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 a small loop placed inside a solenoid, we can follow these steps:

### Step 1: Identify Given Data
- Number of turns per centimeter of the solenoid: \( n = 10 \, \text{turns/cm} \)
- Area of the loop: \( A = 3 \, \text{cm}^2 \)
- Initial current: \( I_1 = 2 \, \text{A} \)
- Final current: \( I_2 = 4 \, \text{A} \)
- Time interval for the change in current: \( \Delta t = 0.2 \, \text{s} \)

### Step 2: Convert Units
- Convert the number of turns per centimeter to turns per meter:
\[ n = 10 \, \text{turns/cm} = 10 \times 100 = 1000 \, \text{turns/m} \]

- Convert the area from cm² to m²:
\[ A = 3 \, \text{cm}^2 = 3 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate Change in Current
- Calculate the change in current (\( \Delta I \)):
\[ \Delta I = I_2 - I_1 = 4 \, \text{A} - 2 \, \text{A} = 2 \, \text{A} \]

### Step 4: Calculate the Rate of Change of Current
- The rate of change of current (\( \frac{\Delta I}{\Delta t} \)):
\[ \frac{\Delta I}{\Delta t} = \frac{2 \, \text{A}}{0.2 \, \text{s}} = 10 \, \text{A/s} \]

### Step 5: Use the Formula for Induced EMF
- The induced electromotive force (EMF) (\( \epsilon \)) in the loop can be calculated using the formula:
\[ \epsilon = -n \cdot A \cdot \frac{dB}{dt} \]
where \( B \) is the magnetic field inside the solenoid given by:
\[ B = \mu_0 n I \]
Thus, the rate of change of magnetic field (\( \frac{dB}{dt} \)):
\[ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \]

### Step 6: Substitute Values
- The permeability of free space (\( \mu_0 \)) is \( 4\pi \times 10^{-7} \, \text{T m/A} \).
- Substitute \( n = 1000 \, \text{turns/m} \) and \( \frac{dI}{dt} = 10 \, \text{A/s} \):
\[ \frac{dB}{dt} = (4\pi \times 10^{-7}) \cdot (1000) \cdot (10) = 4\pi \times 10^{-4} \, \text{T/s} \]

### Step 7: Calculate Induced EMF
- Now substitute \( A \) and \( \frac{dB}{dt} \) into the EMF formula:
\[ \epsilon = -n \cdot A \cdot \frac{dB}{dt} = -1000 \cdot (3 \times 10^{-4}) \cdot (4\pi \times 10^{-4}) \]
\[ \epsilon = -1000 \cdot 3 \times 10^{-4} \cdot 4\pi \times 10^{-4} \]
\[ \epsilon = -12\pi \times 10^{-10} \, \text{V} \approx -3.77 \times 10^{-6} \, \text{V} \]

### Step 8: Final Answer
The induced voltage in the loop is approximately:
\[ \epsilon \approx 3.8 \times 10^{-6} \, \text{V} \]