Holow cylindrical wire carries current I, having inner and outer radii R and 2R respectively. The magnetic field at a point which is `(5R)/4`, distance away from the wire?

Options

A. `(5mu_(0)I)/(18piR)`
B. `(mu_(0)I)/(36piR)`
C. `(5mu_(0)I)/(36piR)`
D. `3/40(mu_(0)I)/(piR)`

Explanation

To solve the problem of finding the magnetic field at a distance of \( \frac{5R}{4} \) from the center of a hollow cylindrical wire with inner radius \( R \) and outer radius \( 2R \) carrying a current \( I \), we can follow these steps:

### Step 1: Identify the parameters
We have:
- Inner radius \( r = R \)
- Outer radius \( R = 2R \)
- Distance from the center \( d = \frac{5R}{4} \)

### Step 2: Determine if the point is inside or outside the hollow cylinder
The distance \( d = \frac{5R}{4} \) is greater than the outer radius \( 2R \) (since \( \frac{5R}{4} = 1.25R < 2R \)). Therefore, the point is outside the hollow cylinder.

### Step 3: Calculate the current density
The current flows through the area between the inner and outer radii. The area \( A \) through which the current flows is given by:
\[ A = \pi (2R)^2 - \pi (R)^2 = \pi (4R^2 - R^2) = \pi (3R^2) = 3\pi R^2 \]
The current density \( J \) is given by:
\[ J = \frac{I}{A} = \frac{I}{3\pi R^2} \]

### Step 4: Calculate the enclosed current
To find the magnetic field outside the hollow cylinder, we need to find the total current \( I \) flowing through the hollow cylinder. Since we are outside the cylinder, we can use the total current \( I \) directly.

### Step 5: Use Ampère's Law to find the magnetic field
For a cylindrical conductor, the magnetic field \( B \) at a distance \( r \) from the axis is given by:
\[ B = \frac{\mu_0 I_{\text{enc}}}{2\pi r} \]
where \( I_{\text{enc}} \) is the total current enclosed by the Amperian loop of radius \( r \).

Since we are outside the hollow cylinder, \( I_{\text{enc}} = I \).

### Step 6: Substitute the values into the equation
Substituting \( I_{\text{enc}} = I \) and \( r = \frac{5R}{4} \):
\[ B = \frac{\mu_0 I}{2\pi \left(\frac{5R}{4}\right)} = \frac{\mu_0 I}{2\pi} \cdot \frac{4}{5R} = \frac{2\mu_0 I}{5\pi R} \]

### Final Answer
Thus, the magnetic field at a distance \( \frac{5R}{4} \) from the center of the hollow cylindrical wire is:
\[ B = \frac{2\mu_0 I}{5\pi R} \]