The magnetic flux linked to a coil of 10 turns changes by 40 mWb in a time of 2 ms. The magnitude of the induced emf produced in the coil is

Options

A. 220 volt
B. 2000 volt
C. 200 volt
D. 0.2 volt

Answer

To solve the problem of finding the induced emf in a coil when the magnetic flux changes, we can use Faraday's law of electromagnetic induction. The formula for the induced emf (ε) is given by:

\[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \]

Where:
- \( \varepsilon \) = induced emf
- \( N \) = number of turns in the coil
- \( \Delta \Phi \) = change in magnetic flux
- \( \Delta t \) = change in time

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns, \( N = 10 \)
- Change in magnetic flux, \( \Delta \Phi = 40 \, \text{mWb} = 40 \times 10^{-3} \, \text{Wb} = 0.04 \, \text{Wb} \)
- Change in time, \( \Delta t = 2 \, \text{ms} = 2 \times 10^{-3} \, \text{s} \)

2. **Substitute the values into the formula:**
\[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \]
\[ \varepsilon = -10 \frac{0.04 \, \text{Wb}}{2 \times 10^{-3} \, \text{s}} \]

3. **Calculate the change in magnetic flux per unit time:**
\[ \frac{\Delta \Phi}{\Delta t} = \frac{0.04}{2 \times 10^{-3}} = \frac{0.04}{0.002} = 20 \, \text{Wb/s} \]

4. **Calculate the induced emf:**
\[ \varepsilon = -10 \times 20 = -200 \, \text{V} \]
The negative sign indicates the direction of the induced emf according to Lenz's law, but we are interested in the magnitude:
\[ |\varepsilon| = 200 \, \text{V} \]

### Final Answer:
The magnitude of the induced emf produced in the coil is **200 V**.