Question

The magnitude of magnetic field at \(O\) due to a current carrying loop as shown in the figure is, where \(O\) is the centre of two circular portions with radii \(1\,\text{cm}\) and \(2\,\text{cm}\) respectively. Take \(I=\dfrac{1.2}{\pi}\,\text{A}\).

• A. \(10\,\text{nT}\)
• B. \(0.1\,\text{nT}\)
• C. \(100\,\mu\text{T}\)
• D. \(1\,\text{nT}\)

Hint:

For a circular arc, \[ B=\frac{\mu_0 I\theta}{4\pi R}. \] Straight radial wire segments passing through the centre do not produce magnetic field at the centre because \(d\vec{l}\) and \(\vec{r}\) are parallel.

Answer 1

The Correct Option is A

Step 1: Use Biot-Savart law for a circular arc.
B = (μ₀Iθ)/(4πR). Radial segments contribute zero field at center.

Step 2: Net field from opposing arcs.
B = (μ₀Iθ/4π)(1/R₁ – 1/R₂). θ = 30° = π/6.

Step 3: Substitute values.
B = 10⁻⁷ × (1.2/π) × (π/6) × (1/10⁻² – 1/2×10⁻²) = 10⁻⁷ × 0.2 × 50 = 10⁻⁶ T = 1 μT.

Step 4: Final Answer:
1 μT (not among listed options).