Question

The current density inside a long, solid, cylindrical wire of radius a = 5.0 mm is in the direction of the central axis and its magnitude varies linearly with radial distance r from the axis according to J = J0r/a, where J0 = 420 A/m2. Find the magnitude of the magnetic field at a distance (a) r=0, (b) r = 2.9 mm and (c) r=5.0 mm from the center.

Answer 1

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

$J_{0}=420\ A/m^2$

Given relation between current density and radial distance

$J=\dfrac{J_{0}r}{a}$

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

$B=\dfrac{\mu_{0}i_{ind}}{2\pi r}$...(I)

We need to calculate the induced current

Using formula of induced current

$i_{ind}=\int_{0}^{r}{J(r)dA}$

$i_{ind}= \int_{0}^{r}{\dfrac{J_{0}r}{a}2\pi r}$

$i_{ind}={\dfrac{2\pi J_{0}}{a}}\int_{0}^{r}{r^2}$

$i_{ind}={\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}$

We need to calculate the magnetic field

Put the value of induced current in equation (I)

$B=\dfrac{\mu_{0}{\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}}{2\pi r}$

$B=\dfrac{\mu_{0}J_{0}r^2}{3a}$

(a). The  magnetic field at a distance r = 0

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times420\times0}{3\times5.0\times10^{-3}}$

$B = 0$

The  magnetic field at a distance 0 is zero.

(b). The  magnetic field at a distance r = 2.9 mm

$B=\dfrac{4\pi\times10^{-7}\times420\times(2.9\times10^{-3})^2}{3\times5.0\times10^{-3}}$

$B = 2.95\times10^{-7}\ T$

The  magnetic field at a distance 2.9 mm is $2.95\times10^{-7}\ T$

(c). The  magnetic field at a distance r = 5.0 mm

$B=\dfrac{4\pi\times10^{-7}\times420\times(5.0\times10^{-3})^2}{3\times5.0\times10^{-3}}$

$B = 8.79\times10^{-7}\ T$

The  magnetic field at a distance 5.0 mm is $8.79\times10^{-7}\ T$

Hence, This is the required solution.