The current density `bar(J)` inside a long, solid cylindrical wire of radius `a = 12 mm` is in the direction of the central axis, and its magnitude varies linearly with radial distance r from the axis according to `J = (J_(0) r)/(a)`, where `J_(0) = (10^(5))/(4 pi) A//m^(2)`. Find the magnitude of the magnetic field at `r = (a)/(2)` in `mu T`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the current density function
The current density \( \bar{J} \) is given by the equation:
\[ J = \frac{J_0 r}{a} \]
where \( J_0 = \frac{10^5}{4\pi} \, \text{A/m}^2 \) and \( a = 12 \, \text{mm} = 12 \times 10^{-3} \, \text{m} \).

### Step 2: Calculate the current flowing through a radius \( r = \frac{a}{2} \)
To find the total current \( I \) flowing through the wire up to radius \( r = \frac{a}{2} \), we use the formula:
\[ I = \int_0^{r} J \, dA \]
where \( dA = 2\pi r \, dr \). Thus, we have:
\[ I = \int_0^{\frac{a}{2}} J \cdot 2\pi r \, dr \]
Substituting \( J \):
\[ I = \int_0^{\frac{a}{2}} \frac{J_0 r}{a} \cdot 2\pi r \, dr = \frac{2\pi J_0}{a} \int_0^{\frac{a}{2}} r^2 \, dr \]

### Step 3: Evaluate the integral
The integral \( \int r^2 \, dr \) from 0 to \( \frac{a}{2} \) is:
\[ \int_0^{\frac{a}{2}} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^{\frac{a}{2}} = \frac{1}{3} \left( \frac{a}{2} \right)^3 = \frac{a^3}{24} \]
Substituting this back into the equation for \( I \):
\[ I = \frac{2\pi J_0}{a} \cdot \frac{a^3}{24} = \frac{\pi J_0 a^2}{12} \]

### Step 4: Substitute the values
Now substituting \( J_0 \) and \( a \):
\[ I = \frac{\pi \left(\frac{10^5}{4\pi}\right) (12 \times 10^{-3})^2}{12} \]
This simplifies to:
\[ I = \frac{10^5 (12 \times 10^{-3})^2}{48} \]

### Step 5: Calculate the magnetic field using Ampere's Law
According to Ampere's Circuital Law:
\[ \oint B \cdot dL = \mu_0 I \]
For a circular path of radius \( r = \frac{a}{2} \):
\[ B \cdot 2\pi \left(\frac{a}{2}\right) = \mu_0 I \]
Thus,
\[ B \cdot \pi a = \mu_0 I \]
Substituting \( I \):
\[ B \cdot \pi a = \mu_0 \cdot \frac{\pi J_0 a^2}{12} \]
Cancelling \( \pi \) and rearranging gives:
\[ B = \frac{\mu_0 J_0 a}{12} \]

### Step 6: Substitute the values for \( B \)
Now substituting \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) and \( J_0 = \frac{10^5}{4\pi} \):
\[ B = \frac{(4\pi \times 10^{-7}) \left(\frac{10^5}{4\pi}\right) (12 \times 10^{-3})}{12} \]
This simplifies to:
\[ B = 10^{-6} \, \text{T} = 10 \, \mu T \]

### Final Answer
The magnitude of the magnetic field at \( r = \frac{a}{2} \) is:
\[ \boxed{10 \, \mu T} \]