To solve the problem, we will follow these steps:
### Step 1: Convert Given Values to SI Units
- Length of the solenoid, \( L = 50 \, \text{cm} = 50 \times 10^{-2} \, \text{m} = 0.5 \, \text{m} \)
- Area of cross-section, \( A = 28 \, \text{cm}^2 = 28 \times 10^{-4} \, \text{m}^2 = 0.00028 \, \text{m}^2 \)
- Number of turns, \( N = 200 \)
- Current, \( I_1 = 5.0 \, \text{A} \)
- Time, \( t = 1.0 \, \text{ms} = 1.0 \times 10^{-3} \, \text{s} \)
### Step 2: Calculate the Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside a solenoid is given by the formula:
\[
B = \mu_0 \cdot \frac{N \cdot I}{L}
\]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
Substituting the values:
\[
B = 4\pi \times 10^{-7} \cdot \frac{200 \cdot 5.0}{0.5}
\]
Calculating:
\[
B = 4\pi \times 10^{-7} \cdot 2000 = 8\pi \times 10^{-4} \, \text{T}
\]
### Step 3: Calculate the Magnetic Flux \( \Phi_1 \) When Current is On
The magnetic flux \( \Phi \) through the solenoid is given by:
\[
\Phi = N \cdot B \cdot A
\]
Substituting the values:
\[
\Phi_1 = 200 \cdot (8\pi \times 10^{-4}) \cdot (0.00028)
\]
Calculating:
\[
\Phi_1 = 200 \cdot 8\pi \times 10^{-4} \cdot 0.00028
\]
\[
\Phi_1 \approx 200 \cdot 8 \cdot 3.14 \cdot 10^{-4} \cdot 0.00028
\]
\[
\Phi_1 \approx 1.4 \times 10^{-6} \, \text{Wb}
\]
### Step 4: Calculate the Magnetic Flux \( \Phi_2 \) When Current is Off
When the current is switched off, the magnetic field becomes zero, hence:
\[
\Phi_2 = 0 \, \text{Wb}
\]
### Step 5: Calculate the Induced EMF
The induced EMF \( \mathcal{E} \) can be calculated using Faraday's law of electromagnetic induction:
\[
\mathcal{E} = -\frac{\Delta \Phi}{\Delta t} = -\frac{\Phi_2 - \Phi_1}{t}
\]
Substituting the values:
\[
\mathcal{E} = -\frac{0 - 1.4 \times 10^{-6}}{1.0 \times 10^{-3}}
\]
Calculating:
\[
\mathcal{E} = \frac{1.4 \times 10^{-6}}{1.0 \times 10^{-3}} = 1.4 \, \text{V}
\]
### Final Answer
The induced EMF is \( 1.4 \, \text{V} \).
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