A coil of area `0.4m^(2)` has 100 turns. A magnetic field of 0.04 Wb `m^(-2)` is acting normal to the coil surface. If this magnetic field is reduced to zero in 0.01 s, then the induced emf in the coil is

Options

A. 160V
B. 250 V
C. 270 V
D. 320 V

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Area of the coil, \( A = 0.4 \, m^2 \)
- Number of turns in the coil, \( N = 100 \)
- Initial magnetic field, \( B_i = 0.04 \, Wb/m^2 \)
- Final magnetic field, \( B_f = 0 \, Wb/m^2 \) (since it is reduced to zero)
- Time duration for the change, \( dt = 0.01 \, s \)

2. **Calculate the Change in Magnetic Field (\( \Delta B \)):**
\[ \Delta B = B_f - B_i = 0 - 0.04 = -0.04 \, Wb/m^2 \]

3. **Calculate the Change in Magnetic Flux (\( \Delta \Phi \)):**
The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = N \cdot A \cdot B \]
Therefore, the change in magnetic flux (\( \Delta \Phi \)) is:
\[ \Delta \Phi = N \cdot A \cdot \Delta B \]
Substituting the values:
\[ \Delta \Phi = 100 \cdot 0.4 \cdot (-0.04) \]
\[ \Delta \Phi = 100 \cdot 0.4 \cdot -0.04 = -1.6 \, Wb \]

4. **Calculate the Induced EMF (\( E \)):**
The induced emf is given by Faraday's law of electromagnetic induction:
\[ E = -\frac{\Delta \Phi}{dt} \]
Substituting the values:
\[ E = -\frac{-1.6}{0.01} = \frac{1.6}{0.01} = 160 \, V \]

5. **Final Result:**
The induced emf in the coil is \( 160 \, V \).