A solid cylindrical conductor of radius `R` carries a current along its length.The current density `J_(1)` however, is not unifrom over the cross section of the conductor but is a function of the radius according to `J=br`, where `b` is a constant.the magnetic field `B` at `r_(1) lt R` is `(mu_(0)b r_(1)^(2))/N`.Then find value of `N`

Explanation

To solve the problem, we need to find the value of \( N \) in the expression for the magnetic field \( B \) at a distance \( r_1 \) from the center of a solid cylindrical conductor carrying a non-uniform current density \( J = br \). Here are the steps to arrive at the solution:

### Step 1: Understand the Current Density
The current density \( J \) is given by:
\[ J = br \]
where \( b \) is a constant and \( r \) is the radial distance from the center of the cylinder.

### Step 2: Calculate the Differential Current Element
To find the total current flowing through a radius \( r_1 \) (where \( r_1 < R \)), we consider a thin cylindrical shell of radius \( r \) and thickness \( dr \). The area \( dA \) of this shell is:
\[ dA = 2\pi r \, dr \]
The differential current \( dI \) flowing through this area is:
\[ dI = J \cdot dA = (br) \cdot (2\pi r \, dr) = 2\pi b r^2 \, dr \]

### Step 3: Integrate to Find the Total Current \( I \)
To find the total current \( I \) flowing through the area up to \( r_1 \), we integrate \( dI \) from \( 0 \) to \( r_1 \):
\[ I = \int_0^{r_1} dI = \int_0^{r_1} 2\pi b r^2 \, dr \]
Calculating the integral:
\[ I = 2\pi b \int_0^{r_1} r^2 \, dr = 2\pi b \left[ \frac{r^3}{3} \right]_0^{r_1} = 2\pi b \frac{r_1^3}{3} \]
Thus, the total current \( I \) is:
\[ I = \frac{2\pi b r_1^3}{3} \]

### Step 4: Apply Ampère's Circuital Law
According to Ampère's law, the magnetic field \( B \) at a distance \( r_1 \) from the center of the cylinder can be expressed as:
\[ \oint B \cdot dl = \mu_0 I \]
The left side of the equation simplifies to \( B \cdot 2\pi r_1 \) (since \( B \) is constant along the circular path):
\[ B \cdot 2\pi r_1 = \mu_0 I \]
Substituting \( I \):
\[ B \cdot 2\pi r_1 = \mu_0 \left( \frac{2\pi b r_1^3}{3} \right) \]
Dividing both sides by \( 2\pi r_1 \):
\[ B = \frac{\mu_0 b r_1^2}{3} \]

### Step 5: Compare with Given Expression
The problem states that the magnetic field \( B \) at \( r_1 \) is also given by:
\[ B = \frac{\mu_0 b r_1^2}{N} \]
By comparing the two expressions for \( B \):
\[ \frac{\mu_0 b r_1^2}{3} = \frac{\mu_0 b r_1^2}{N} \]
We can cancel \( \mu_0 b r_1^2 \) (assuming \( r_1 \neq 0 \)):
\[ \frac{1}{3} = \frac{1}{N} \]
Thus, we find:
\[ N = 3 \]

### Final Answer
The value of \( N \) is:
\[ \boxed{3} \]