A straight wire carrying a current of 14 A is bent into a semi-circular arc of radius 2.2 cm as shown in the figure. The magnetic field produced by the current at the centre (O) of the arc. is ____________ × 10⁻⁴ T

Solution

Physics sir · 4 steps

The straight parts point straight at O, so they give nothing. Only the half-circle counts.

1. Straight parts. Both straight wires lie on the line through O. For every element, dl⃗ is parallel to r⃗, so dl⃗ × r⃗ = 0 and their field at O is zero.
2. Semicircular arc. A full circular loop gives (μ₀ I)/2R at its centre. Every element of the arc adds its field in the same direction (into the page), so half a circle gives half of that.

   B = (μ₀ I)/4R

3. Put in numbers. B = (4π × 10⁻⁷ × 14)/(4 × 0.022) = (14π × 10⁻⁷)/0.022. With π = 22/7, this is (44 × 10⁻⁷)/0.022 = 2 × 10⁻⁴ T.
4. So. The field at O is 2 × 10⁻⁴ T, so the answer is 2.

The common slip: 4. Using the full-loop formula (μ₀ I)/2R instead of (μ₀ I)/4R for the semicircle gives 4, not 2.

The idea to keep. Any straight wire whose line passes through the point gives zero field there. A circular arc of angle θ gives (μ₀ I θ)/(4π R) at its centre, so a semicircle gives (μ₀ I)/4R.

Answer from the NTA final answer key. Solution by PrepWiser.