A long solenoid with 10 turns per cm has a small loop of area `3cm^(2)` placed inside, normal to the axis of the solenoid. If the currnet carried by the solenoid changes steadily from 2 A to 4 A in 0.2 s, what is the induced voltage in the loop, while the current is changing?

Options

A. `4.2xx10^(-8)V`
B. `2.8xx10^(-8)V`
C. `7.3xx10^(-6)V`
D. `3.8xx10^(-6)V`

Explanation

To solve the problem of finding the induced voltage in the loop placed inside the solenoid, we can follow these steps:

### Step 1: Understand the given data
- Turns per cm of the solenoid: \( n = 10 \) turns/cm
- Convert to turns/m:
\[ n = 10 \, \text{turns/cm} \times 100 \, \text{cm/m} = 1000 \, \text{turns/m} \]
- Area of the loop:
\[ A = 3 \, \text{cm}^2 = 3 \times 10^{-4} \, \text{m}^2 \]
- Initial current (\( I_1 \)): 2 A
- Final current (\( I_2 \)): 4 A
- Time interval (\( \Delta t \)): 0.2 s

### Step 2: Calculate the change in current per unit time
- Change in current (\( \Delta I \)):
\[ \Delta I = I_2 - I_1 = 4 \, \text{A} - 2 \, \text{A} = 2 \, \text{A} \]
- Rate of change of current (\( \frac{dI}{dt} \)):
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{2 \, \text{A}}{0.2 \, \text{s}} = 10 \, \text{A/s} \]

### Step 3: Calculate the magnetic field inside the solenoid
- The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 n I \]
- Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 4: Calculate the change in magnetic flux
- The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
- Therefore, the induced emf (\( \mathcal{E} \)) can be expressed as:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -\frac{d(B \cdot A)}{dt} = A \cdot \frac{dB}{dt} \]
- Since \( B = \mu_0 n I \), we have:
\[ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \]

### Step 5: Substitute the values into the equation
- Substitute \( n = 1000 \, \text{turns/m} \), \( A = 3 \times 10^{-4} \, \text{m}^2 \), and \( \frac{dI}{dt} = 10 \, \text{A/s} \):
\[ \frac{dB}{dt} = (4\pi \times 10^{-7}) \cdot (1000) \cdot (10) = 4\pi \times 10^{-4} \, \text{T/s} \]
- Now, calculate the induced emf:
\[ \mathcal{E} = A \cdot \frac{dB}{dt} = (3 \times 10^{-4}) \cdot (4\pi \times 10^{-4}) = 12\pi \times 10^{-8} \, \text{V} \]
- Approximating \( \pi \approx 3.14 \):
\[ \mathcal{E} \approx 3.77 \times 10^{-7} \, \text{V} \approx 3.8 \times 10^{-6} \, \text{V} \]

### Final Answer
The induced voltage in the loop is approximately \( 3.8 \times 10^{-6} \) volts.