A 100 turns closed-packed coil of diameter 0.20m is placed with its plane perpendicular to a uniform magnetic field. The field value then varies at a uniform rate `0.10 Wb//m^2` to `0.30 Wb//m^2` in `5xx10^(-2)s`. Find the emf induced in the coil.

Explanation

To find the induced electromotive force (emf) in the coil, we will follow these steps:

### Step 1: Identify the given values
- Number of turns (N) = 100
- Diameter of the coil (D) = 0.20 m
- Radius of the coil (r) = D/2 = 0.20 m / 2 = 0.10 m
- Initial magnetic field (B1) = 0.10 Wb/m²
- Final magnetic field (B2) = 0.30 Wb/m²
- Time interval (Δt) = 5 × 10^(-2) s

### Step 2: Calculate the change in magnetic field (ΔB)
\[ \Delta B = B2 - B1 = 0.30 \, \text{Wb/m}^2 - 0.10 \, \text{Wb/m}^2 = 0.20 \, \text{Wb/m}^2 \]

### Step 3: Calculate the area of the coil (A)
The area (A) of the coil can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.10 \, \text{m})^2 = \pi (0.01 \, \text{m}^2) = 0.0314 \, \text{m}^2 \]

### Step 4: Use the formula for induced emf (ε)
The induced emf (ε) can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -N \frac{d\Phi}{dt} \]
Where \(\Phi\) (magnetic flux) is given by:
\[ \Phi = B \cdot A \]
Thus,
\[ \epsilon = -N A \frac{dB}{dt} \]

### Step 5: Calculate \(\frac{dB}{dt}\)
\[ \frac{dB}{dt} = \frac{\Delta B}{\Delta t} = \frac{0.20 \, \text{Wb/m}^2}{5 \times 10^{-2} \, \text{s}} = 4 \, \text{Wb/m}^2/\text{s} \]

### Step 6: Substitute values into the emf formula
Now substituting the values into the emf formula:
\[ \epsilon = -100 \times 0.0314 \, \text{m}^2 \times 4 \, \text{Wb/m}^2/\text{s} \]
\[ \epsilon = -100 \times 0.0314 \times 4 = -12.56 \, \text{V} \]

### Step 7: Final result
The magnitude of the induced emf is:
\[ \epsilon = 12.56 \, \text{V} \]