A coil having `500` square loops each of side `10 cm` is placed normal to a magnetic flux which increase at the rate of `1.0 (tesla)//(second)`. The induced r.m.f. in volts is

Options

A. `0.1`
B. `0.5`
C. `1`
D. `5`

Explanation

To solve the problem, we need to calculate the induced electromotive force (emf) in a coil with the given parameters. We will use Faraday's law of electromagnetic induction, which states that the induced emf (E) in a coil is equal to the negative rate of change of magnetic flux (Φ) through the coil.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Number of loops (N) = 500
- Side length of each loop (l) = 10 cm = 0.1 m
- Rate of change of magnetic field (dB/dt) = 1.0 T/s

2. **Calculate the area (A) of one loop:**
\[ A = l^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

3. **Calculate the total magnetic flux (Φ) through the coil:**
The magnetic flux through one loop is given by:
\[ \Phi = B \cdot A \]
Since the magnetic field is increasing, we can express the change in magnetic flux as:
\[ \Delta \Phi = A \cdot \Delta B \]

4. **Calculate the induced emf (E) using Faraday's law:**
According to Faraday's law:
\[ E = -N \frac{d\Phi}{dt} \]
Substituting the expression for magnetic flux:
\[ E = -N \cdot A \cdot \frac{dB}{dt} \]

5. **Substitute the known values:**
\[ E = -500 \cdot 0.01 \, \text{m}^2 \cdot 1.0 \, \text{T/s} \]
\[ E = -500 \cdot 0.01 \cdot 1 = -5 \, \text{V} \]

6. **Final result:**
The induced emf is:
\[ E = 5 \, \text{V} \quad (\text{taking magnitude}) \]

### Final Answer:
The induced r.m.f. in volts is **5 V**.