Question

In a long cylindrical wire of radius R, magnetic induction varies with the distance from axis as B = cr". Find the function of current density in wire with the distance from axis of wire.

• A. J(r) = \frac{c(n+1)}{\mu_0} r^{n-1}
• B. J(r) = \frac{c(n-1)}{\mu_0} r^{n+1}
• C. J(r) = \frac{c}{\mu_0} r^{n}
• D. J(r) = \frac{cn}{\mu_0} r^{n-1}

Answer 1

Answer: (A)

J(r) = \frac{c(n+1)}{\mu_0} r^{n-1}

Answer 2

To find the current density $J(r)$ inside the wire, we use Ampere's Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$.

1. Amperian Loop: Choose a circular Amperian loop of radius $r$ concentric with the wire.
2. Calculate $\oint \vec{B} \cdot d\vec{l}$: The magnetic induction is given as $B = cr^n$. For a circular loop of radius $r$, the line integral is $\oint \vec{B} \cdot d\vec{l} = B(r) \cdot (2\pi r) = (cr^n)(2\pi r) = 2\pi c r^{n+1}$.
3. Relate $I_{enc}$ to $J(r')$: The enclosed current $I_{enc}$ is found by integrating the current density $J(r')$ over the area enclosed by the Amperian loop: $I_{enc} = \int_{0}^{r} J(r') (2\pi r') dr'$.
4. Apply Ampere's Law: Equating the two sides of Ampere's Law: $2\pi c r^{n+1} = \mu_0 \int_{0}^{r} J(r') (2\pi r') dr'$.
5. Simplify: $c r^{n+1} = \mu_0 \int_{0}^{r} J(r') r' dr'$.
6. Differentiate to find $J(r)$: To isolate $J(r)$, we differentiate both sides with respect to $r$. Using the Leibniz integral rule: $\frac{d}{dr} \left( \mu_0 \int_{0}^{r} J(r') r' dr' \right) = \mu_0 J(r) r$. Differentiating the left side: $\frac{d}{dr} (c r^{n+1}) = c(n+1)r^n$. Equating the derivatives: $c(n+1)r^n = \mu_0 J(r) r$.
7. Solve for $J(r)$: $J(r) = \frac{c(n+1)r^n}{\mu_0 r} = \frac{c(n+1)}{\mu_0} r^{n-1}$.