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 follow these steps:

### Step 1: Understand the given parameters
- Cross-sectional area of the solenoid, \( A = 10^{-4} \, m^2 \)
- Number of turns per unit length of the solenoid, \( n = 500 \, \text{turns/m} \)
- Number of turns in the coil wrapped around the solenoid, \( N = 100 \)
- Change in current, \( \Delta I = I_2 - I_1 = 2 \, A - 0 \, A = 2 \, A \)
- Time interval for the change in current, \( \Delta t = 3.14 \, ms = 3.14 \times 10^{-3} \, s \)

### Step 2: Calculate the magnetic field inside the solenoid
The magnetic field \( B \) inside an ideal solenoid is given by the formula:
\[ B = \mu_0 n I \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) is the permeability of free space.

### Step 3: Find the change in magnetic field due to the change in current
The change in magnetic field \( \Delta B \) when the current changes from \( 0 \) to \( 2 \, A \) is:
\[ \Delta B = \mu_0 n \Delta I \]

### Step 4: Calculate the magnetic flux \( \Phi \) through the coil
The magnetic flux \( \Phi \) linked with the coil is given by:
\[ \Phi = N \cdot B \cdot A \]
Thus, the change in magnetic flux \( \Delta \Phi \) is:
\[ \Delta \Phi = N \cdot \Delta B \cdot A \]

### Step 5: Apply Faraday's Law to find the induced EMF
According to Faraday's Law, the induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Since we are interested in the average EMF during the time interval \( \Delta t \):
\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \]

### Step 6: Substitute the values and calculate
1. Calculate \( \Delta B \):
\[ \Delta B = \mu_0 n \Delta I = (4\pi \times 10^{-7}) \cdot (500) \cdot (2) = 4\pi \times 10^{-7} \cdot 1000 \]
\[ \Delta B = 4\pi \times 10^{-4} \, T \]

2. Calculate \( \Delta \Phi \):
\[ \Delta \Phi = N \cdot \Delta B \cdot A = 100 \cdot (4\pi \times 10^{-4}) \cdot (10^{-4}) = 4\pi \times 10^{-6} \, Wb \]

3. Calculate the induced EMF:
\[ \mathcal{E} = -\frac{4\pi \times 10^{-6}}{3.14 \times 10^{-3}} = -\frac{4\pi}{3.14} \times 10^{-3} \, V \]
Since \( \pi \approx 3.14 \), we find that:
\[ \mathcal{E} \approx -4 \times 10^{-3} \, V = -4 \, mV \]

### Final Answer
The induced EMF developed in the second coil is approximately \( 4 \, mV \).

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