Question

In the above problem, what will be the magnetic field B inside the wire at a distance r from its axis, if the current density J is uniform across the cross section of the wire?

Answer 1

Figure shows the cross section of a long straight wire of radius a that carries a current I out of the page. Because the current is uniformly distributed over the cross section of the wire, the magnet ic field \(\vec{B}\) due to the current must be cylindrically symmetrical. Thus, along the Amperian loop of radius r(r < a), symmetry suggests that \(\vec{B}\) is tangent to the loop, as shown in the figure.
\(\oint \vec{B} \cdot \overrightarrow{d l}=B \oint d l\) = B(2πr) ……….. (1)
Because the current is uniformly distributed, the current I_encl enclosed by the loop is proportional to the area encircled by the loop; that is,
I_encl = Jπr²
By right-hand rule, the sign of ‘d is positive. Then, by Ampere’s law,
B (2πr) = µ₀ I_encl = µ₀ Jπr² ……………….. (2)
∴ B = \(\frac{\mu_{0} J}{2} r\) ……………… (3)
OR
I_encl = I\(\frac{\pi r^{2}}{\pi a^{2}}\)
By right-hand rule, the sign of I\frac{\pi r^{2}}{\pi a^{2}} is positive. Then, by Ampere’s law,
\(\oint B d l\) = µ₀ I_encl
∴ B(2πr) = µ₀I \(\frac{r^{2}}{a^{2}}\) …………. (4)
∴ B = (\(\frac{\mu_{0} I}{2 \pi a^{2}}\) )r ………… (5)