Question

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

• A. $10\mu T$
• B. $4\mu T$
• C. $5\mu T$
• D. $3\mu T$

Answer 1

The correct option is A $10\mu T$

Ampere's law says that $\oint B\cdot dl=\mu \cdot I$,
Here $B.dl$ is equal to $\oint B\cdot dl$
Here, dl is $2\pi r$
And $I=\int J.da$
Finally, $B\times 2\pi a/2={\mu }_0\times {\int }_0^{\frac{a}{2}}{J}_0\times \frac{r}{a}\times 2\pi rdr$
$\pi aB=\frac{2\pi {\mu }_o{J}_o}{a}\times \frac{(a/2{)}^3}{3}$

$B=\frac{{\mu }_o{J}_{o}a}{12}$

$B=\frac{4\pi \times {10}^{-7}\times {10}^5/4\pi \times 0.012}{12}$

Putting values
B is equal to 10$\mu T$