A long solenoid has 1000 turns per metre. It has a small loop of area `10^(-4)m^2` placed inside it with the normal to the loop and parallel to the axis. Calculate the induced emf across the loop if the current in the solenoid is the rate of 10 A/s.

Explanation

To solve the problem, we need to calculate the induced electromotive force (emf) across a small loop placed inside a long solenoid. The key steps are outlined below:

### Step 1: Identify the given values
- Number of turns per meter of the solenoid, \( n = 1000 \, \text{turns/m} \)
- Area of the loop, \( A = 10^{-4} \, \text{m}^2 \)
- Rate of change of current in the solenoid, \( \frac{di}{dt} = 10 \, \text{A/s} \)

### Step 2: Calculate the mutual inductance \( M \)
The formula for the mutual inductance \( M \) of a solenoid with a small loop inside it is given by:
\[ M = \mu_0 n A \]
where:
- \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)
- \( n \) is the number of turns per meter
- \( A \) is the area of the loop

Substituting the values:
\[ M = (4\pi \times 10^{-7}) \times (1000) \times (10^{-4}) \]

### Step 3: Simplify the expression for \( M \)
Calculating \( M \):
\[ M = 4\pi \times 10^{-7} \times 1000 \times 10^{-4} = 4\pi \times 10^{-8} \, \text{H} \]

### Step 4: Calculate the induced emf \( \mathcal{E} \)
The induced emf \( \mathcal{E} \) can be calculated using the formula:
\[ \mathcal{E} = M \frac{di}{dt} \]
Substituting the values of \( M \) and \( \frac{di}{dt} \):
\[ \mathcal{E} = (4\pi \times 10^{-8}) \times (10) \]

### Step 5: Calculate the final value of \( \mathcal{E} \)
Calculating \( \mathcal{E} \):
\[ \mathcal{E} = 4\pi \times 10^{-7} \, \text{V} \]
Using \( \pi \approx 3.14 \):
\[ \mathcal{E} \approx 4 \times 3.14 \times 10^{-7} \approx 1.256 \times 10^{-6} \, \text{V} \approx 1.26 \, \mu\text{V} \]

### Final Answer
The induced emf across the loop is approximately \( 1.26 \, \mu\text{V} \).

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