A straight section `PQ` of a circuit lies 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. Proportional to `a`
B. Proportional to `a^(2)`
C. Proprotional to `1//a`
D. Zero

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. Here’s a step-by-step solution:

### Step 1: Understand the Setup
The straight section \( PQ \) of the circuit lies along the X-axis from \( x = -\frac{a}{2} \) to \( x = +\frac{a}{2} \). The current \( i \) flows through this section.

### Step 2: Identify the Point of Interest
We need to calculate the magnetic field at the point \( X = +a \), which is located to the right of the wire segment \( PQ \).

### Step 3: Apply the Biot-Savart Law
The Biot-Savart law states that the magnetic field \( dB \) at a point due to a current element \( Idl \) is given by:
\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \times \hat{r}}{r^2} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( dl \) is the current element,
- \( \hat{r} \) 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 4: Determine the Geometry
For a small segment \( dl \) at position \( x \) along the wire, the distance \( r \) to the point \( X = +a \) is:
\[ r = a - x \]
The angle \( \theta \) between \( dl \) and \( \hat{r} \) can be determined from the geometry of the setup.

### Step 5: Calculate the Angle
Since \( dl \) is along the X-axis and \( \hat{r} \) points from the wire to the point \( X = +a \), the angle \( \theta \) between \( dl \) and \( \hat{r} \) is \( 0^\circ \) for all points along the wire (since both are along the same line).

### Step 6: Evaluate \( dl \times \hat{r} \)
Since \( \theta = 0^\circ \), we have:
\[ dl \times \hat{r} = dl \cdot \sin(0^\circ) = 0 \]
This implies that the magnetic field contribution \( dB \) from each segment \( dl \) is zero.

### Step 7: Conclude the Result
Since the magnetic field contributions from all segments of the wire add up to zero, the total magnetic field \( B \) at point \( X = +a \) is:
\[ B = 0 \]

### Final Answer
The magnetic field at the point \( X = +a \) due to the straight section \( PQ \) is \( 0 \).

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