Question

Two infinitely long wires are placed at \((1\text{ cm},1\text{ cm})\) and \((+1\text{ cm},-1\text{ cm})\) with \(1\text{ A}\) current in each and in the same directions perpendicular to the \(x-y\) plane. Let the magnetic field due to these current carrying wires at the origin be \(\vec{B}\). If \(B_0\) is the magnitude of the field if only one of them was present, then \(\dfrac{|\vec{B}|}{B_0}\) is:

• A. \(\sqrt{2}\)
• B. \(1\)
• C. \(\dfrac{1}{\sqrt{2}}\)
• D. \(\dfrac{1}{2\sqrt{2}}\)

Hint:

For magnetic fields due to multiple long straight current carrying wires, first find the direction of each field using the right-hand thumb rule, then add the fields vectorially.

Answer 1

The Correct Option is A

Step 1: Find the distance of each wire from the origin.
The two wires are at $(1,1)\,\text{cm}$ and $(1,-1)\,\text{cm}$. The distance of each from the origin $(0,0)$ is: \[ r = \sqrt{1^2 + 1^2} = \sqrt{2}\,\text{cm} \] Since both wires are at equal distance from the origin and carry equal current ($1\,\text{A}$), the magnitude of the magnetic field at the origin due to each wire is the same, and we call this $B_0$.
Step 2: Recall the magnetic field due to an infinite straight wire.
For a long straight wire carrying current $I$, the magnetic field at distance $r$ is: \[ B_0 = \frac{\mu_0 I}{2\pi r} \] The field is tangential (perpendicular to the line joining the wire to the field point).
Step 3: Find the direction of $\vec{B}_1$ due to wire at $(1,1)\,\text{cm}$.
The current is in the $+z$ direction. Using the right-hand thumb rule, the magnetic field circles counterclockwise around the wire (when viewed from $+z$). At the origin, the position vector from the wire at $(1,1)$ to the origin is $(-1,-1)$. The field direction is perpendicular to this, pointing in the direction $(-1,+1)/\sqrt{2}$ (rotated $90^\circ$ counterclockwise from $(-1,-1)$). So $\vec{B}_1$ has components $(-1/\sqrt{2}, +1/\sqrt{2}) \times B_0$.
Step 4: Find the direction of $\vec{B}_2$ due to wire at $(1,-1)\,\text{cm}$.
Similarly, the position vector from wire at $(1,-1)$ to origin is $(-1,+1)$. The field direction is perpendicular: $(-1,-1)/\sqrt{2}$ (rotated $90^\circ$ counterclockwise). So $\vec{B}_2$ has components $(-1/\sqrt{2}, -1/\sqrt{2}) \times B_0$.
Step 5: Add the two field vectors.
\[ \vec{B} = \vec{B}_1 + \vec{B}_2 = B_0\left(\frac{-1}{\sqrt{2}} + \frac{-1}{\sqrt{2}},\; \frac{+1}{\sqrt{2}} + \frac{-1}{\sqrt{2}}\right) = B_0\left(\frac{-2}{\sqrt{2}},\; 0\right) \] The $y$-components cancel and the $x$-components add: \[ |\vec{B}| = B_0 \times \frac{2}{\sqrt{2}} = B_0\sqrt{2} \]
Step 6: State the final answer.
\[ \frac{|\vec{B}|}{B_0} = \boxed{\sqrt{2}} \]