Question

The magnetic induction inside a long cylinder of radius R vary as $B = ar^2$, where a is constant of proper unit. The current density in wire as function of distance (r) from centre of cylinder is expressed as $\frac{xa}{\mu_0}r^n$. Find (x + n).

Answer 1

4

Answer 2

To find the current density $J(r)$ inside the cylinder, 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 $r$ inside the cylinder.

2. Calculate $\oint \vec{B} \cdot d\vec{l}$: $\oint \vec{B} \cdot d\vec{l} = B(r) \cdot (2\pi r) = (ar^2) (2\pi r) = 2\pi a r^3$

3. Relate $I_{enc}$ to $J(r')$: $I_{enc} = \int_0^r J(r') dA' = \int_0^r J(r') (2\pi r') dr'$

4. Apply Ampere's Law: $2\pi a r^3 = \mu_0 \int_0^r J(r') (2\pi r') dr'$ $a r^3 = \mu_0 \int_0^r J(r') r' dr'$

5. Differentiate to find $J(r)$: $\frac{d}{dr} (a r^3) = \frac{d}{dr} \left( \mu_0 \int_0^r J(r') r' dr' \right)$ $3a r^2 = \mu_0 J(r) r$

6. Solve for $J(r)$: $J(r) = \frac{3a r^2}{\mu_0 r} = \frac{3a}{\mu_0} r$

7. Compare with the given form: $J = \frac{xa}{\mu_0}r^n$. Comparing with $J(r) = \frac{3a}{\mu_0} r$, we get $x=3$ and $n=1$.

8. Calculate (x + n): $x+n = 3+1 = 4$