This question involves the concepts of Biot-Savart Law and magnetic field. It can be solved by the application of Biot-Savart Law on a current-carrying wire.
The magnitude of the magnetic field is "5.28 x 10⁻⁶ T".
Using the Biot-Savart Law to find out the magnitude of the magnetic field produced by the wire at a point 5 cm above it, directly:
[tex]B = \frac{\mu_o I}{2\pi r}\\\\[/tex]
where,
B = magnetic field = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = [tex]\frac{ne}{t} = (8.25\ x\ 10^{18}\ elctrons/sec)(1.6\ x\ 10^{-19}\ C /electron)[/tex] = 1.32 A
r = radius = distance above wire = 5 cm = 0.05 m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1.32\ A)}{2\pi(0.05\ m)}[/tex]
B = 5.28 x 10⁻⁶ T
Learn more about magnetic field here:
https://brainly.com/question/23096032?referrer=searchResults
The attached picture shows the illustration of magnetic field due to a wire.
