Respuesta :
To solve this problem we must simply consider the concept given for the magnetic field from the Ampere law.
For which you have to
[tex]B = \frac{\mu_0 I}{2\pi r}[/tex]
Where,
[tex]\mu_0 =[/tex]Permeability constant
I = Current
r = Distance
Re-organizing the equation so that it can be expressed in terms of the magnetic field and the radius we have to
[tex]B = \frac{\mu_0 I}{2\pi} * \frac{1}{r}[/tex]
Replacing the change of current and the permeability constant
[tex]B = \frac{(4\pi * 10^{-7}) (8.7-4)}{2\pi} * \frac{1}{r}[/tex]
[tex]B = (2* 10^{-7}) (8.7-4) * \frac{1}{r}[/tex]
[tex]B = (9.4* 10^{-7})* \frac{1}{r}[/tex]
Therefore the approximate value for the magnitude of the magnetic field a large distance r from both wires is [tex](9.4* 10^{-7})* \frac{1}{r} T\cdot m[/tex]
From two straight wires, the magnitude of the magnetic field a large distance r is,
[tex]9.4\times10^{-7}\times\dfrac{1}{r}[/tex]
What is magnetic field?
The magnetic field is the field in the space and around the magnet in which the magnetic field can be fill.
The magnetic filed between two parallel wires can be given as,
[tex]B=\dfrac{\mu_oI}{2\pi r}[/tex]
Here, ([tex]\mu_o[/tex]) is the permeability constant, (r) is the radius of the wire and (I) is the current carried by it.
The current carried by wire one is 4.0 A to the right and the current carried by wire two is 8.7 A to the left.
Put the values as,
[tex]B=\dfrac{4\pi\times10^{-7}\times(8.7-4)}{2\pi r}\\B=9.4\times10^{-7}\times\dfrac{1}{r}[/tex]
Thus, the magnitude of the magnetic field a large distance r from both wires is,
[tex]9.4\times10^{-7}\times\dfrac{1}{r}[/tex]
Learn more about magnetic field here;
https://brainly.com/question/7802337
