An infinitely long cylinderical wire of radius R is carrying a current with current density `j=alphar^(3)` (where `alpha` is constant and r is the distance from the axis of the wire). If the magnetic fixed at `r=(R)/(2)` is `B_(1)` and at r=2R is `B_(2)` then the ratio `(B_(2))/(B_(1))` is

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the problem, we need to find the ratio of the magnetic fields \( B_2 \) and \( B_1 \) at two different distances from the axis of an infinitely long cylindrical wire carrying a current with a given current density.

### Step-by-Step Solution:

1. **Understanding Current Density**:
The current density \( j \) is given by:
\[ j = \alpha r^3 \]
where \( \alpha \) is a constant and \( r \) is the distance from the axis of the wire.

2. **Finding Total Current Enclosed**:
To find the magnetic field inside the wire at \( r = \frac{R}{2} \), we will first calculate the total current \( I \) enclosed within a radius \( r \):
\[ I = \int_0^r j \cdot dA = \int_0^r j \cdot 2\pi r' \, dr' \]
Substituting \( j \):
\[ I = \int_0^r \alpha (r')^3 \cdot 2\pi r' \, dr' = 2\pi \alpha \int_0^r (r')^4 \, dr' \]
Evaluating the integral:
\[ I = 2\pi \alpha \left[ \frac{(r')^5}{5} \right]_0^r = \frac{2\pi \alpha r^5}{5} \]

3. **Applying Ampere's Law**:
According to Ampere's Law, the magnetic field \( B \) at a distance \( r \) from the center is given by:
\[ B \cdot 2\pi r = \mu_0 I \]
Therefore,
\[ B = \frac{\mu_0 I}{2\pi r} \]
Substituting for \( I \):
\[ B = \frac{\mu_0}{2\pi r} \cdot \frac{2\pi \alpha r^5}{5} = \frac{\mu_0 \alpha r^4}{5} \]

4. **Finding \( B_1 \) at \( r = \frac{R}{2} \)**:
Now, substituting \( r = \frac{R}{2} \):
\[ B_1 = \frac{\mu_0 \alpha \left(\frac{R}{2}\right)^4}{5} = \frac{\mu_0 \alpha R^4}{5 \cdot 16} = \frac{\mu_0 \alpha R^4}{80} \]

5. **Finding \( B_2 \) at \( r = 2R \)**:
For \( r = 2R \), we consider the total current \( I \) flowing through the entire wire:
\[ I = \int_0^R j \cdot 2\pi r' \, dr' = 2\pi \alpha \int_0^R (r')^4 \, dr' = 2\pi \alpha \left[ \frac{(r')^5}{5} \right]_0^R = \frac{2\pi \alpha R^5}{5} \]
Now applying Ampere's Law for \( r = 2R \):
\[ B_2 \cdot 2\pi (2R) = \mu_0 I \]
Thus,
\[ B_2 = \frac{\mu_0 I}{4\pi R} = \frac{\mu_0}{4\pi R} \cdot \frac{2\pi \alpha R^5}{5} = \frac{\mu_0 \alpha R^4}{10} \]

6. **Finding the Ratio \( \frac{B_2}{B_1} \)**:
Now we can find the ratio:
\[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 \alpha R^4}{10}}{\frac{\mu_0 \alpha R^4}{80}} = \frac{80}{10} = 8 \]

### Final Answer:
The ratio \( \frac{B_2}{B_1} \) is \( 8 \).