The problem asks for an analysis of the magnetic field produced by two infinite straight wires and a circular loop. The total magnetic field at any point is the vector sum of the fields from these three sources: $\overrightarrow B = {\overrightarrow B _1} + {\overrightarrow B _2} + {\overrightarrow B _{{\rm{loop}}}}$ Let's analyze the magnetic field at the two points of interest: the origin $(0,0,0)$ and the center of the loop $(0,0,\sqrt 3 R)$.

1. Magnetic Field at the Origin (0,0,0)

• Field from Wire 1 (${\overrightarrow B _1}$): The wire is at $x = +R$ and carries current $I_1$ along the y-axis. The distance from the wire to the origin is $R$. According to the right-hand rule, if $I_1 > 0$ (current in $+\hat{j}$ direction), the magnetic field at the origin points in the $+\hat{k}$ direction. The magnitude is $B_1 = \frac{{{\mu _0}{I_1}}}{{2\pi R}}$. So, ${\overrightarrow B _1}(0,0,0) = \frac{{{\mu _0}{I_1}}}{{2\pi R}}\widehat k$

• Field from Wire 2 (${\overrightarrow B _2}$): The wire is at $x = -R$ and carries current $I_2$ along the y-axis. The distance from the wire to the origin is $R$. According to the right-hand rule, if $I_2 > 0$ (current in $+\hat{j}$ direction), the magnetic field at the origin points in the $-\hat{k}$ direction. So, ${\overrightarrow B _2}(0,0,0) = - \frac{{{\mu _0}{I_2}}}{{2\pi R}}\widehat k$

• Field from the Loop (${\overrightarrow B _{{\rm{loop}}}}$): The loop has radius $R$ and its center is at $(0, 0, \sqrt{3}R)$. The origin is on the axis of the loop at a distance $z' = \sqrt{3}R$ from its center. The magnetic field on the axis of a circular loop is given by B = \frac{{{\mu _0}I{R^2}}}{{2{{({R^2} + z{'^2})}^{3/2}}}}}. The current $I$ is clockwise as seen from above, so its magnetic moment points in the $-\hat{k}$ direction. ${\overrightarrow B _{{\rm{loop}}}}(0,0,0) = \frac{{{\mu _0}I{R^2}}}{{2{{({R^2} + {{(\sqrt 3 R)}^2})}^{3/2}}}}( - \widehat k) = \frac{{{\mu _0}I{R^2}}}{{2{{(4{R^2})}^{3/2}}}}( - \widehat k) = \frac{{{\mu _0}I{R^2}}}{{2(8{R^3})}}( - \widehat k) = - \frac{{{\mu _0}I}}{{16R}}\widehat k$

• Total Field at the Origin: $\overrightarrow B (0,0,0) = {\overrightarrow B _1} + {\overrightarrow B _2} + {\overrightarrow B _{{\rm{loop}}}} = \left( {\frac{{{\mu _0}{I_1}}}{{2\pi R}} - \frac{{{\mu _0}{I_2}}}{{2\pi R}} - \frac{{{\mu _0}I}}{{16R}}} \right)\widehat k = \frac{{{\mu _0}}}{R}\left( {\frac{{{I_1} - {I_2}}}{{2\pi }} - \frac{I}{{16}}} \right)\widehat k$

Evaluation of Options A, B, C:

• A: If ${I_1} = {I_2},$ then $\overrightarrow B$ cannot be equal to zero at the origin. If $I_1 = I_2$, the expression for the field becomes: $\overrightarrow B (0,0,0) = \frac{{{\mu _0}}}{R}\left( {0 - \frac{I}{{16}}} \right)\widehat k = - \frac{{{\mu _0}I}}{{16R}}\widehat k$ Since the loop current $I$ is constant and non-zero, $\overrightarrow B (0,0,0) \ne 0$. Thus, statement A is true.

• B: If ${I_1} > 0$ and ${I_2} < 0,$ then $\overrightarrow B$ can be equal to zero at the origin. For $\overrightarrow B(0,0,0) = 0$, we need: $\frac{{{I_1} - {I_2}}}{{2\pi }} - \frac{I}{{16}} = 0 \implies {I_1} - {I_2} = \frac{{\pi I}}{8}$ Given $I_1 > 0$ and $I_2 < 0$, let $I_2 = -|I_2|$. Then $I_1 - (-|I_2|) = I_1 + |I_2|$. Since $I_1 > 0$ and $|I_2| > 0$, their sum is positive. The right side, $\frac{{\pi I}}{8}$, is also positive (assuming $I>0$). This condition can be satisfied by choosing appropriate values for the currents. Thus, statement B is true.

• C: If ${I_1} < 0$ and ${I_2} > 0,$ then $\overrightarrow B$ can be equal to zero at the origin. The condition for zero field is still ${I_1} - {I_2} = \frac{{\pi I}}{8}$. Given $I_1 < 0$ and $I_2 > 0$, the left side $(I_1 - I_2)$ is a negative number minus a positive number, which is always negative. The right side $\frac{{\pi I}}{8}$ is positive. A negative number cannot equal a positive number. Therefore, the field cannot be zero. Thus, statement C is false.

2. Magnetic Field at the Center of the Loop $(0,0,\sqrt 3 R)$

Let's find the z-component of the magnetic field, $B_z$, at the loop's center $P_c(0, 0, \sqrt{3}R)$.

• Field from the Loop ($B_{z, \text{loop}}$): The field at the center of the loop itself is $B = \frac{{{\mu _0}I}}{{2R}}$. Since the current is clockwise from above, the field is in the $-\hat{k}$ direction. $B_{z, \text{loop}} = -\frac{\mu_0 I}{2R}$

• Field from Wire 1 ($B_{z,1}$): The distance from wire 1 (at $x=R, z=0$) to $P_c$ is $d_1 = \sqrt{{(0-R)^2} + {{(\sqrt 3 R - 0)}^2}} = 2R$. The magnetic field has magnitude $B_1 = \frac{{{\mu _0}{I_1}}}{{2\pi d_1}} = \frac{{{\mu _0}{I_1}}}{{4\pi R}}$. The field lines are circles in the xz-plane. The vector from the wire to the point $P_c$ is $\vec{r} = -R\hat{i} + \sqrt{3}R\hat{k}$. The field direction is tangential, given by $\hat{j} \times \vec{r}$, which results in a direction vector $\frac{{\sqrt 3 }}{2}\widehat i + \frac{1}{2}\widehat k$. The z-component of the field is $B_{z,1} = B_1 \cos\theta$, where $\cos\theta = \frac{R}{2R} = \frac{1}{2}$. More formally, $B_{z,1} = \frac{{\mu _0}{I_1}}{4\pi R} \times (\frac{1}{2}) = \frac{{\mu _0}{I_1}}{{8\pi R}}$.

• Field from Wire 2 ($B_{z,2}$): By symmetry, the distance is $d_2 = 2R$ and the magnitude is $B_2 = \frac{{{\mu _0}{I_2}}}{{4\pi R}}$. The vector from the wire to the point is $\vec{r} = R\hat{i} + \sqrt{3}R\hat{k}$. The field direction is given by $\hat{j} \times \vec{r}$, resulting in a direction vector $\frac{{\sqrt 3 }}{2}\widehat i - \frac{1}{2}\widehat k$. The z-component is $B_{z,2} = -\frac{{\mu _0}{I_2}}{{8\pi R}}$.

• Total z-component at the loop's center: $B_z(P_c) = B_{z,1} + B_{z,2} + B_{z, \text{loop}} = \frac{{{\mu _0}{I_1}}}{{8\pi R}} - \frac{{{\mu _0}{I_2}}}{{8\pi R}} - \frac{{{\mu _0}I}}{{2R}} = \frac{{{\mu _0}}}{R}\left( {\frac{{{I_1} - {I_2}}}{{8\pi }} - \frac{I}{2}} \right)$

Evaluation of Option D:

• D: If ${I_1} = {I_2},$ then the $z$-component of the magnetic field at the center of the loop is $\left( { - {{{\mu _0}I} \over {2R}}} \right)$. If $I_1 = I_2$, the expression for the z-component becomes: $B_z(P_c) = \frac{{{\mu _0}}}{R}\left( {\frac{{{I_1} - {I_1}}}{{8\pi }} - \frac{I}{2}} \right) = - \frac{{{\mu _0}I}}{{2R}}$ This matches the value given in the statement. Thus, statement D is true.

Conclusion: Statements A, B, and D are true. Statement C is false.