Ampere's law provides us an easy way to calculate the magnetic field due to a symmetrical distribution of current. Its mathemfield expression is `ointvecB.dl=mu_0I_("in")`.
The quantity on the left hand side is known as line as integral of magnetic field over a closed Ampere's loop.
In the above question, find the magnetic field induction at a point distance r from the axis when rgta. Assume relative permeability of the medium surrounding the conductor to be unity.

Options

A. `(mu_0ka^4)/(4r)`
B. `(mu_0kr^3)/2`
C. `(mu_0kpia^4)/(2r)`
D. `(mu_0ka^3)/4`

Explanation

To find the magnetic field induction at a point distance \( r \) from the axis of a cylindrical conductor when \( r > A \), we can follow these steps:

### Step 1: Identify the Given Information
We know that:
- The current density \( J = k R^2 \)
- We need to find the magnetic field \( B \) at a distance \( r \) from the axis when \( r > A \).
- The relative permeability of the surrounding medium is unity, which means \( \mu = \mu_0 \).

### Step 2: Calculate the Enclosed Current \( I_{\text{in}} \)
To find the enclosed current \( I_{\text{in}} \) within a cylindrical conductor of radius \( A \), we use the formula:
\[ I_{\text{in}} = \int J \cdot dA \]
Here, \( dA \) for a cylindrical shell at radius \( R \) is given by \( dA = 2\pi R \, dR \).

Substituting \( J \):
\[ I_{\text{in}} = \int_0^A k R^2 (2\pi R \, dR) = 2\pi k \int_0^A R^3 \, dR \]

### Step 3: Perform the Integral
Now, we calculate the integral:
\[ \int_0^A R^3 \, dR = \left[ \frac{R^4}{4} \right]_0^A = \frac{A^4}{4} \]
Thus, substituting back:
\[ I_{\text{in}} = 2\pi k \cdot \frac{A^4}{4} = \frac{\pi k A^4}{2} \]

### Step 4: Apply Ampere's Law
According to Ampere's Law:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{in}} \]
For a circular path of radius \( r \) (where \( r > A \)), \( d\vec{l} = 2\pi r \). Thus:
\[ B \cdot (2\pi r) = \mu_0 \cdot \frac{\pi k A^4}{2} \]

### Step 5: Solve for the Magnetic Field \( B \)
Now, we can solve for \( B \):
\[ B \cdot (2\pi r) = \mu_0 \cdot \frac{\pi k A^4}{2} \]
Dividing both sides by \( 2\pi r \):
\[ B = \frac{\mu_0 k A^4}{4r} \]

### Final Answer
The magnetic field induction at a point distance \( r \) from the axis when \( r > A \) is:
\[ B = \frac{\mu_0 k A^4}{4r} \]

---