Question

The current density $\vec{J}$ inside a solid long cylindrical wire having a radius $a=12mm$ is in the direction of the central axis and its magnitude is changing linearly with radial distance $r$ from the axis according to $J=\dfrac{{{J}_{0}}r}{a}$ ​ where ${{J}_{0}}=\dfrac{{{10}^{5}}}{4\pi }\dfrac{A}{{{m}^{2}}}$. What will be the magnitude of the magnetic field at $r=\dfrac{a}{2}$ in $\mu T$?
$\begin{align}
  & A.10\mu T \\
 & B.4\mu T \\
 & C.5\mu T \\
 & D.3\mu T \\
\end{align}$

Answer 1

Hint: According to ampere’s circuital law, the closed integral of the dot product of magnetic field and the length element will be equivalent to the product of the permeability and the current. The closed integral of the dot product of magnetic field and the length element will be equivalent to the product of magnetic field and length element here. The current will be the integral of the product of current density and area element. This will help you in answering this question.

Complete step by step answer:
According to Ampere's circuital law, the closed integral of the dot product of magnetic field and the length element will be equivalent to the product of the permeability and the current. This can be written as an equation,
$\oint{B\cdot dl=\mu \cdot I}$
The closed integral of the dot product of magnetic field and the length element will be equivalent to the product of magnetic field and length element here.
$\oint{B\cdot dl=B\cdot dl}$
The length element can be written as,
$dl=2\pi r$
The current will be the integral of the product of current density and area element.
$I=\int{J\cdot da}$
Finally we can write that,
$B\times 2\pi \dfrac{a}{2}={{\mu }_{0}}\times \int\limits_{0}^{\dfrac{a}{2}}{{{J}_{0}}}\times \dfrac{r}{a}\times 2\pi rdr$
That is,
$\begin{align}
  & B\pi a=\dfrac{2\pi {{\mu }_{0}}{{J}_{0}}}{a}\times \dfrac{{{\left( \dfrac{a}{2} \right)}^{3}}}{3} \\
 & \Rightarrow B=\dfrac{{{\mu }_{0}}{{J}_{0}}a}{12} \\
\end{align}$
It has been already mentioned in the question that the radius of the cylindrical wire is,
$a=12mm$
The current density is given as,
${{J}_{0}}=\dfrac{{{10}^{5}}}{4\pi }\dfrac{A}{{{m}^{2}}}$
Substituting these values in the equation can be written as,
$B=\dfrac{4\pi \times {{10}^{-7}}\times \dfrac{{{10}^{5}}}{4\pi }\times 0.012}{12}$
Simplifying this equation will give,
$B=10\mu T$
Therefore the magnetic field has been obtained as $10\mu T$.

So, the correct answer is “Option A”.

Note: Current density is a common parameter used in electromagnetism. It is defined as the amount of charge per unit time which is flowing through a unit area of a particular area cross section. The current density has been expressed in the unit of ampere per square metre.