Question

We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a total current of 3.00 A. Find the magnitude of the magnetic field at a distance of 0.72 cm from the center axis. Treat the wire as a cylinder.

Answer 1

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

$B=(\frac{\mu i}{2\pi R^{2} } )r$

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

$B=(\frac{4\pi x10^{-7} *3}{2\pi *0.024^{2} } )*7.2x10^{-3}= 7.49x10^{-6} T$