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

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

Step-by-step explanation:

The magnetic field is:

`B=((\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=((4\pi x10^(-7) *3)/(2\pi *0.024^(2) ) )*7.2x10^(-3)= 7.49x10^(-6) T`