Answer:
magnitude of net magnetic field at given point is
[tex]B = 5 \times 10^{-6} T[/tex]
Explanation:
As we know that magnetic field due to a long current carrying wire is given as
[tex]B = \frac{\mu_o i}{2\pi r}[/tex]
here we we will find the magnetic field due to wire which is along x axis is given as
[tex]i = 30 A[/tex]
r = 2 m
now we have
[tex]B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}[/tex]
[tex]B_1 = 3\times 10^{-6} T [/tex] into the plane
Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as
[tex]i = 40 A[/tex]
r = 2 m
now we have
[tex]B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}[/tex]
[tex]B_2 = 4\times 10^{-6} T [/tex] along + x direction
Since the two magnetic field is perpendicular to each other
So here net magnetic field is given as
[tex]B = \sqrt{B_1^2 + B_2^2}[/tex]
[tex]B = 5 \times 10^{-6} T[/tex]
