Question

A cylindrical conductor of radius R carries a current along its length. The current density $J$ varies according to the distance $r$ from axis as: $J = J_0 \left[1-\frac{r}{2R}\right]$, where $J_0$ is a constant. Then

• A. Maximum magnetic field is at $r = R$
• B. Maximum magnetic field is at $r = 2R/3$
• C. Maximum magnetic field is $\frac{\mu_0J_0R}{6}$
• D. Maximum magnetic field is $\frac{\mu_0J_0R}{3}$

Answer 1

Answer: (A), (D)

A, D

Answer 2

The problem asks us to find the maximum magnetic field due to a cylindrical conductor carrying a current with a non-uniform current density.

The current density is given by $J = J_0 \left[1-\frac{r}{2R}\right]$, where $R$ is the radius of the conductor and $r$ is the distance from the axis.

We will use Ampere's Law to find the magnetic field $B(r)$. Ampere's Law states $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$, where $I_{enc}$ is the current enclosed by the Amperian loop. Due to the cylindrical symmetry, the magnetic field lines are concentric circles, and $B$ is constant along such a loop. So, $B(r) (2\pi r) = \mu_0 I_{enc}$.

1. Magnetic Field inside the conductor ($r \le R$)

For an Amperian loop of radius $r$ inside the conductor, the enclosed current $I_{enc}$ is found by integrating the current density over the area of the loop: $I_{enc} = \int_0^r J(r') dA'$ where $dA' = (2\pi r') dr'$ is the area of an elemental ring. $I_{enc} = \int_0^r J_0 \left[1-\frac{r'}{2R}\right] (2\pi r') dr'$ $I_{enc} = 2\pi J_0 \int_0^r \left[r'-\frac{(r')^2}{2R}\right] dr'$ $I_{enc} = 2\pi J_0 \left[\frac{(r')^2}{2}-\frac{(r')^3}{6R}\right]_0^r$ $I_{enc} = 2\pi J_0 \left[\frac{r^2}{2}-\frac{r^3}{6R}\right]$ $I_{enc} = \pi J_0 r^2 \left[1-\frac{r}{3R}\right]$

Now, apply Ampere's Law: $B(r) (2\pi r) = \mu_0 \pi J_0 r^2 \left[1-\frac{r}{3R}\right]$ $B(r) = \frac{\mu_0 J_0 r}{2} \left[1-\frac{r}{3R}\right]$ $B(r) = \frac{\mu_0 J_0}{2} \left[r-\frac{r^2}{3R}\right]$ for $r \le R$.

To find the maximum magnetic field inside the conductor, we differentiate $B(r)$ with respect to $r$ and set it to zero: $\frac{dB(r)}{dr} = \frac{\mu_0 J_0}{2} \left[1-\frac{2r}{3R}\right]$ Setting $\frac{dB(r)}{dr} = 0$: $1-\frac{2r}{3R} = 0 \implies r = \frac{3R}{2}$

This value $r = \frac{3R}{2}$ is greater than $R$, meaning the maximum of the function $B(r)$ occurs outside the range $r \le R$. Since the second derivative $\frac{d^2B(r)}{dr^2} = -\frac{\mu_0 J_0}{3R}$ is negative, the function is concave down. This implies that within the range $0 \le r \le R$, $B(r)$ is monotonically increasing. Therefore, the maximum magnetic field inside the conductor occurs at $r=R$.

Let's calculate $B(R)$: $B(R) = \frac{\mu_0 J_0 R}{2} \left[1-\frac{R}{3R}\right] = \frac{\mu_0 J_0 R}{2} \left[1-\frac{1}{3}\right]$ $B(R) = \frac{\mu_0 J_0 R}{2} \left[\frac{2}{3}\right] = \frac{\mu_0 J_0 R}{3}$

2. Magnetic Field outside the conductor ($r > R$)

For an Amperian loop of radius $r$ outside the conductor, the enclosed current $I_{enc}$ is the total current flowing through the conductor. $I_{total} = \int_0^R J(r') (2\pi r') dr'$ Using the expression for $I_{enc}$ derived above and setting $r=R$: $I_{total} = \pi J_0 R^2 \left[1-\frac{R}{3R}\right] = \pi J_0 R^2 \left[\frac{2}{3}\right]$ $I_{total} = \frac{2\pi J_0 R^2}{3}$

Now, apply Ampere's Law for $r > R$: $B(r) (2\pi r) = \mu_0 I_{total}$ $B(r) (2\pi r) = \mu_0 \frac{2\pi J_0 R^2}{3}$ $B(r) = \frac{\mu_0 J_0 R^2}{3r}$ for $r > R$.

For $r > R$, the magnetic field $B(r)$ is inversely proportional to $r$, which means it continuously decreases as $r$ increases. Therefore, the maximum magnetic field outside the conductor occurs at $r=R$. At $r=R$, $B(R) = \frac{\mu_0 J_0 R^2}{3R} = \frac{\mu_0 J_0 R}{3}$.

3. Overall Maximum Magnetic Field

Comparing the magnetic field behavior:

• For $r \le R$, $B(r)$ increases from $0$ at $r=0$ to $\frac{\mu_0 J_0 R}{3}$ at $r=R$.
• For $r > R$, $B(r)$ decreases from $\frac{\mu_0 J_0 R}{3}$ at $r=R$ towards $0$ as $r \to \infty$.

Thus, the maximum magnetic field occurs at the surface of the conductor, i.e., at $r=R$, and its value is $\frac{\mu_0 J_0 R}{3}$.

Evaluating the options: (A) Maximum magnetic field is at $r = R$. (True) (B) Maximum magnetic field is at $r = 2R/3$. (False, as $B(R) > B(2R/3)$) (C) Maximum magnetic field is $\frac{\mu_0J_0R}{6}$. (False, the value is $\frac{\mu_0J_0R}{3}$) (D) Maximum magnetic field is $\frac{\mu_0J_0R}{3}$. (True)

Both statements (A) and (D) are correct.