Question

A 1 cm straight segment of a conductor carrying 1 A current in \( x \)-direction lies symmetrically at the origin of Cartesian coordinate system. The magnetic field due to this segment at point (1m, 1m, 0) is:

• A. \( 1.0 \times 10^{-9} \, \text{T} \)
• B. \( -1.0 \times 10^{-9} \, \text{T} \)
• C. \( \frac{5.0}{\sqrt{2}} \times 10^{-10} \, \text{T} \)
• D. \( -\frac{5.0}{\sqrt{2}} \times 10^{-10} \, \text{T} \)

Hint:

The Biot-Savart law provides a way to calculate the magnetic field produced by a current-carrying element.

Answer 1

The Correct Option is C

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.