Aconductor of lengfht 0.4 m is moving with a speed of 7 m/s perpendicular to a magnetic field of intensity` 0.9 Wb//m^(2)` .The induced emf across the coduct is

Options

A. `1.26 V`
B. `2.52 V`
C. `5.04 V`
D. `25.2 V`

Explanation

To find the induced electromotive force (emf) across a conductor moving in a magnetic field, we can use the formula:

\[ \text{Induced EMF} (E) = B \cdot L \cdot V \]

where:
- \(E\) is the induced emf,
- \(B\) is the magnetic field intensity (in Weber/m²),
- \(L\) is the length of the conductor (in meters),
- \(V\) is the speed of the conductor (in meters/second).

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the conductor, \(L = 0.4 \, \text{m}\)
- Speed of the conductor, \(V = 7 \, \text{m/s}\)
- Magnetic field intensity, \(B = 0.9 \, \text{Wb/m}^2\)

2. **Substitute the values into the formula:**
\[ E = B \cdot L \cdot V \]
\[ E = 0.9 \, \text{Wb/m}^2 \cdot 0.4 \, \text{m} \cdot 7 \, \text{m/s} \]

3. **Calculate the product:**
- First, calculate \(0.4 \cdot 7\):
\[ 0.4 \cdot 7 = 2.8 \]
- Now multiply by \(0.9\):
\[ E = 0.9 \cdot 2.8 = 2.52 \, \text{V} \]

4. **Final Result:**
The induced emf across the conductor is:
\[ E = 2.52 \, \text{V} \]

### Conclusion:
The induced emf across the conductor is **2.52 volts**.

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