A closely wounded circular coil of radius 5 cm produces a magnetic field of 37.68 × 10⁻⁴ T at its center. The current through the coil is _________A.
[Given, number of turns in the coil is 100 and π=3.14]

Solution

Physics sir · 3 steps

At the centre of a coil every turn adds the same field, so multiply the single-loop formula by N.

1. Field at centre. Biot-Savart gives (μ₀ I)/2R at the centre of one circular loop. A closely wound coil has all N turns at the same radius, so their fields add up.

   B = (μ₀ N I)/2R ⇒ I = 2RB/(μ₀ N)

2. Put in numbers. With R = 0.05 m, the top is 2 × 0.05 × 37.68 × 10⁻⁴ = 3.768 × 10⁻⁴. The bottom is 4 × 3.14 × 10⁻⁷ × 100 = 1.256 × 10⁻⁴.
3. So. I = (3.768 × 10⁻⁴)/(1.256 × 10⁻⁴) = 3 A, so the answer is 3.

The common slip: 300. Putting R = 5 in centimetres instead of 0.05 m makes the current 100 times too large and gives 300, not 3.

The idea to keep. For a tightly wound coil at its centre, use B = (μ₀ N I)/2R. Convert the radius to metres before you substitute.

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