Question

The current density $J$ inside a long,solid, cylindrical wire of radius $a=12\text{mm}$ is in the direction of the central axis, and its magnitude varies lenearly with radial distance $r$ from the axis according to $J=\frac{J_{0}r}{a}$ where $J_0=\frac{{10}^5}{4pi}{\text{A/m}}^2$ .Find the magnitude of the magnetic field at $r=\frac{a}{2}$ in $\mu \text{T}$.

Answer 1

using ampere's circuital law

$B2\pi \left(\frac{a}{2}\right)={\mu }_0{I}_{in}$

$B2\pi \left(\frac{a}{2}\right)={\mu }_0{\int }_O^{q/2}2\pi rdr\times \frac{J_{0}r}{a}$

$B2\pi \left(\frac{a}{2}\right)={\mu }_0\frac{2\pi }{3}\frac{J_0}{a}\left(\frac{a^3}{8}\right)$

$B=\frac{{\mu }_0{J}_0}{3a^2}\frac{a^3}{4}=\frac{4\pi \times {10}^{-7}}{3}\times \frac{{10}^5}{4\pi }\times \frac{12\times {10}^{-3}}{4}$

$B=10\mu \text{T}$

Ans $10$