We will calculate the magnetic field for each of the supplied currents. Recall that the magnetic field is equivalent to the product between the permeability of free space, the current, and the distance. Both have their respective vector direction, so their final magnitude will be the scalar form of the values found,
For the current of 30 A
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Here,
[tex]\mu[/tex] = Permeability at free space
I = Current
r = Distance between them
[tex]B = \frac{(4\pi *10^{-7})(30)}{2\pi (2)}[/tex]
[tex]B = 3*10^{-6}T[/tex]
For the second current of 40A,
[tex]B = \frac{(4\pi *10^{-7})(40)}{2\pi (2)}[/tex]
[tex]B = 4*10^{-6}T[/tex]
The wire goes through z-direction, then the Magnetic field is in x-direction positive.
Finally the magnitude will be
[tex]|B| = \sqrt{(3*10^{-6})^2+(4*10^{-6})^2}[/tex]
[tex]|B| = 5*10^{-6}T[/tex]
[tex]|B| = 5\mu T[/tex]
