A long, straight wire carrying a 30 A current is situated in a...

The magnetic field produced by the long straight wire at a distance $r = 2.0 \text{ cm}$ is given by:

$B_{wire} = \dfrac{\mu_0 I}{2 \pi r}$

Substituting the given values:

$B_{wire} = \dfrac{4 \pi \times 10^{-7} \times 30}{2 \pi \times 0.02}$

$B_{wire} = 2 \times 10^{-7} \times 1500 = 3.0 \times 10^{-4} \text{ T}$

The direction of this magnetic field is perpendicular to the wire. The external magnetic field is parallel to the current in the wire, so it is perpendicular to the magnetic field produced by the wire.

The magnitude of the external magnetic field is $B_{ext} = 4.0 \times 10^{-4} \text{ T}$.

Since the two magnetic fields are perpendicular to each other, the resultant magnetic field is:

$B_{net} = \sqrt{B_{wire}^2 + B_{ext}^2}$

$B_{net} = \sqrt{(3.0 \times 10^{-4})^2 + (4.0 \times 10^{-4})^2}$

$B_{net} = \sqrt{9.0 \times 10^{-8} + 16.0 \times 10^{-8}}$

$B_{net} = \sqrt{25.0 \times 10^{-8}} = 5.0 \times 10^{-4} \text{ T}$