A small, metal wire loop is dragged across the gap between the poles of a magnet in 0.4 s. If the change in magnetic flux for the wire is `8xx10^(-4)Wb`, then the emf induced in the wire is

Options

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

Answer

To find the induced electromotive force (emf) in the wire loop, we can use Faraday's law of electromagnetic induction, which states:

\[ \text{emf} (E) = -\frac{d\Phi}{dt} \]

Where:
- \(d\Phi\) is the change in magnetic flux,
- \(dt\) is the change in time.

### Step-by-Step Solution:

1. **Identify the given values**:
- Change in magnetic flux, \(d\Phi = 8 \times 10^{-4} \, \text{Wb}\)
- Change in time, \(dt = 0.4 \, \text{s}\)

2. **Apply Faraday's Law**:
According to Faraday's law, the induced emf can be calculated as:
\[ E = -\frac{d\Phi}{dt} \]

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

4. **Calculate the induced emf**:
\[ E = -\frac{8 \times 10^{-4}}{0.4} = -2 \times 10^{-3} \, \text{V} \]

5. **Interpret the negative sign**:
The negative sign indicates the direction of the induced emf (Lenz's law), but the magnitude of the induced emf is:
\[ E = 2 \times 10^{-3} \, \text{V} \text{ or } 2 \, \text{mV} \]

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

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