same direction.Determine the magnetic field vector at a point 10.0 cmfrom one wire and 6.0 cmfrom the other.

13.0 cm 10,0 cm 10.0 cm +

Answer 1

Use the expression for the magnetic field due to wire to calculate the magnetic field at point P.

Here, is the permeability of free space, I is the current passing through the wire, and r is the distance from the wire to the point P.

The following figure shows the magnetic fields produced by the two wires at point P.

Here, is the current produced by the wire 1, is the current produced by the wire 2, is the magnetic field due to wire 1, is the magnetic field due to wire 2, is the angle made by the magnetic field with the line joining the two wires, and is the angle made by the magnetic field with the line joining the two wires.

According to law of cosines, the cosine of the angle made by the magnetic field vector produced by the wire 1 is given by,

Here, is the distance between the two wires, is the distance from the wire 1 to the point P, and is the distance from the wire 2 to the point P.

Substitute 13 cm for , 10 cm for , and 6 cm for in above equation and solve it as follows:

According to law of cosines, the cosine of the angle made by the magnetic field vector produced by the wire 2 is given by,

Here, is the distance between the two wires, is the distance from the wire 1 to the point P, and is the distance from the wire 2 to the point P.

Substitute 13 cm for , 10 cm for , and 6 cm for in above equation and solve it as follows:

The magnetic field due to wire 1 is given by,

Substitute for , 35 A for , and 10 cm for in above equation as follows:

The magnetic field vector due to wire 1 is given by,

Substitute for and for in above equation as follows:

The magnetic field due to wire 2 is given by,

Substitute for , 35 A for , and 6 cm for in above equation as follows:

The magnetic field vector due to wire 2 is given by,

Substitute for and for in above equation as follows:

The net magnetic field at point P is given by,

Substitute for and for in above equation as follows:

Therefore, the magnetic field vector is .

The magnitude of the net magnetic field is given by,

Here, is the x component of the magnetic field and is the y- component of the magnetic field.

Substitute for and for in above equation as follows:

Therefore, the magnitude of the magnetic field at point P is .

The direction of the magnetic field is given by,

Substitute for and for in above equation as follows:

Therefore, the magnitude of the magnetic field at a point P is and is directed at an angle of above the horizontal.

Answer 2

Given data: Length of parallel wires,1 current ,I = 35A 13.0cm p 13cm 0.13m a=10cm = 0.10m b-6.0cm 0.06m Using law of cosines,

cos 2ab 0.12+0.062-0.132 2×0.10×0.06 =-0.275 cos-0.275) = 105.96' θ=106. Magneticfield of one wire 2π(0.10) 4π× 10-X35 0.628 = 7.0x10-5T Magneticfield of anotherwire, 4π×10-, x35 2n× 006 =1.16×10-4 T =1.2×10 Resultantfield 17.0x10-5)"+(1.2x10")"-2x(7.0x10-5)x(1.2x10-4)cos 106. = 1.2x10