To solve the problem, we need to calculate the magnetic field at point P(2a, 0) due to a thin sheet of width 2a carrying a uniformly distributed current I in the +y direction. Here’s a step-by-step breakdown of the solution:
### Step 1: Understanding the Setup
We have a thin sheet of width 2a placed in the x-y plane between the lines x = -a and x = a. The current I flows in the +y direction, uniformly distributed across the width of the sheet.
### Step 2: Define the Current Density
The current density (λ) can be defined as the current per unit length across the width of the sheet:
\[
\lambda = \frac{I}{2a}
\]
### Step 3: Consider an Element of Current
Consider a differential element of current (di) at a position x within the sheet. The width of this element is dx, and the current through this element is given by:
\[
di = \lambda \, dx = \frac{I}{2a} \, dx
\]
### Step 4: Calculate the Magnetic Field Contribution (dB)
The magnetic field contribution (dB) at point P due to the current element di can be calculated using the formula for the magnetic field around a long straight wire:
\[
dB = \frac{\mu_0 \, di}{2\pi r}
\]
where \( r \) is the distance from the current element to point P. Since point P is at (2a, 0), the distance \( r \) from the current element at position x is:
\[
r = 2a - x
\]
Thus, we have:
\[
dB = \frac{\mu_0}{2\pi (2a - x)} \cdot \frac{I}{2a} \, dx
\]
### Step 5: Integrate to Find the Total Magnetic Field (B)
To find the total magnetic field at point P, we need to integrate dB from x = -a to x = a:
\[
B = \int_{-a}^{a} dB = \int_{-a}^{a} \frac{\mu_0 I}{4\pi a (2a - x)} \, dx
\]
### Step 6: Change of Variables
Let \( u = 2a - x \), then \( du = -dx \). The limits change accordingly:
- When \( x = -a \), \( u = 3a \)
- When \( x = a \), \( u = a \)
Thus, the integral becomes:
\[
B = \frac{\mu_0 I}{4\pi a} \int_{3a}^{a} \frac{-du}{u} = \frac{\mu_0 I}{4\pi a} \left[ \ln u \right]_{3a}^{a}
\]
This simplifies to:
\[
B = \frac{\mu_0 I}{4\pi a} \left( \ln a - \ln 3a \right) = \frac{\mu_0 I}{4\pi a} \ln \frac{a}{3a} = \frac{\mu_0 I}{4\pi a} \ln \frac{1}{3} = -\frac{\mu_0 I}{4\pi a} \ln 3
\]
### Step 7: Final Expression for B
The magnitude of the magnetic field at point P is:
\[
B = \frac{\mu_0 I}{4\pi a} \ln 3
\]
### Step 8: Relate to Given Expression
The problem states that the magnetic field can be expressed as:
\[
B = \frac{1}{n} \cdot \frac{\mu_0 I}{\pi a} \ln 3
\]
By comparing the two expressions for B, we find:
\[
\frac{\mu_0 I}{4\pi a} \ln 3 = \frac{1}{n} \cdot \frac{\mu_0 I}{\pi a} \ln 3
\]
Cancelling common terms, we get:
\[
\frac{1}{4} = \frac{1}{n}
\]
Thus, \( n = 4 \).
### Conclusion
The value of n is:
\[
\boxed{4}
\]
