A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.1 s. The magnetic flux between the pole pieces is `8 xx 10^(-4)` Wb. Find the magnitude of induced e.m.f.

Options

A. `4xx10^(-3)` V
B. `8xx10^(-3)` V
C. `2xx10^(-3)V`
D. `6xx10^(-3)V`

Explanation

To solve the problem of finding the magnitude of induced e.m.f. when a metal wire is dragged across a magnetic field, we can use Faraday's law of electromagnetic induction. According to this law, the induced e.m.f. (electromotive force) in a circuit is equal to the rate of change of magnetic flux through the circuit.

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic flux (Φ) = \(8 \times 10^{-4}\) Wb (Weber)
- Time interval (Δt) = 0.1 s

2. **Understand the formula for induced e.m.f.:**
The induced e.m.f. (ε) can be calculated using the formula:
\[ \epsilon = -\frac{d\Phi}{dt} \]
where \(d\Phi\) is the change in magnetic flux and \(dt\) is the time interval.

3. **Substitute the known values into the formula:**
Since the problem states that the wire is dragged across the magnetic field, we can assume that the change in magnetic flux is equal to the given magnetic flux (since it starts from zero). Thus, we have:
\[ d\Phi = 8 \times 10^{-4} \text{ Wb} \]
Now substituting the values into the formula:
\[ \epsilon = -\frac{8 \times 10^{-4}}{0.1} \]

4. **Calculate the induced e.m.f.:**
\[ \epsilon = -\frac{8 \times 10^{-4}}{0.1} = -8 \times 10^{-3} \text{ V} \]
The negative sign indicates the direction of the induced e.m.f. according to Lenz's law, but we are interested in the magnitude:
\[ |\epsilon| = 8 \times 10^{-3} \text{ V} \]

5. **Final answer:**
The magnitude of the induced e.m.f. is:
\[ 8 \times 10^{-3} \text{ V} \text{ or } 8 \text{ mV} \]