A straight section `PQ` of a circuit lise along the `X`-axis from `x=-a/2` to `x=a/2` and carriers a steady current `i`. The magnetic field due to the section `PQ` at a point `X=+a` will be

Options

A. `propa`
B. zero
C. `propa^(2)`
D. `prop1/a`

Explanation

To find the magnetic field at point \( X = +a \) due to the straight section \( PQ \) of the circuit carrying a steady current \( i \), we can use the Biot-Savart Law. The magnetic field \( \mathbf{B} \) due to a current-carrying conductor can be calculated using the formula:

\[ \mathbf{B} = \frac{\mu_0}{4\pi} \int \frac{i \, d\mathbf{l} \times \hat{\mathbf{r}}}{r^2} \]

where:
- \( \mu_0 \) is the permeability of free space,
- \( i \) is the current,
- \( d\mathbf{l} \) is the differential length element of the conductor,
- \( \hat{\mathbf{r}} \) is the unit vector from the current element to the point where the field is being calculated,
- \( r \) is the distance from the current element to the point.

### Step 1: Define the Geometry
The section \( PQ \) lies along the x-axis from \( x = -\frac{a}{2} \) to \( x = \frac{a}{2} \). We need to find the magnetic field at the point \( x = a \).

### Step 2: Set Up the Integral
The differential length element \( d\mathbf{l} \) can be expressed as \( dx \) along the x-axis. The position vector from a point on the conductor to the point \( x = a \) is given by \( \mathbf{r} = a - x \).

### Step 3: Calculate the Distance and Unit Vector
The distance \( r \) from the current element at position \( x \) to the point \( x = a \) is:
\[ r = a - x \]

The unit vector \( \hat{\mathbf{r}} \) pointing from the current element to the point \( x = a \) is:
\[ \hat{\mathbf{r}} = \frac{(a - x)\hat{\mathbf{i}}}{|a - x|} = \hat{\mathbf{i}} \quad (\text{since } a - x > 0 \text{ for } x < a) \]

### Step 4: Apply the Biot-Savart Law
Now, substituting into the Biot-Savart Law, we have:
\[ \mathbf{B} = \frac{\mu_0 i}{4\pi} \int_{-\frac{a}{2}}^{\frac{a}{2}} \frac{dx \, \hat{\mathbf{i}} \times \hat{\mathbf{i}}}{(a - x)^2} \]

Since \( \hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0 \), the entire integral evaluates to zero:
\[ \mathbf{B} = 0 \]

### Conclusion
Thus, the magnetic field at the point \( x = a \) due to the straight section \( PQ \) of the circuit carrying a steady current \( i \) is:
\[ \mathbf{B} = 0 \]

### Final Answer
The magnetic field at the point \( X = +a \) is \( 0 \).

---