Two concentric circular coils with radii 1 cm and 1000 cm, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ × 10⁻⁸ H. (Take, π²=10 )

Solution

Physics sir · 4 steps

Put the current in the big coil: its centre field is uniform over the tiny coil, so flux is B times area.

1. Pick the big coil. Let current I flow in the large coil (N₂ = 200, R = 10 m). Its field at the centre is B = (μ₀ N₂ I)/2R. The small coil has radius r = 0.01 m, far smaller than R, so this field is practically uniform over it.
2. Flux through small coil. The small coil has N₁ = 10 turns, each of area π r², so the total flux linked is N₁ B π r². Dividing this by I gives the mutual inductance.

   M = (μ₀ N₁ N₂ π r²)/2R

3. Put in numbers. M = (4π × 10⁻⁷ × 10 × 200 × π × 10⁻⁴)/(2 × 10). With π² = 10, the numerator is 40 × 10⁻⁷ × 2000 × 10⁻⁴ = 8 × 10⁻⁷, and dividing by 20 gives 4 × 10⁻⁸ H.
4. So. The mutual inductance is 4 × 10⁻⁸ H, so the answer is 4.

The common slip: 8. Writing the centre field as (μ₀ N I)/R and dropping the 2 gives 8, not 4.

The idea to keep. For mutual inductance, let the current flow in the coil whose field is easy to find over the other coil. Here that is the big coil, whose centre field is uniform over the small one.

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