Respuesta :
Explanation:
Given that,
Radius a= 5.0 mm
Radial distance r= 0, 2.9 mm, 5.0 mm
Current density at the center of the wire is given by
[tex]J_{0}=420\ A/m^2[/tex]
Given relation between current density and radial distance
[tex]J=\dfrac{J_{0}r}{a}[/tex]
We know that,
When the current passing through the wire changes with radial distance,
then the magnetic field is induced in the wire.
The induced magnetic field is
[tex]B=\dfrac{\mu_{0}i_{ind}}{2\pi r}[/tex]...(I)
We need to calculate the induced current
Using formula of induced current
[tex]i_{ind}=\int_{0}^{r}{J(r)dA}[/tex]
[tex]i_{ind}= \int_{0}^{r}{\dfrac{J_{0}r}{a}2\pi r}[/tex]
[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}\int_{0}^{r}{r^2}[/tex]
[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}[/tex]
We need to calculate the magnetic field
Put the value of induced current in equation (I)
[tex]B=\dfrac{\mu_{0}{\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}}{2\pi r}[/tex]
[tex]B=\dfrac{\mu_{0}J_{0}r^2}{3a}[/tex]
(a). The magnetic field at a distance r = 0
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times0}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 0[/tex]
The magnetic field at a distance 0 is zero.
(b). The magnetic field at a distance r = 2.9 mm
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(2.9\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 2.95\times10^{-7}\ T[/tex]
The magnetic field at a distance 2.9 mm is [tex]2.95\times10^{-7}\ T[/tex]
(c). The magnetic field at a distance r = 5.0 mm
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(5.0\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 8.79\times10^{-7}\ T[/tex]
The magnetic field at a distance 5.0 mm is [tex]8.79\times10^{-7}\ T[/tex]
Hence, This is the required solution.
