Two concentric circular coils, C₁ and C₂ are
placed in the XY plane. C₁ has 500 turns, and
a radius of 1 cm. C₂ has 200 turns and radius
of 20 cm. C₂ carries a time dependent current
I(t) = (5t² – 2t + 3) A where t is in s. The emf
induced in C₁ (in mV), at the instant t = 1 s is
4/x. The value of x is ___ .

Solution

Physics sir · 5 steps

The small coil sits at the big coil's centre, so its flux is just the centre field times its own area.

1. Field from C2. C₁ (radius 1 cm) is tiny compared with C₂ (radius 20 cm). So the field over all of C₁ is the field at the centre of C₂: B = (μ₀ N₂ I)/2R₂.
2. Mutual inductance. The total flux through C₁ is N₁ B π r₁². Write this as M I to read off M.

   M = (μ₀ N₁ N₂ π r₁²)/2R₂ = (4π × 10⁻⁷ × 500 × 200 × π × 10⁻⁴)/(2 × 0.2) = π² × 10⁻⁵ H

3. Rate of current. dI/dt = 10t - 2. At t = 1 s this is 8 A s⁻¹.
4. Induced emf. ε = MdI/dt = 8π² × 10⁻⁵ V. Taking π² ≈ 10 gives 8 × 10⁻⁴ V = 0.8 mV.
5. So. 0.8 = 4/x, so x = 5.

The common slip: 2.5. Using B = μ₀ N I/R without the 2 in the denominator doubles the emf to 1.6 mV and gives 2.5.

The idea to keep. For a tiny coil at the centre of a big coil, treat the big coil's centre field as uniform over the small one. Then M = μ₀ N₁ N₂ π r²/2R, and the emf comes from dI/dt, not from I.

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