Consider a long cylindrical wire carrying current along the axis of wire. The current distributed non-uniformly on cross section. The magnetic field B inside the wire varies with distance `r` from the axis as `B=kr^(4)` where k is a cosntant. The current density `J` varies with distance `r` as `J=cr^(n)` then the value of `n` is (where c is another constant).

Explanation

To solve the problem, we need to find the value of \( n \) given the magnetic field \( B \) inside a cylindrical wire and the current density \( J \).

### Step-by-Step Solution:

1. **Given Information**:
- The magnetic field inside the wire is given by:
\[ B = k r^4 \]
- The current density is given by:
\[ J = c r^n \]

2. **Apply Ampere's Law**:
Ampere's Law states that:
\[ \oint B \cdot dl = \mu_0 I_{\text{enclosed}} \]
For a cylindrical wire, we can choose an Amperian loop of radius \( r \).

3. **Calculate the Left Side of Ampere's Law**:
The line integral of \( B \) around the Amperian loop is:
\[ \oint B \cdot dl = B \cdot (2\pi r) = (k r^4) \cdot (2\pi r) = 2\pi k r^5 \]

4. **Calculate the Right Side of Ampere's Law**:
The enclosed current \( I_{\text{enclosed}} \) can be calculated using the current density \( J \):
\[ I_{\text{enclosed}} = \int J \, dA \]
The differential area element \( dA \) for a cylindrical shell is:
\[ dA = 2\pi r \, dr \]
Therefore:
\[ I_{\text{enclosed}} = \int_0^r J \, dA = \int_0^r (c r^n) (2\pi r \, dr) = 2\pi c \int_0^r r^{n+1} \, dr \]

5. **Evaluate the Integral**:
\[ \int_0^r r^{n+1} \, dr = \frac{r^{n+2}}{n+2} \]
Thus, we have:
\[ I_{\text{enclosed}} = 2\pi c \cdot \frac{r^{n+2}}{n+2} \]

6. **Set the Left Side Equal to the Right Side**:
Now we equate the two sides of Ampere's Law:
\[ 2\pi k r^5 = \mu_0 \left(2\pi c \cdot \frac{r^{n+2}}{n+2}\right) \]

7. **Cancel Common Factors**:
Dividing both sides by \( 2\pi \):
\[ k r^5 = \mu_0 c \cdot \frac{r^{n+2}}{n+2} \]

8. **Equate the Powers of \( r \)**:
Since this equation must hold for all \( r \), we can equate the powers of \( r \):
\[ 5 = n + 2 \]

9. **Solve for \( n \)**:
Rearranging gives:
\[ n = 5 - 2 = 3 \]

### Final Answer:
The value of \( n \) is:
\[ \boxed{3} \]