A current of 5 A exists in a square loop of side 1/√2 m. Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10⁻⁶ T. where, value of p is ________ [ Take μ₀=4 π × 10⁻⁷ T mA⁻¹ ].

Solution

Physics sir · 4 steps

Treat the square as four finite wires. Find one side's field at the centre, then multiply by four.

1. Field of one side. Each side is a finite straight wire. The centre is a perpendicular distance d = a/2 from it, and the two ends make angles of 45° on either side. All four sides give fields in the same direction, so they add.

   B_(side) = (μ₀ I)/(4π d)(sin 45° + sin 45°) = (μ₀ I)/(4π d) √2

2. Put in numbers. Here d = 1/2√2 m and (μ₀ I)/4π = 10⁻⁷ × 5 = 5 × 10⁻⁷. So B_(side) = 5 × 10⁻⁷ × 2√2 × √2 = 2 × 10⁻⁶ T.
3. Add four sides. B = 4 × 2 × 10⁻⁶ = 8 × 10⁻⁶ T. This matches the standard result B = (2√2 μ₀ I)/(π a).
4. So. p = 8.

The common slip: 2. Stopping at one side gives 2, not 8. All four sides add their fields in the same direction at the centre.

The idea to keep. For a polygon loop, find the field of one side using the finite-wire formula with the correct perpendicular distance and angles. Then multiply by the number of sides.

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