A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is `30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin.

Options

A. Zero
B. `(mu_(0)I(b-a))/(24ab)`
C. `(mu_(0)I)/(4pi)[(b-a)/(ab)]`
D. `(mu_(0)I)/(4pi)[2(b-a)+(pi)/3(a+b)]`

Explanation

To find the magnitude of the magnetic field \( B \) at the origin \( O \) due to the current loop ABCD, we will analyze the contributions from different sections of the loop.

### Step-by-Step Solution:

**Step 1: Analyze the Current Loop**
- The current loop consists of two arcs (BC and DA) and two straight wires (AB and CD).
- The angle between the wires AB and CD at the origin \( O \) is \( 30^\circ \).

**Step 2: Magnetic Field Due to Arc DA**
- The magnetic field \( B_{DA} \) at the origin due to the arc DA can be calculated using the formula:
\[ B_{DA} = \frac{\mu_0 I}{4\pi a} \cdot \theta \]
where \( \theta \) is the angle in radians.
- Convert \( 30^\circ \) to radians:
\[ \theta = \frac{30 \times \pi}{180} = \frac{\pi}{6} \text{ radians} \]
- Substitute \( \theta \) into the formula:
\[ B_{DA} = \frac{\mu_0 I}{4\pi a} \cdot \frac{\pi}{6} = \frac{\mu_0 I}{24 a} \]
- The direction of \( B_{DA} \) is upward.

**Step 3: Magnetic Field Due to Arc BC**
- Similarly, the magnetic field \( B_{BC} \) at the origin due to the arc BC is given by:
\[ B_{BC} = \frac{\mu_0 I}{4\pi b} \cdot \theta \]
- Using the same angle \( \theta = \frac{\pi}{6} \):
\[ B_{BC} = \frac{\mu_0 I}{4\pi b} \cdot \frac{\pi}{6} = \frac{\mu_0 I}{24 b} \]
- The direction of \( B_{BC} \) is downward.

**Step 4: Magnetic Field Due to Straight Wires AB and CD**
- The straight wires AB and CD do not contribute to the magnetic field at the origin because they are aligned along the line \( OA \) and do not create a magnetic field at that point.

**Step 5: Calculate the Net Magnetic Field**
- The net magnetic field \( B_{net} \) at the origin is the vector sum of the contributions from the arcs:
\[ B_{net} = B_{DA} - B_{BC} = \frac{\mu_0 I}{24 a} - \frac{\mu_0 I}{24 b} \]
- Factor out \( \frac{\mu_0 I}{24} \):
\[ B_{net} = \frac{\mu_0 I}{24} \left( \frac{1}{a} - \frac{1}{b} \right) \]

**Step 6: Final Expression**
- Thus, the final expression for the magnetic field at the origin is:
\[ B_{net} = \frac{\mu_0 I}{24} \left( \frac{b - a}{ab} \right) \]
- The direction of the net magnetic field is upward (out of the plane of the paper).