Question

An infinitely long wire carrying \(1\ \text{A}\) current in the \(+Z\) direction is placed at \((1\ \text{cm},1\ \text{cm})\). Another wire carrying \(1\ \text{A}\) in \(+X\) direction is placed at \(y=1\ \text{cm}\). 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 \((1\ \text{cm},1\ \text{cm})\) was present, then \(\frac{\vec{B}}{B_0}\) is

• A. \(\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},-\sqrt2\right)\)
• B. \(\left(\frac{1}{2},\frac{1}{2},-1\right)\)
• C. \(\left(\sqrt2,\sqrt2,-\sqrt2\right)\)
• D. \(\left(\frac{1}{2\sqrt2},\frac{1}{2\sqrt2},-\frac{1}{2}\right)\)

Hint:

For an infinitely long straight current-carrying wire, \[ B=\frac{\mu_0 I}{2\pi r}. \] The direction of magnetic field is found using the right-hand thumb rule.

Answer 1

The Correct Option is A

Step 1: Field from wire at (1,1) cm carrying current in +Z at origin.
The wire is at perpendicular distance \( r = \sqrt{2}\text{ cm} \) from origin. Direction: \( \hat{l}\times\hat{r} \) where \( \hat{r} \) points from wire to origin \( = (-\hat{x}-\hat{y})/\sqrt{2} \). \( \hat{z}\times(-\hat{x}-\hat{y})/\sqrt{2} = (-\hat{y}+\hat{x})/\sqrt{2} \). Field magnitude \( B_0 = \frac{\mu_0 I}{2\pi r} \). Component: \( B_0(\hat{x}-\hat{y})/\sqrt{2} \).

Step 2: Field from wire at y=1 cm carrying current in +X at origin.
Wire along x at y=1 cm: distance = 1 cm from origin. \( \hat{x}\times(-\hat{y}) = -\hat{z} \). Field \( = B_1(-\hat{z}) \), where \( B_1 = \frac{\mu_0}{2\pi(0.01)} \). With \( r = \sqrt{2}\) cm for first wire, \( B_0 = \frac{B_1}{\sqrt{2}} \). Net \( \vec{B}/B_1 = \frac{1}{\sqrt{2}}\hat{x} - \frac{1}{\sqrt{2}}\hat{y} - \hat{z} \propto (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, -\sqrt{2}) \) after normalizing by \( 1/\sqrt{2} \).

\[ \boxed{\vec{B}/B_0 = \left(\frac{1}{\sqrt{2}},\,-\frac{1}{\sqrt{2}},\,-\sqrt{2}\right)} \]