A long cylindrical wire (radius = 2.0 cm) carries a current of 40 a that is uniformly distributed over a cross section of the wire. what is the magnitude of the magnetic field at a point which is 1.5 cm from the axis of the wire?

Respuesta :

Magnetic field inside the wire will be calculated by Ampere's law

Ampere's law is given by

[tex] \int B.dl = \mu_o i[/tex]

[tex] B\int dl = \mu_o *\frac{i}{\pi R^2}*\pi r^2[/tex]

[tex] B.2\pi r = \mu_o *\frac{i}{R^2} r^2[/tex]

[tex] B = \frac{\mu_o i* r}{2\pi R^2}[/tex]

[tex] B = \frac{4\pi * 10^{-7}* 40* 0.015}{2\pi 0.02^2}[/tex]

[tex] B = 3 * 10^{-4} T{/tex}