Question

An infinitely long wire carrying 1 A current in the +Z direction is placed at (1 cm, 1 cm). Another wire carrying 1 A in +X direction is placed at y = 1 cm. If the magnetic field due to this configuration at the origin is $\vec{B$. Let $B_0$ be the magnitude of the field if only the wire at (1 cm, 1 cm) was present, then $\frac{\vec{B}}{B_0}$ is}

• A. $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, -\sqrt{2})$
• B. $(\frac{1}{2}, \frac{1}{2}, -1)$
• C. $(\sqrt{2}, \sqrt{2}, -\sqrt{2})$
• D. $(\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}}, -\frac{1}{2})$

Answer 1

The Correct Option is A

The magnetic field due to an infinitely long straight wire carrying current $I$ at a perpendicular distance $d$ is $B_{wire} = \frac{\mu_0 I}{2\pi d}$. The direction is given by the right-hand rule. Wire 1 (W1): Current $I_1 = 1$ A in +Z direction. Position $x_1=1$ cm, $y_1=1$ cm. (This means it's parallel to Z-axis passing through $(0.01 \text{m}, 0.01 \text{m}, z)$). We need the field at the origin $(0,0,0)$. The distance $d_1$ from origin to W1 (which is in the xy-plane) is $d_1 = \sqrt{(1\text{cm})^2 + (1\text{cm})^2} = \sqrt{1^2+1^2} \text{ cm} = \sqrt{2} \text{ cm} = \sqrt{2} \times 10^{-2} \text{ m}$. Magnitude of field due to W1 at origin: $B_{W1} = \frac{\mu_0 I_1}{2\pi d_1} = \frac{\mu_0 (1)}{2\pi (\sqrt{2} \times 10^{-2})}$. This is $B_0$. So, $B_0 = \frac{\mu_0}{2\sqrt{2}\pi \times 10^{-2}}$. Direction of $\vec{B}_{W1}$: Wire is at $(1,1)$ in xy-plane, current is $+\hat{k}$. Vector from wire to origin is $(0-1)\hat{i} + (0-1)\hat{j} = -\hat{i}-\hat{j}$. $\vec{dl} \times \vec{r}$ direction: Current is $\hat{k}$. Position vector $\vec{r}_{wire} = \hat{i}+\hat{j}$ (in cm). Point is origin. Vector from point on wire (0,0,z_p) to origin is $-(x_1 \hat{i} + y_1 \hat{j})$. More simply, the vector $\vec{d_1}$ from wire to origin is $(-\hat{i}-\hat{j})/\sqrt{2}$. The field lines are circles in xy-plane. Current in $+\hat{k}$. At origin relative to wire at $(1,1,z)$, the direction of field is perpendicular to vector from $(1,1)$ to $(0,0)$. Vector from $(1,1)$ to $(0,0)$ is $(-\hat{i}-\hat{j})$. Field is tangential, so perpendicular to this, which would be proportional to $(-\hat{i}+\hat{j})$ or $(\hat{i}-\hat{j})$. Using right-hand rule: Thumb along $+\hat{k}$. Origin is to the "south-west" of $(1,1)$. Field lines are counter-clockwise. The tangent at origin will be in direction $(\hat{i}-\hat{j})/\sqrt{2}$. $\vec{B}_{W1} = B_0 \frac{(\hat{i}-\hat{j})}{\sqrt{2}}$. Wire 2 (W2): Current $I_2 = 1$ A in +X direction. Position: $y_2=1$ cm. (This means it's parallel to X-axis, passing through $(x, 0.01 \text{m}, 0)$). We need field at origin $(0,0,0)$. The wire is on the line $(x, 1\text{cm}, 0)$. The perpendicular distance from origin to this line is $d_2 = 1 \text{ cm} = 10^{-2} \text{ m}$ (along y-axis). Magnitude of field due to W2 at origin: $B_{W2} = \frac{\mu_0 I_2}{2\pi d_2} = \frac{\mu_0 (1)}{2\pi (1 \times 10^{-2})} = \frac{\mu_0}{2\pi \times 10^{-2}}$. Compare with $B_0$: $B_{W2} = \sqrt{2} B_0$. Direction of $\vec{B}_{W2}$: Current is $\hat{i}$. Wire is above origin along y-axis. Point $(0,0,0)$ is below the wire $(x, 1\text{cm}, 0)$. Vector from wire to origin is $-\hat{j}$. Field lines circle the wire. Thumb along $+\hat{i}$. Below the wire (negative y relative to wire), field is in $-\hat{k}$ direction. So, $\vec{B}_{W2} = B_{W2} (-\hat{k}) = \sqrt{2} B_0 (-\hat{k})$. Total magnetic field $\vec{B} = \vec{B}_{W1} + \vec{B}_{W2}$. $\vec{B} = B_0 \frac{(\hat{i}-\hat{j})}{\sqrt{2}} - \sqrt{2} B_0 \hat{k}$. $\vec{B} = B_0 \left( \frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} - \sqrt{2}\hat{k} \right)$. The question asks for $\frac{\vec{B}}{B_0}$. $\frac{\vec{B}}{B_0} = \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, -\sqrt{2} \right)$. This matches option (a). \[ \boxed{\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, -\sqrt{2}\right)} \]