Question

Two long, straight wires, one above the other, are separated by a distance 2a and are parallel to the x-axis. Let the be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current I in the What are the magnitude and direction of the net magnetic field of the two wires at a point in the plane of the wires (a) midway between them; (b) at a distance a above the upper wire; (c) at a distance a below the lower wire?

Answer 1

Answer:

field B = µ₀c I / 2πr

The field in the xy plane due to the fact that the two wires are perpendicular to the plane Bx and By are everywhere 0 on the plane.

a) Midway between, the Bz components cancel, so <0, 0, 0> T

b) Bz = µ₀ x I / 2πa + µ₀ x I / 2π(3a) = (µ₀ x I / 2π)(1/a + 1/3a)

Bz = (µ₀ x I / 2πr)(3/3a + 1/3a) = (µ₀x I / 2πr)(4 / 3a) = 2µ₀ x I / 3πa

c) By symmetry, Bz = -2µ₀ x I / 3πa (that is, down into the plane)

Answer 2

Ampere's law allows to find the results for the resulting magnetic field due to two cables carrying current in the same direction are;

     a) Point between cables   B = 0 T

     b) Point at the top of the cables  B = $2 \frac{\mu_o I}{2 \pi r}$

     c) Point at the bottom of the cables  B =  $-2 \frac{\mu_o I }{2 \pi r}$  

Given parameters

• Distance between the two cables d = 2a

• Current I

To find

• Magnetic field

   a) between the cables

   b) Distance above upper cable

   c) Distance below the lower cable

Ampere's law relates the magnetic field and the current flowing through a wire.

       ∫ B . ds = μ₀ I  

Where B is the magnetic field, ds is the length differential, μ₀ is the permeability of the vacuum and I is the current in the wire.

For a point on the outside of the cable we create a surface circulating around the cable and its length is

            s = 2π r

t

The magnetic field at any point for a distance (r) greater than the wire radius (a).

           B = $\frac{\mu_o I}{2 \pi r}$  

Where the direction of the magnetic field is given by the rule that the thumb points in the direction of the current and the closed fingers point in the direction of the magnetic field.

It indicates that we have two cables, therefore the total magnetic field is the vector sum of the field created by each cable.

Let's apply each situation:

a) At a point in the middle of the two cables.

You see attached, the lower cable creates a field with direction entering the leaf and the upper cable creates a field in the opposite direction leaving the leaf, therefore the two fields are in the opposite direction and are subtracted, consequently the total field is zero.

           B = 0 T

b) at a point above the cables.

In this case, the cable creates a magnetic field with the same direction, therefore they add up.

Let's write the expression for the magnetic field taking into account the separation of the two wires is 2a for the distance

       $B= \mu_o \ I \frac{1}{2\pi \ r} + \mu_o \ I \frac{1}{2\pi \ (r+2a)} \\B=\frac{ \mu_o \ I}{2\pi \ r} ( 1 + \frac{1}{1+ \frac{2a}{r} } ) = \frac{ \mu_o \ I}{2\pi \ r} \ \frac{(1+2a/r) +1)}{1+ 2a/r} \\B= \frac{ \mu_o \ I}{2\pi \ r} 2$

c) Point at the bottom of the cables.

In this case, the fields are also added, but it goes in the opposite direction or to the previous on.

       $B= - 2 \frac{\mu_o I}{2\pi r}$  

In conclusion using Ampere's law we can find the results for the resulting magnetic field due to two cables carrying current in the same direction are;

       a) Point between cables B = 0 T

       b) Point at the top of the cables B = $2 \frac{\mu_o I}{2 \pi r}$

       c) Point at the bottom of the cables B = $-2 \frac{\mu_o I}{2 \pi r}$  

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