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. The formula for induced emf (\(E\)) is given by:

\[ E = -\frac{\Delta \Phi}{\Delta t} \]

where:
- \(\Delta \Phi\) is the change in magnetic flux,
- \(\Delta t\) is the change in time.

### Step-by-step Solution:

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

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

3. **Calculate the induced emf**:
- First, calculate the division:
\[ E = -\frac{8 \times 10^{-4}}{0.4} \]
- To simplify, we can multiply the numerator and denominator by 10 to eliminate the decimal:
\[ E = -\frac{8 \times 10^{-3}}{4} \]
- Now, perform the division:
\[ E = -2 \times 10^{-3} \, \text{V} \]

4. **Final result**:
The induced emf in the wire is:
\[ E = 2 \times 10^{-3} \, \text{V} \quad (\text{ignoring the negative sign as it indicates direction}) \]