A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.4 sec. If magnetic flux between the pole pieces is known to be `8 xx 10^(-4)`Wb, then induced emf in the wire, is

Options

A. `4 xx 10^(-3)` V
B. `8 xx 10^(-3)` V
C. `2 xx 10^(-3)` V
D. `6 xx 10^(-3)` V

Answer

To find the induced electromotive force (emf) in the wire, we can use Faraday's law of electromagnetic induction, which states that the induced emf (ε) in a closed circuit is equal to the negative rate of change of magnetic flux (Φ) through the circuit. The formula is given by:

\[ \varepsilon = -\frac{d\Phi}{dt} \]

Where:
- \( \varepsilon \) = induced emf
- \( d\Phi \) = change in magnetic flux
- \( dt \) = change in time

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic flux (Φ) = \( 8 \times 10^{-4} \, \text{Wb} \)
- Time (t) = \( 0.4 \, \text{s} \)

2. **Calculate the rate of change of magnetic flux:**
Since the wire is dragged across the magnetic field, we can assume that the change in magnetic flux is equal to the magnetic flux itself (as it goes from 0 to \( 8 \times 10^{-4} \, \text{Wb} \)). Therefore, we have:
- \( d\Phi = 8 \times 10^{-4} \, \text{Wb} \)

3. **Substitute the values into the formula:**
\[ \varepsilon = -\frac{d\Phi}{dt} = -\frac{8 \times 10^{-4} \, \text{Wb}}{0.4 \, \text{s}} \]

4. **Perform the calculation:**
\[ \varepsilon = -\frac{8 \times 10^{-4}}{0.4} = -2 \times 10^{-3} \, \text{V} \]

5. **Determine the magnitude of the induced emf:**
Since we are interested in the magnitude, we take the absolute value:
\[ \varepsilon = 2 \times 10^{-3} \, \text{V} \]

### Final Answer:
The induced emf in the wire is \( 2 \times 10^{-3} \, \text{V} \) or \( 2 \, \text{mV} \).

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