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Chapter 26: Problem 64

A long conducting rod of radius \(R\) carries a nonuniform current density \(J=J_{0} r / R,\) where \(J_{0}\) is a constant and \(r\) is the radial distance from the rod's axis. Find expressions for the magnetic field strength (a) inside and (b) outside the rod.

Short Answer

Expert verified
The magnetic field strength inside the rod is \(B = \frac {\mu_0 J_0 r^2}{3R}\) and outside the rod is \(B = \frac {\mu_0 J_0 R^2}{2r}\)

Step by step solution

01

- Applying Ampère’s Law for the Inside of the Rod

The current enclosed by a hypothetical path of radius r inside the rod can be calculated by integrating the current density over the enclosed circular cross-section. Therefore, \(I = \int_0^{r} J 2\pi r' dr' = \int_0^{r} J_0 \frac {r'}{R} 2\pi r' dr' = 2\pi J_0 \int_0^{r} (\frac {r'^2}{R}) dr'\). Step 1 gives: \(I = 2\pi J_0 \frac {r^3}{3R}\). Now apply Ampère’s law, \(B 2\pi r = \mu_0 I \Rightarrow B = \frac {\mu_0 I}{2\pi r}\). Substituting the value of I from step 1, the magnetic field inside the rod is given by \(B = \frac {\mu_0 J_0 r^2}{3R}\).
02

- Applying Ampère’s Law for the Outside of the Rod

For a path of radius r outside the rod, the entire current of the rod is enclosed. Therefore, \(I = \int_0^{R} J 2\pi r' dr' = \int_0^{R} J_0 \frac {r'}{R} 2\pi r' dr' = 2\pi J_0 \int_0^{R} (\frac {r'^2}{R}) dr'\). Step 2 gives: \(I = 2\pi J_0 \frac {R^2}{2}\). Now apply Ampère’s law, \(B 2\pi r = \mu_0 I \Rightarrow B = \frac {\mu_0 I}{2\pi r}\). Substituting the value of I from step 2, the magnetic field outside the rod is given by \(B = \frac {\mu_0 J_0 R^2}{2r}\)

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