Question: A 1 cm straight segment of a conductor carrying 1 A current in the \( x \)-direction lies symmetrically at the origin of the Cartesian coordinate system. The magnetic field due to this segment at point \( (1\,\text{m}, 1\,\text{m}, 0) \) is:
1. Understanding the Setup:
- The conductor is 1 cm long (i.e., 0.01 m), symmetrically placed along the x-axis centered at the origin. That means it extends from \( -0.005\,\text{m} \) to \( +0.005\,\text{m} \).
- Current \( I = 1\,\text{A} \)
- Point of observation is at \( (1, 1, 0) \)
2. Magnetic Field Due to a Finite Current Segment:
We use the Biot–Savart law for a finite straight conductor:
\[
\vec{B} = \frac{\mu_0 I}{4\pi r} (\sin\theta_1 + \sin\theta_2)\hat{n}
\]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \)
- \( r \) = perpendicular distance from wire to point = \( \sqrt{(1)^2 + (1)^2} = \sqrt{2} \, \text{m} \)
- Since the wire is symmetric about the origin, \( \theta_1 = \theta_2 \), and they are measured from the center of the wire to the ends.
- \( \tan\theta = \frac{L/2}{r} = \frac{0.005}{\sqrt{2}} \Rightarrow \theta \approx \tan^{-1}(0.0035) \approx 0.2^\circ \) (very small)
- \( \sin\theta \approx \theta \text{ (in radians)} \), so \( \sin\theta_1 + \sin\theta_2 \approx 2\theta \)
So the approximate magnetic field is:
\[
B = \frac{4\pi \times 10^{-7} \times 1}{4\pi \times \sqrt{2}} \times 2 \times \frac{0.005}{\sqrt{2}} = \frac{10^{-7} \cdot 2 \cdot 0.005}{2} = \frac{10^{-7} \cdot 0.005}{\sqrt{2}} = \frac{5.0 \times 10^{-10}}{\sqrt{2}} \, \text{T}
\]
3. Final Answer:
Option (C) \( \frac{5.0 \times 10^{-10}}{\sqrt{2}} \, \text{T} \) is correct.