A field of `2xx10^(-2)`T acts at right angles to a coil of 50 turns of area `1000 cm hat 2`. The coil is removed from the field in 0.1 second . Then the induced emf in the coil is __

Options

A. `1 V`
B. 80 KV
C. 7.96 V
D. none of the above

Answer

To solve the problem, we need to calculate the induced electromotive force (emf) in the coil when it is removed from the magnetic field. Here are the steps to arrive at the solution:

### Step 1: Identify the Given Values
- Magnetic field strength (B) = \(2 \times 10^{-2}\) T
- Number of turns in the coil (N) = 50 turns
- Area of the coil (A) = 1000 cm² = \(1000 \times 10^{-4}\) m² = \(10^{-2}\) m²
- Time taken to remove the coil from the field (Δt) = 0.1 s

### Step 2: Calculate the Magnetic Flux (Φ)
The magnetic flux (Φ) through the coil when it is in the magnetic field is given by the formula:
\[ \Phi = B \cdot A \]
Substituting the values:
\[ \Phi = (2 \times 10^{-2} \, \text{T}) \cdot (10^{-2} \, \text{m}^2) = 2 \times 10^{-4} \, \text{Wb} \]

### Step 3: Determine the Change in Magnetic Flux (ΔΦ)
When the coil is removed from the magnetic field, the final magnetic flux becomes zero (Φ = 0). Therefore, the change in magnetic flux (ΔΦ) is:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - (2 \times 10^{-4} \, \text{Wb}) = -2 \times 10^{-4} \, \text{Wb} \]

### Step 4: Calculate the Induced EMF (ε)
The induced emf (ε) in the coil can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Substituting the values:
\[ \epsilon = -\frac{\Delta \Phi}{\Delta t} = -\frac{-2 \times 10^{-4} \, \text{Wb}}{0.1 \, \text{s}} = \frac{2 \times 10^{-4}}{0.1} = 2 \times 10^{-3} \, \text{V} = 0.002 \, \text{V} \]

### Step 5: Consider the Total Induced EMF for the Coil
Since the coil has 50 turns, the total induced emf (ε_total) is:
\[ \epsilon_{\text{total}} = N \cdot \epsilon = 50 \cdot (2 \times 10^{-3} \, \text{V}) = 0.1 \, \text{V} \]

### Final Answer
The induced emf in the coil is **0.1 V**.

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