A coil of circular cross - section having 1000 turns and `4cm^(2)` face area is placed with its axis parallel to a magnetic field which decreases by `10^(-2)"Wb m"^(-2)` in 0.01 s. the e.m.f induced in the coil is :

Options

A. 0.4 mV
B. 4 mV
C. 200 mV
D. 400 mV

Answer

To solve the problem, we will use Faraday's law of electromagnetic induction, which states that the induced electromotive force (e.m.f) in a coil is equal to the rate of change of magnetic flux through the coil.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns in the coil, \( n = 1000 \)
- Area of the coil, \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- Change in magnetic field, \( \Delta B = 10^{-2} \, \text{Wb/m}^2 \)
- Time interval, \( \Delta t = 0.01 \, \text{s} \)

2. **Calculate the change in magnetic flux (\( \Delta \Phi \)):**
The magnetic flux (\( \Phi \)) through the coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Since the axis of the coil is parallel to the magnetic field, \( \theta = 0^\circ \) and \( \cos(0) = 1 \). Thus, the change in magnetic flux (\( \Delta \Phi \)) when the magnetic field changes is:
\[ \Delta \Phi = \Delta B \cdot A \]

3. **Substitute the values to find \( \Delta \Phi \):**
\[ \Delta \Phi = (10^{-2} \, \text{Wb/m}^2) \cdot (4 \times 10^{-4} \, \text{m}^2) = 4 \times 10^{-6} \, \text{Wb} \]

4. **Calculate the induced e.m.f (\( E \)):**
According to Faraday's law, the induced e.m.f is given by:
\[ E = -\frac{n \Delta \Phi}{\Delta t} \]
Substitute the values:
\[ E = -\frac{1000 \cdot (4 \times 10^{-6} \, \text{Wb})}{0.01 \, \text{s}} = -\frac{4 \times 10^{-3} \, \text{Wb}}{0.01 \, \text{s}} = -0.4 \, \text{V} \]
The negative sign indicates the direction of the induced e.m.f, but we are interested in the magnitude, which is:
\[ E = 0.4 \, \text{V} \]

5. **Convert the result to millivolts (if needed):**
\[ E = 0.4 \, \text{V} = 400 \, \text{mV} \]

### Final Answer:
The induced e.m.f in the coil is **0.4 V** or **400 mV**.