An ideal solenoid of cross-sectional area `10^(-4)m^(2)` has 500 turns per metre. At the centre of this solenoid, another coil of 100 turns is wrapped closely around it. If the current in the coil changes from 0 to 2 A in 3.14 ms, the emf developed in the second coil is

Options

A. 1 mv
B. 2 mv
C. 3 mv
D. 4 mv

Answer

To solve the problem, we will use the concept of mutual inductance and the formula for induced electromotive force (emf) in the secondary coil due to a changing current in the primary coil.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cross-sectional area of the solenoid, \( A = 10^{-4} \, m^2 \)
- Number of turns per meter of the solenoid, \( n_1 = 500 \, \text{turns/m} \)
- Number of turns in the secondary coil, \( n_2 = 100 \, \text{turns} \)
- Change in current, \( \Delta I = 2 \, A - 0 \, A = 2 \, A \)
- Time interval, \( \Delta t = 3.14 \, ms = 3.14 \times 10^{-3} \, s \)

2. **Calculate the Rate of Change of Current:**
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{2 \, A}{3.14 \times 10^{-3} \, s} \approx 637.76 \, A/s \]

3. **Calculate the Mutual Inductance \( M \):**
The formula for mutual inductance \( M \) between the solenoid and the coil is given by:
\[ M = \mu_0 \cdot n_1 \cdot n_2 \cdot A \]
where \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) (permeability of free space).

Substituting the values:
\[ M = (4\pi \times 10^{-7}) \cdot (500) \cdot (100) \cdot (10^{-4}) \]

Calculating this:
\[ M = (4\pi \times 10^{-7} \cdot 500 \cdot 100 \cdot 10^{-4}) = 4\pi \times 5 \times 10^{-9} = 20\pi \times 10^{-9} \, H \]

Approximating \( \pi \approx 3.14 \):
\[ M \approx 20 \cdot 3.14 \times 10^{-9} \approx 62.8 \times 10^{-9} \, H \approx 6.28 \times 10^{-8} \, H \]

4. **Calculate the Induced EMF in the Secondary Coil:**
The induced emf \( \mathcal{E} \) in the secondary coil is given by:
\[ \mathcal{E} = -M \cdot \frac{dI}{dt} \]
Substituting the values:
\[ \mathcal{E} = - (6.28 \times 10^{-8}) \cdot (637.76) \]

Calculating this:
\[ \mathcal{E} \approx - (6.28 \times 10^{-8} \cdot 637.76) \approx -4.0 \times 10^{-5} \, V \]

Since we are interested in the magnitude:
\[ |\mathcal{E}| \approx 4.0 \times 10^{-5} \, V = 4.0 \, mV \]

5. **Final Answer:**
The induced emf in the secondary coil is approximately \( 4 \, mV \).