A thin sheet of width 2a is placed in the x-y plane between the lines `x=-a` and `x=a`. The sheet is very long in the -y and +y directions. It carries a current l towards the +y direction. This current is uniformly distributed over the width of the sheet. The magnitude of magnetic fiedl at the point `P(2a,0)` due to the sheet is `(1/n)((mu_(0)I)/(pia))log_(e)3`. Here the value of n is ______________.

Explanation

To solve the problem, we need to find the magnetic field at the point \( P(2a, 0) \) due to a thin sheet of current that is uniformly distributed between \( x = -a \) and \( x = a \). The sheet carries a current \( I \) in the positive \( y \)-direction.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The sheet is located between \( x = -a \) and \( x = a \) in the \( x-y \) plane.
- The point \( P(2a, 0) \) is located at \( x = 2a \).

2. **Setting Up the Current Element**:
- Consider a small strip of the sheet at position \( x \) with a width \( dx \).
- The current \( I \) is uniformly distributed across the width of the sheet, so the current density \( J \) can be defined as:
\[ J = \frac{I}{2a} \]
- The small current element \( dI \) through the strip of width \( dx \) is:
\[ dI = J \cdot dx = \frac{I}{2a} \cdot dx \]

3. **Calculating the Magnetic Field Contribution**:
- The magnetic field \( dB \) at point \( P \) due to the small current element \( dI \) can be calculated using the formula for the magnetic field due to a straight current-carrying wire:
\[ dB = \frac{\mu_0 dI}{2 \pi r} \]
- Here, \( r \) is the distance from the current element to the point \( P \). The distance \( r \) can be expressed as:
\[ r = 2a - x \]

4. **Substituting \( dI \) into the Magnetic Field Equation**:
- Substituting \( dI \) into the equation for \( dB \):
\[ dB = \frac{\mu_0}{2 \pi (2a - x)} \cdot \left( \frac{I}{2a} \cdot dx \right) \]
- This simplifies to:
\[ dB = \frac{\mu_0 I}{4 \pi a (2a - x)} \cdot dx \]

5. **Integrating to Find Total Magnetic Field**:
- To find the total magnetic field \( B \) at point \( P \), integrate \( dB \) from \( x = -a \) to \( x = a \):
\[ B = \int_{-a}^{a} \frac{\mu_0 I}{4 \pi a (2a - x)} \, dx \]

6. **Evaluating the Integral**:
- The integral can be evaluated as:
\[ B = \frac{\mu_0 I}{4 \pi a} \int_{-a}^{a} \frac{1}{2a - x} \, dx \]
- The integral \( \int \frac{1}{2a - x} \, dx \) can be solved to yield:
\[ \int_{-a}^{a} \frac{1}{2a - x} \, dx = \left[-\ln(2a - x)\right]_{-a}^{a} = -\ln(2a - a) + \ln(2a + a) = -\ln(a) + \ln(3a) = \ln(3) \]

7. **Final Expression for Magnetic Field**:
- Substituting back into the expression for \( B \):
\[ B = \frac{\mu_0 I}{4 \pi a} \cdot \ln(3) \]

8. **Comparing with Given Expression**:
- The problem states that the magnetic field at point \( P \) is given by:
\[ B = \frac{1}{n} \cdot \frac{\mu_0 I}{\pi a} \ln(3) \]
- Comparing both expressions, we find:
\[ \frac{\mu_0 I}{4 \pi a} \ln(3) = \frac{1}{n} \cdot \frac{\mu_0 I}{\pi a} \ln(3) \]
- This implies:
\[ \frac{1}{n} = \frac{1}{4} \implies n = 4 \]

### Conclusion:
The value of \( n \) is \( \boxed{4} \).