A heart pacing device consists of a coil of `50` turns `&` radius `1 mm` just inside the body with a coil of `1000` turns `&` radius `2 cm` placed concentrically and co axially just outside the body.Calculate the average induced `EMF` in the internal coil, if a current of `1A` in the external coil collapses in `10` milliseconds.

Answer

To solve the problem of calculating the average induced EMF in the internal coil of a heart pacing device, we will follow these steps:

### Step 1: Identify the parameters given in the problem
- Number of turns in the internal coil (N₁) = 50 turns
- Radius of the internal coil (r₁) = 1 mm = 1 × 10⁻³ m
- Number of turns in the external coil (N₂) = 1000 turns
- Radius of the external coil (r₂) = 2 cm = 2 × 10⁻² m
- Current in the external coil (I) = 1 A
- Time for the current to collapse (Δt) = 10 ms = 10 × 10⁻³ s

### Step 2: Calculate the magnetic field (B) produced by the external coil
The magnetic field (B) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 N I}{2r} \]
where:
- \(\mu_0\) = permeability of free space = \(4\pi \times 10^{-7} \, \text{T m/A}\)
- N = number of turns in the external coil = 1000
- I = current = 1 A
- r = radius of the external coil = 2 × 10⁻² m

Substituting the values:
\[ B = \frac{4\pi \times 10^{-7} \times 1000 \times 1}{2 \times 2 \times 10^{-2}} = \frac{4\pi \times 10^{-4}}{4 \times 10^{-2}} = \pi \times 10^{-2} \, \text{T} \]

### Step 3: Calculate the change in magnetic field (ΔB)
Since the current collapses from 1 A to 0 A, the change in magnetic field (ΔB) is equal to the initial magnetic field (B):
\[ \Delta B = B - 0 = \pi \times 10^{-2} \, \text{T} \]

### Step 4: Calculate the area (A) of the internal coil
The area (A) of the internal coil is given by:
\[ A = \pi r_1^2 \]
Substituting the radius of the internal coil:
\[ A = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Calculate the average induced EMF (E)
The average induced EMF (E) in the internal coil can be calculated using Faraday's law of electromagnetic induction:
\[ E = -N_1 \frac{d\Phi}{dt} \]
where \(\Phi = B \cdot A\) is the magnetic flux.

The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = A \Delta B = \pi \times 10^{-6} \times \pi \times 10^{-2} = \pi^2 \times 10^{-8} \, \text{Wb} \]

Now, the average induced EMF is:
\[ E = -N_1 \frac{\Delta \Phi}{\Delta t} = -50 \frac{\pi^2 \times 10^{-8}}{10 \times 10^{-3}} = -50 \frac{\pi^2 \times 10^{-8}}{10^{-2}} = -5 \pi^2 \times 10^{-6} \, \text{V} \]

### Final Result
The average induced EMF in the internal coil is:
\[ E = 5 \pi^2 \times 10^{-6} \, \text{V} \quad (\text{approximately } 4.93 \, \mu V) \]

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