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.
If the current density in a linear conductor of radius a varies with r according to relation `J=kr^2`, where k is a constant and r is the distance of a point from the axis of conductor, find the magnetic field induction at a point distance r from the axis when rlta. Assume relative permeability of 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 linear conductor with a varying current density, we can follow these steps:

### Step 1: Understand the Current Density
The current density \( J \) is given by the relation:
\[ J = k r^2 \]
where \( k \) is a constant and \( r \) is the distance from the axis of the conductor.

### Step 2: Define the Geometry
Consider a linear conductor with radius \( a \). We are interested in finding the magnetic field at a point where \( r < a \).

### Step 3: Calculate the Current in a Differential Element
To find the total current flowing through a circular shell of thickness \( dr \) at a distance \( r \) from the center, we can express the differential current \( dI \) as:
\[ dI = J \cdot dA \]
where \( dA \) is the area of the differential shell. The area \( dA \) can be calculated as:
\[ dA = 2 \pi r \, dr \]
Thus, substituting for \( J \):
\[ dI = (k r^2) (2 \pi r \, dr) = 2 \pi k r^3 \, dr \]

### Step 4: Integrate to Find the Total Current \( I \)
To find the total current \( I \) flowing through the conductor up to radius \( r \), we integrate \( dI \) from \( 0 \) to \( r \):
\[ I = \int_0^r dI = \int_0^r 2 \pi k r^3 \, dr \]
Calculating the integral:
\[ I = 2 \pi k \left[ \frac{r^4}{4} \right]_0^r = 2 \pi k \frac{r^4}{4} = \frac{\pi k r^4}{2} \]

### Step 5: 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 \( r \), the left-hand side simplifies to:
\[ B \cdot (2 \pi r) \]
Thus, we have:
\[ B (2 \pi r) = \mu_0 I \]
Substituting for \( I \):
\[ B (2 \pi r) = \mu_0 \left( \frac{\pi k r^4}{2} \right) \]

### Step 6: Solve for \( B \)
Now, we can solve for \( B \):
\[ B (2 \pi r) = \frac{\mu_0 \pi k r^4}{2} \]
Dividing both sides by \( 2 \pi r \):
\[ B = \frac{\mu_0 k r^3}{4} \]

### Final Result
The magnetic field induction at a point distance \( r \) from the axis of the conductor (where \( r < a \)) is:
\[ B = \frac{\mu_0 k r^3}{4} \]

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