A magnetic field of `2xx10^(-2)T` acts at right angles to a coil of area `100cm^(2)` with `50` turns. The average emf induced in the coil is `0.1V`, when it is removed from the field in time `t`. The value of `t` is

Options

A. 10 s
B. 0.1 s
C. 0.01 s
D. 1 s

Explanation

To solve the problem, we need to find the time \( t \) it takes for the average emf induced in the coil to be \( 0.1 \, V \) when it is removed from a magnetic field. We will use the formula for induced emf and magnetic flux.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Magnetic field strength, \( B = 2 \times 10^{-2} \, T \)
- Area of the coil, \( A = 100 \, cm^2 = 100 \times 10^{-4} \, m^2 = 10^{-2} \, m^2 \)
- Number of turns, \( N = 50 \)
- Induced emf, \( E = 0.1 \, V \)

2. **Calculate the Initial Magnetic Flux**:
The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Since the magnetic field acts at right angles to the coil, \( \theta = 0 \) and \( \cos(0) = 1 \):
\[ \Phi = N \cdot B \cdot A = 50 \cdot (2 \times 10^{-2}) \cdot (10^{-2}) = 50 \cdot 2 \cdot 10^{-4} = 10^{-2} \, Wb \]

3. **Use the Formula for Induced EMF**:
The average induced emf \( E \) is related to the change in magnetic flux \( \Delta \Phi \) over time \( t \):
\[ E = -\frac{\Delta \Phi}{\Delta t} \]
When the coil is completely removed from the magnetic field, the final flux \( \Phi_f = 0 \). Therefore, the change in flux \( \Delta \Phi \) is:
\[ \Delta \Phi = \Phi - \Phi_f = 10^{-2} - 0 = 10^{-2} \, Wb \]
Substituting this into the emf equation gives:
\[ E = \frac{10^{-2}}{t} \]

4. **Rearranging to Find Time \( t \)**:
Rearranging the equation to solve for \( t \):
\[ t = \frac{10^{-2}}{E} \]
Substituting the value of \( E = 0.1 \, V \):
\[ t = \frac{10^{-2}}{0.1} = 10^{-1} \, s = 0.1 \, s \]

5. **Final Answer**:
The value of \( t \) is \( 0.1 \, s \).