Magnetic Field from Parallel Wires and Circular Loop

To determine the correct statements, we analyze the magnetic field produced by each component of the system: the two infinite straight wires and the circular loop.

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

• Wire 1 is along $x = +R, z = 0$ with current $\vec{I}_1 = I_1 \hat{j}$. The position vector from wire 1 to the origin is $\vec{r}_1 = -R\hat{i}$. Using the Biot-Savart law for an infinite straight wire: $\vec{B}_1(0,0,0) = \frac{\mu_0 I_1}{2\pi R} (\hat{j} \times (-\hat{i})) = \frac{\mu_0 I_1}{2\pi R} \hat{k}$

• Wire 2 is along $x = -R, z = 0$ with current $\vec{I}_2 = I_2 \hat{j}$. The position vector from wire 2 to the origin is $\vec{r}_2 = +R\hat{i}$. $\vec{B}_2(0,0,0) = \frac{\mu_0 I_2}{2\pi R} (\hat{j} \times \hat{i}) = -\frac{\mu_0 I_2}{2\pi R} \hat{k}$

• Circular Loop of radius $R$ is centered at $(0, 0, \sqrt{3}R)$ in a plane parallel to the $xy$-plane. The current $I$ circulates clockwise when viewed from above (i.e., looking in the $-\hat{k}$ direction). By the right-hand rule, the magnetic field along the axis of the loop points in the $-\hat{k}$ direction. At the origin, the axial distance from the centre of the loop is $z_0 = \sqrt{3}R$. The magnetic field at the origin due to the loop is: $\vec{B}_{\text{loop}}(0,0,0) = -\frac{\mu_0 I R^2}{2(R^2 + z_0^2)^{3/2}} \hat{k} = -\frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} \hat{k} = -\frac{\mu_0 I}{16R}\hat{k}$

The total magnetic field at the origin is: $\vec{B}(0,0,0) = \vec{B}_1 + \vec{B}_2 + \vec{B}_{\text{loop}} = \left[ \frac{\mu_0}{2\pi R}(I_1 - I_2) - \frac{\mu_0 I}{16R} \right] \hat{k}$

For $\vec{B}(0,0,0) = \vec{0}$, we must have: $I_1 - I_2 = \frac{\pi I}{8}$

• Option A: If $I_1 = I_2$, then $I_1 - I_2 = 0$, giving: $\vec{B}(0,0,0) = -\frac{\mu_0 I}{16R}\hat{k} \neq \vec{0}$ Thus, $\vec{B}$ cannot be equal to zero. (Statement A is correct)

• Option B: If $I_1 > 0$ and $I_2 < 0$, then $(I_1 - I_2) > 0$. Since $\frac{\pi I}{8} > 0$, it is possible to choose currents such that $I_1 - I_2 = \frac{\pi I}{8}$, yielding $\vec{B}(0,0,0) = \vec{0}$. (Statement B is correct)

• Option C: If $I_1 < 0$ and $I_2 > 0$, then $(I_1 - I_2) < 0$. Since $\frac{\pi I}{8} > 0$, the condition $I_1 - I_2 = \frac{\pi I}{8}$ can never be satisfied. Hence, $\vec{B}$ cannot be zero. (Statement C is incorrect)

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

• Wire 1 at $(R, y, 0)$: The displacement vector to the centre of the loop is $\vec{r}_1' = -R\hat{i} + \sqrt{3}R\hat{k}$. The distance is: $d = \sqrt{(-R)^2 + (\sqrt{3}R)^2} = 2R$ The magnetic field produced by wire 1 is: $\vec{B}_1 = \frac{\mu_0 I_1}{2\pi (2R)^2} (\hat{j} \times (-R\hat{i} + \sqrt{3}R\hat{k})) = \frac{\mu_0 I_1}{4\pi R} \left(\sqrt{3}\hat{i} + \hat{k}\right)$

• Wire 2 at $(-R, y, 0)$: The displacement vector to the centre of the loop is $\vec{r}_2' = R\hat{i} + \sqrt{3}R\hat{k}$. The distance is: $d = 2R$ The magnetic field produced by wire 2 is: $\vec{B}_2 = \frac{\mu_0 I_2}{2\pi (2R)^2} (\hat{j} \times (R\hat{i} + \sqrt{3}R\hat{k})) = \frac{\mu_0 I_2}{4\pi R} \left(\sqrt{3}\hat{i} - \hat{k}\right)$

The combined $z$-component from the two straight wires is: $B_{z,\text{wires}} = \frac{\mu_0}{4\pi R}(I_1 - I_2)$

If $I_1 = I_2$, then: $B_{z,\text{wires}} = 0$

• Circular Loop: At the centre of the loop, the magnetic field is entirely along the axis ($z$-axis). Because the current circulates clockwise when viewed from above: $\vec{B}_{\text{loop}} = -\frac{\mu_0 I}{2R}\hat{k}$

Therefore, when $I_1 = I_2$, the net $z$-component of the magnetic field at the centre of the loop is: $B_z = -\frac{\mu_0 I}{2R}$ (Statement D is correct)

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Conclusion

The correct statements are A, B, and D.