Answer:
[tex]B_{net} = 3.3 \times 10^{-5} T[/tex]
Explanation:
Magnetic field due to straight long wire is always perpendicular to the line joining the wire and the point
it is given by the equation
[tex]B = \frac{\mu_0 i}{2\pi r}[/tex]
now the magnetic field due to 30 A current is given as
[tex]B_1 = \frac{(2\times 10^{-7})(30)}{0.15}[/tex]
[tex]B_1 = 4\times 10^{-5} T[/tex]
Now magnetic field due to other wire having current 40 A is given as
[tex]B_2 = \frac{(2 \times 10^{-7})(40)}{0.25}[/tex]
[tex]B_2 = 3.2 \times 10^{-5} T[/tex]
now net field along Y direction is given as
[tex]B_y = B_2cos37[/tex]
[tex]B_y = 2.56 \times 10^{-5} T[/tex]
now for X direction magnetic field we know
[tex]B_x = B_1 - B_2sin37[/tex]
[tex]B_x = (4\times 10^{-5}) - (3.2 \times 10^{-5}sin37)[/tex]
[tex]B_x = 2.07 \times 10^{-5} T[/tex]
Now net magnetic field is given as
[tex]B_{net} = \sqrt{B_x^2 + B_y^2}[/tex]
[tex]B_{net} = \sqrt{2.56^2 + 2.07^2} \times 10^{-5} T[/tex]
[tex]B_{net} = 3.3 \times 10^{-5} T[/tex]