The rate of change of magnetic flux density through a circular coil of area `10^(-2)m^(2)` and number of turns `100` is `10^(3) Wb//m^(2)s`. The value of induced e.m.f. will be

Options

A. `10^(-2)V`
B. `10^(-3)V`
C. `10`V
D. `10^(3)` V

Answer

To solve the problem, we need to find the induced electromotive force (e.m.f.) in a circular coil when there is a change in magnetic flux density. We can use Faraday's law of electromagnetic induction, which states that the induced e.m.f. (ε) in a closed loop is equal to the negative rate of change of magnetic flux through the loop.

### Given Data:
- Area of the coil (A) = \(10^{-2} \, m^2\)
- Number of turns (N) = 100
- Rate of change of magnetic flux density (dB/dt) = \(10^{3} \, Wb/m^2/s\)

### Step-by-Step Solution:

1. **Calculate the Change in Magnetic Flux (Φ)**:
The magnetic flux (Φ) through one turn of the coil is given by:
\[ Φ = B \cdot A \]
where B is the magnetic flux density and A is the area of the coil.

2. **Calculate the Rate of Change of Magnetic Flux**:
The rate of change of magnetic flux (dΦ/dt) through one turn can be expressed as:
\[ \frac{dΦ}{dt} = A \cdot \frac{dB}{dt} \]
Substituting the values:
\[ \frac{dΦ}{dt} = 10^{-2} \, m^2 \cdot 10^{3} \, Wb/m^2/s = 10^{-2} \cdot 10^{3} = 10^{1} \, Wb/s \]

3. **Calculate the Total Rate of Change of Magnetic Flux for the Coil**:
Since there are N turns in the coil, the total rate of change of magnetic flux (dΦ_total/dt) is:
\[ \frac{dΦ_{total}}{dt} = N \cdot \frac{dΦ}{dt} = 100 \cdot 10^{1} \, Wb/s = 1000 \, Wb/s \]

4. **Calculate the Induced e.m.f. (ε)**:
According to Faraday's law, the induced e.m.f. is given by:
\[ ε = -\frac{dΦ_{total}}{dt} \]
Therefore:
\[ ε = -1000 \, V \]
Since we are interested in the magnitude, we can write:
\[ ε = 1000 \, V \]

### Final Answer:
The value of the induced e.m.f. is \(1000 \, V\).

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