The current of 5 A flows in a square loop of sides 1 m is placed in air. The magnetic field at the centre of the loop is X √2 × 10⁻⁷ T. The value of X is _________.

Solution

Physics sir · 4 steps

Treat one side as a finite straight wire, then multiply by four, since all four sides add up.

1. One side. Each side is a finite straight wire. The centre is at perpendicular distance d = a/2 = 0.5 m from it, and each end of the side is seen at 45° from that perpendicular.

   B_(side) = (μ₀ I)/(4π d)(sin 45° + sin 45°) = 10⁻⁷ × 5/0.5 × √2

2. Put in numbers. μ₀/4π = 10⁻⁷ T m A⁻¹ and 5/0.5 = 10, so B_(side) = 10√2 × 10⁻⁷ T.
3. Add four sides. The current goes the same way round the loop, so all four sides give a field at the centre in the same direction. They add directly: B = 4 × 10√2 × 10⁻⁷ = 40√2 × 10⁻⁷ T.
4. So. Comparing with X√2 × 10⁻⁷ T gives X = 40.

The common slip: 20. Taking the perpendicular distance as the full side, 1 m, instead of half the side, 0.5 m, halves every term and gives 20.

The idea to keep. For a polygon loop, treat each side as a finite wire, (μ₀ I)/(4π d)(sin θ₁ + sin θ₂), with d from the centre to the side, then multiply by the number of sides.

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