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 will use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a coil is equal to the rate of change of magnetic flux through the coil.

### Step-by-Step Solution:

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

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

3. **Calculate the magnetic flux (Φ) through one loop:**
Since the coil is placed normal to the magnetic field, the magnetic flux through one loop is given by:
\[ \Phi = B \cdot A \]
Here, B is the magnetic field and A is the area of the loop. The change in magnetic flux (dΦ/dt) can be expressed as:
\[ \frac{d\Phi}{dt} = A \cdot \frac{dB}{dt} \]

4. **Substituting the values into the equation for dΦ/dt:**
\[ \frac{d\Phi}{dt} = 0.01 \, \text{m}^2 \cdot 1.0 \, \text{T/s} = 0.01 \, \text{Wb/s} \]

5. **Calculate the induced emf (E) using Faraday's law:**
\[ E = -N \cdot \frac{d\Phi}{dt} \]
Since we are interested in the magnitude, we can ignore the negative sign:
\[ E = 500 \cdot 0.01 \, \text{V} = 5 \, \text{V} \]

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