A coil of area `100 cm^(2)` has `500` turns. Magnetic field of `0.1 "weber"//"meter"^(2)` is perpendicular to the coil. The field is reduced to zero in `0.1` second. The induced `e.m.f.` in the coil is

Options

A. `1 V`
B. `5 V`
C. `50 V`
D. zero

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Area of the coil, \( A = 100 \, \text{cm}^2 \)
- Number of turns, \( N = 500 \)
- Initial magnetic field, \( B_i = 0.1 \, \text{weber/m}^2 \)
- Final magnetic field, \( B_f = 0 \, \text{weber/m}^2 \)
- Time taken for the change, \( \Delta t = 0.1 \, \text{s} \)

### Step 2: Convert the area from cm² to m²
To use SI units, we need to convert the area from cm² to m²:
\[ A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the change in magnetic flux
The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = B \cdot A \]
Initially, the magnetic flux \( \Phi_i \) is:
\[ \Phi_i = B_i \cdot A = 0.1 \, \text{weber/m}^2 \cdot 0.01 \, \text{m}^2 = 0.001 \, \text{weber} \]
Final magnetic flux \( \Phi_f \) when the magnetic field is reduced to zero:
\[ \Phi_f = B_f \cdot A = 0 \cdot 0.01 = 0 \, \text{weber} \]
Thus, the change in magnetic flux \( \Delta \Phi \) is:
\[ \Delta \Phi = \Phi_f - \Phi_i = 0 - 0.001 = -0.001 \, \text{weber} \]

### Step 4: Calculate the induced e.m.f. using Faraday's law
According to Faraday's law of electromagnetic induction, the induced e.m.f. \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \mathcal{E} = -500 \cdot \frac{-0.001}{0.1} \]
Calculating this gives:
\[ \mathcal{E} = 500 \cdot \frac{0.001}{0.1} = 500 \cdot 0.01 = 5 \, \text{volts} \]

### Step 5: Consider Lenz's Law
The negative sign in Faraday's law indicates the direction of the induced e.m.f. according to Lenz's law, which states that the induced current will oppose the change in magnetic flux. Thus, we have:
\[ \mathcal{E} = 5 \, \text{volts} \quad (\text{indicating the magnitude}) \]

### Final Answer
The induced e.m.f. in the coil is \( 5 \, \text{volts} \).

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