Question

Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of 4 A and the other is carrying a current of 9 A. The magnitude of the magnetic field exactly in between the two wires is ____ T.

Answer 1

The value is
`B = 3.33 *10^(-6) \ T`

Step-by-step explanation:

From the question we are told that

The distance of separation is
`d = 0.6 \ m`

The current on the one wire is
`I_1 = 9 \ A`

The current on the second wire is
`I_2 = 4 \ A`

Generally the magnitude of the field exerted between the current carrying wire is

`B = B_1 - B_2`

Here
`B_1` is the magnetic field due to the first wire which is mathematically represented as

`B_1 = (\mu_o * I_1 )/(2 \pi * d_1)`

Here
`d_1` is the distance to the half way point of the separation and the value is

`d_1 = 0.3 \ m`

`B_2` is the magnetic field due to the first wire which is mathematically represented as

`B_2 = (\mu_o * I_2 )/(2 \pi * d_2)`

Here
`d_2` is the distance to the half way point of the separation and the value is

`d_2 = 0.3 \ m`

This means that
`d_1 = d_2 = a = 0.3`

So

`B = (\mu_o * I_1 )/(2 \pi * d_1) - (\mu_o * I_2 )/(2 \pi * d_2)`

=>
`B = (\mu_o * (I_1 - I_2))/(2 \pi *0.3 )`

=>
`B = ( 4\pi * 10^(-7) * (9- 4))/(2 * 3.142 *0.3 )`

=>
`B = 3.33 *10^(-6) \ T`