Question

A long straight cylindrical region of radius $a$ carries a current along its length. The current density $\left(J\right)$ varies from the axis to the edge of the cylindrical region according to $J=J_0(1-\frac{r}{a})$ , where $r$ is distance from the axis $(0\le r\le a)$. Choose the correct answer (s) among the following.

• A. Mean density $\overline{J}=\frac{J_0}{3}$
• B. At $r=0;B=0$
• C. At $r=a;B=\frac{{\mu }_0{J}_{0}a}{6}$
• D. The $B-r$ graph is parabolic with maxima at distance $r=\frac{a}{2}$

Answer 1

Answer: (A), (B), (C)

At $r=a;B=\frac{{\mu }_0{J}_{0}a}{6}$

Given, $J=J_0(1-\frac{r}{a})$

$\text{STEP-I}:-$

We know by definition, current density is current per unit area.

Thus, current $I=\underset{0}{\overset{a}{\int }}\left(2\pi rdr\right)J$

Substituting $J$ we get,

$I=2\pi J_0\underset{0}{\overset{a}{\int }}(r-\frac{r^2}{a})dr$

$\Rightarrow I=\frac{1}{3}\pi a^2{J}_0$

Mean current density is $\overline{J}=\frac{I}{\pi a^2}=\frac{J_0}{3}$

Option (a) is correct.

$\text{STEP-II}:-$


Current enclosed within distance $r$ from the axis is $I_r=\underset{0}{\overset{r}{\int }}\left(2\pi rdr\right)J$

By substituting the value of $J$ and solving the integration we get,

$I_r=\pi J_0{r}^2(1-\frac{2r}{3a})$

Using ampere's law on a circle of radius $r$ gives,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}I$

$\Rightarrow B\cdot 2\pi r={\mu }_0{I}_r$

Substituting the value of $I_r$ obtained in the above equation gives,

$B=\frac{{\mu }_0{J}_{o}r}{2}(1-\frac{2r}{3a})$


At $r=0$ we can see that, $B=0$

Option (b) is correct.

At $r=a\to B=\frac{{\mu }_0{J}_{0}a}{6}$

Option (c) is correct.

From the equation of $B$ we can say that, the graph of $B\text{Vs}r$ is a parabola.

The graph is parabolic with maxima at a distance given by, $\frac{dB}{dr}=0\Rightarrow 1-\frac{4r}{3a}=0$

$\Rightarrow r=\frac{3a}{4}$

Hence, options (a) , (b) and (c) are the correct answers.