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. `0.1` sec
B. `0.01` sec
C. `1` sec
D. `20` sec

Explanation

To find the time \( t \) during which the coil is removed from the magnetic field, we can use Faraday's law of electromagnetic induction, which states that the induced emf (\( \varepsilon \)) in a coil is given by:

\[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \]

Where:
- \( \varepsilon \) is the induced emf,
- \( N \) is the number of turns in the coil,
- \( \Delta \Phi \) is the change in magnetic flux,
- \( \Delta t \) is the time interval over which the change occurs.

### Step 1: Calculate the initial magnetic flux (\( \Phi_i \))

The magnetic flux (\( \Phi \)) through the coil is given by:

\[ \Phi = B \cdot A \]

Where:
- \( B \) is the magnetic field strength,
- \( A \) is the area of the coil.

Given:
- \( B = 2 \times 10^{-2} \, T \)
- Area \( A = 100 \, cm^2 = 100 \times 10^{-4} \, m^2 = 0.01 \, m^2 \)

Now, calculate \( \Phi_i \):

\[ \Phi_i = B \cdot A = (2 \times 10^{-2}) \cdot (0.01) = 2 \times 10^{-4} \, Wb \]

### Step 2: Calculate the final magnetic flux (\( \Phi_f \))

When the coil is removed from the magnetic field, the final magnetic flux (\( \Phi_f \)) becomes zero:

\[ \Phi_f = 0 \, Wb \]

### Step 3: Calculate the change in magnetic flux (\( \Delta \Phi \))

The change in magnetic flux (\( \Delta \Phi \)) is given by:

\[ \Delta \Phi = \Phi_f - \Phi_i = 0 - (2 \times 10^{-4}) = -2 \times 10^{-4} \, Wb \]

### Step 4: Substitute values into Faraday's law

Using the average induced emf (\( \varepsilon = 0.1 \, V \)) and the number of turns (\( N = 50 \)), we can substitute into the formula:

\[ 0.1 = -50 \frac{-2 \times 10^{-4}}{\Delta t} \]

### Step 5: Solve for \( \Delta t \)

Rearranging the equation gives:

\[ \Delta t = 50 \cdot \frac{-2 \times 10^{-4}}{0.1} \]

Calculating the right side:

\[ \Delta t = 50 \cdot \frac{2 \times 10^{-4}}{0.1} = 50 \cdot 2 \times 10^{-3} = 0.1 \, s \]

### Final Answer

The value of \( t \) is \( 0.1 \, s \).

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