Zero Magnetic Field on Loop Axis Due to Loop and Parallel Wires

To find the direction of the currents and the height $h$ at which the total magnetic field vanishes, we set up a Cartesian coordinate system:

• Let the plane of the paper be the $xy$-plane, with the center of the circular loop at the origin $(0, 0, 0)$.
• The axis of the loop lies along the $z$-axis, where "above the plane" corresponds to the positive $z$-direction ($+\hat{k}$).
• Wire 1 lies along the line $x = -d$, parallel to the $y$-axis (directed from $\text{P}$ to $\text{Q}$).
• Wire 2 lies along the line $x = +d$, parallel to the $y$-axis (with points $\text{R}$ and $\text{S}$ along it).

1. Magnetic Field Due to the Circular Loop

The current $I$ in the loop flows counterclockwise as viewed from above (looking down along the $-z$ direction). By the right-hand rule, the magnetic field along the loop's axis at a height $h$ ($z = h$) points in the $+\hat{k}$ direction:

$$\vec{B}_{\text{loop}} = \frac{\mu_0 I a^2}{2(a^2 + h^2)^{3/2}} \hat{k}$$

2. Magnetic Field Due to the Parallel Wires

For the net magnetic field on the axis at $(0, 0, h)$ to be zero, the combined magnetic field of the two wires, $\vec{B}_{\text{wires}}$, must be directed entirely along $-\hat{k}$ to cancel $\vec{B}_{\text{loop}}$, meaning any horizontal component must sum to zero.

Let the position vectors of the field point $(0, 0, h)$ relative to Wire 1 and Wire 2 be:

$$\vec{r}_1 = d\hat{i} + h\hat{k}, \quad r_1 = \sqrt{d^2 + h^2}$$ $$\vec{r}_2 = -d\hat{i} + h\hat{k}, \quad r_2 = \sqrt{d^2 + h^2}$$

• If the current in Wire 1 flows along $\text{PQ}$ (the $+\hat{j}$ direction), the magnetic field is:

$$\vec{B}_1 = \frac{\mu_0 I}{2\pi r_1^2} (\hat{j} \times \vec{r}_1) = \frac{\mu_0 I}{2\pi (d^2 + h^2)} \left[ \hat{j} \times (d\hat{i} + h\hat{k}) \right] = \frac{\mu_0 I}{2\pi (d^2 + h^2)} (h\hat{i} - d\hat{k})$$

• For the $\hat{i}$-components to cancel and the $\hat{k}$-components to reinforce in the $-\hat{k}$ direction, the current in Wire 2 must be directed along $\text{SR}$ (the $-\hat{j}$ direction):

$$\vec{B}_2 = \frac{\mu_0 I}{2\pi r_2^2} (-\hat{j} \times \vec{r}_2) = \frac{\mu_0 I}{2\pi (d^2 + h^2)} \left[ -\hat{j} \times (-d\hat{i} + h\hat{k}) \right] = \frac{\mu_0 I}{2\pi (d^2 + h^2)} (-h\hat{i} - d\hat{k})$$

The resultant magnetic field of the two wires is:

$$\vec{B}_{\text{wires}} = \vec{B}_1 + \vec{B}_2 = -\frac{\mu_0 I d}{\pi (d^2 + h^2)} \hat{k}$$

Thus, the currents in Wire 1 and Wire 2 must be directed along $\text{PQ}$ and $\text{SR}$, respectively.

3. Calculating the Height $h$

Equating the magnitude of the magnetic field from the loop to the magnitude of the field from the wires:

$$|\vec{B}_{\text{loop}}| = |\vec{B}_{\text{wires}}|$$ $$\frac{\mu_0 I a^2}{2(a^2 + h^2)^{3/2}} = \frac{\mu_0 I d}{\pi (d^2 + h^2)}$$

Given $d \approx a$:

$$\frac{\mu_0 I a^2}{2(a^2 + h^2)^{3/2}} = \frac{\mu_0 I a}{\pi (a^2 + h^2)}$$

Simplifying:

$$\frac{a}{2\sqrt{a^2 + h^2}} = \frac{1}{\pi}$$ $$\sqrt{a^2 + h^2} = \frac{\pi}{2} a \approx \frac{3.1416}{2} a \approx 1.57 a$$

Squaring both sides:

$$a^2 + h^2 \approx (1.57)^2 a^2 \approx 2.467 a^2$$ $$h^2 \approx 1.467 a^2 \implies h \approx \sqrt{1.467} a \approx 1.21 a \approx 1.2a$$

Therefore, the currents in wire 1 and wire 2 are directed along $\text{PQ}$ and $\text{SR}$, respectively, and $h \approx 1.2a$.

Correct Option: C