To find the magnetic field \( B \) generated by a cylindrical conductor with a non-uniform current density, we can apply Ampère's Law. This law relates the magnetic field around a closed loop to the current passing through that loop. Given that the current density \( J \) varies with the radius as \( J = br \), where \( b \) is a constant, we can derive the magnetic field both inside and outside the conductor.
Magnetic Field Inside the Conductor (r1 < R)
For a point inside the conductor at a distance \( r_1 \) from the axis, we first need to determine the total current \( I_{\text{enc}} \) enclosed by a circular loop of radius \( r_1 \). The current density \( J \) is given as \( J = br \), so we can express the current through a differential area element \( dA \) as:
- Area element: \( dA = 2\pi r \, dr \)
- Current through \( dA \): \( dI = J \cdot dA = br \cdot (2\pi r \, dr) = 2\pi b r^2 \, dr \)
To find the total current \( I_{\text{enc}} \) enclosed by the radius \( r_1 \), we integrate from 0 to \( r_1 \):
\( I_{\text{enc}} = \int_0^{r_1} 2\pi b r^2 \, dr = 2\pi b \left[ \frac{r^3}{3} \right]_0^{r_1} = \frac{2\pi b r_1^3}{3} \)
Now, applying Ampère's Law, which states that the line integral of the magnetic field \( B \) around a closed loop is equal to the permeability of free space \( \mu_0 \) times the enclosed current:
\( \oint B \cdot dl = \mu_0 I_{\text{enc}} \)
For a circular path of radius \( r_1 \), the left side simplifies to \( B(2\pi r_1) \). Thus, we have:
\( B(2\pi r_1) = \mu_0 \left( \frac{2\pi b r_1^3}{3} \right) \)
Solving for \( B \), we find:
\( B = \frac{\mu_0 b r_1^2}{3} \)
Magnetic Field Outside the Conductor (r2 > R)
For a point outside the conductor at a distance \( r_2 \) from the axis, the total current \( I \) flowing through the entire conductor must be calculated. We can find this by integrating the current density over the entire cross-section of the conductor:
\( I = \int_0^R 2\pi b r^2 \, dr = 2\pi b \left[ \frac{r^3}{3} \right]_0^R = \frac{2\pi b R^3}{3} \)
Now, applying Ampère's Law again for a circular path of radius \( r_2 \):
\( \oint B \cdot dl = \mu_0 I \)
Thus, we have:
\( B(2\pi r_2) = \mu_0 \left( \frac{2\pi b R^3}{3} \right) \)
Solving for \( B \) gives us:
\( B = \frac{\mu_0 b R^3}{3 r_2} \)
Summary of Results
In summary, the expressions for the magnetic field \( B \) are:
- For \( r_1 < R \): \( B = \frac{\mu_0 b r_1^2}{3} \)
- For \( r_2 > R \): \( B = \frac{\mu_0 b R^3}{3 r_2} \)
This analysis shows how the magnetic field behaves differently inside and outside the cylindrical conductor due to the non-uniform current density. The dependence on the radius in both cases illustrates the influence of the current distribution on the magnetic field generated.
