Two infinitely long straight wires lie in the xy-plane along the lines `x =pm R`. The wire located at `x =+R` carries a constant current `I_(1)` and the wire located at ` x = -R` carries a constant current `I_(2)`. A circular loop of radius R is suspended with its centre at `(0,0, sqrt(3)R)` and in a plane parallel to the xy-plane. This loop carries a constant current ? in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the `+hat J` direction. Which of the following statements regarding the magnetic field ` vec B` is (are) true?

Options

A. If `I_(1)=I_(2)`, then `vecB`cannot be equal to zero at the origin (0, 0, 0)
B. If `I_(1)gt0` and `I_(2)lt0`, then `vecB` can be equal to zero at the origin (0, 0, 0)
C. If `I_(1)lt0` and `I_(2)gt0` then `vecB`can be equal to zero at the origin (0, 0, 0)
D. If `I_(1)=I_(2)`, then the z-component of the magnetic field at the centre of the loop is `(-(mu_(0)I)/(2R))`

Explanation

To solve the problem regarding the magnetic field generated by two infinitely long straight wires and a circular loop, we will follow these steps:

### Step 1: Understand the Configuration
We have two infinitely long straight wires located at \( x = +R \) and \( x = -R \) in the xy-plane. The wire at \( x = +R \) carries a current \( I_1 \) in the positive y-direction, and the wire at \( x = -R \) carries a current \( I_2 \) in the negative y-direction. A circular loop of radius \( R \) is centered at \( (0, 0, \sqrt{3}R) \) and carries a current \( I \) in the clockwise direction as viewed from above.

### Step 2: Determine the Magnetic Field Due to Each Wire
Using the formula for the magnetic field around a long straight wire:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( r \) is the distance from the wire.

- For wire 1 at \( x = +R \):
- The distance from the center of the loop to wire 1 is \( \sqrt{R^2 + ( \sqrt{3}R)^2} = \sqrt{4R^2} = 2R \).
- The magnetic field \( B_1 \) due to wire 1 at the center of the loop is:
\[ B_1 = \frac{\mu_0 I_1}{2 \pi (2R)} = \frac{\mu_0 I_1}{4 \pi R} \]
- The direction of \( B_1 \) is determined using the right-hand rule and points in the positive z-direction (out of the page).

- For wire 2 at \( x = -R \):
- The distance from the center of the loop to wire 2 is also \( 2R \).
- The magnetic field \( B_2 \) due to wire 2 at the center of the loop is:
\[ B_2 = \frac{\mu_0 I_2}{2 \pi (2R)} = \frac{\mu_0 I_2}{4 \pi R} \]
- The direction of \( B_2 \) points in the negative z-direction (into the page).

### Step 3: Calculate the Net Magnetic Field Due to the Wires
- If \( I_1 = I_2 \), then:
\[ B_{\text{net}} = B_1 - B_2 = \frac{\mu_0 I_1}{4 \pi R} - \frac{\mu_0 I_2}{4 \pi R} = 0 \]
Thus, the magnetic field due to the wires cancels out at the center of the loop.

### Step 4: Determine the Magnetic Field Due to the Loop
The magnetic field at the center of the loop due to the loop itself is given by:
\[ B_{\text{loop}} = \frac{\mu_0 I}{2R} \]
Since the current in the loop is clockwise, the direction of \( B_{\text{loop}} \) is in the negative z-direction (into the page).

### Step 5: Combine the Magnetic Fields
- If \( I_1 = I_2 \), the total magnetic field at the center of the loop is:
\[ B_{\text{total}} = B_{\text{loop}} + B_{\text{net}} = -\frac{\mu_0 I}{2R} + 0 = -\frac{\mu_0 I}{2R} \]

### Conclusion
The magnetic field at the center of the loop is non-zero and directed into the page when \( I_1 = I_2 \).